【集训】最短路！
最短路
P4779 【模板】单源最短路径（标准版）
#include <bits/stdc++.h>
using namespace std;
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define DEBUG(x) cerr << #x << " = " << x << endl
#define ll long long
typedef pair <int, int> PII;
typedef unsigned int uint;
typedef unsigned long long ull;
#define i128 __int128
#define fi first
#define se second
mt19937 rnd(chrono::system_clock::now().time_since_epoch().count());
#define ClockA clock_t start, end; start = clock()
#define ClockB end = clock(); cerr << "time = " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;
//#define int long long
inline int rd(){
int x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9'){
if (ch == '-')
f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
return x * f;
}
#define rd rd()
void wt(int x){
if (x < 0)
putchar('-'), x = -x;
if (x > 9)
wt(x / 10);
putchar(x % 10 + '0');
return;
}
void wt(char x){
putchar(x);
}
void wt(int x, char k){
wt(x),putchar(k);
}
namespace Star_F{
const int N = 200005;
int n, m, s;
int h[N], e[N << 1], ne[N << 1], w[N << 1], idx;
vector<PII> v[N];
int dis[N];
struct node{
int d, id;
bool friend operator<(node a,node b){
return a.d > b.d;
}
};
void add(int u,int v,int W){
e[++idx] = v, ne[idx] = h[u], w[idx] = W, h[u] = idx;
}
priority_queue<node> q;
void dij(){
memset(dis, 0x3f, sizeof(dis));
dis[s] = 0;
q.push({0, s});
while(!q.empty()){
int d = q.top().d, u = q.top().id;
q.pop();
if(d>dis[u])
continue;
for (int i = h[u]; i;i=ne[i]){
int v = e[i];
if(dis[v]>dis[u]+w[i]){
dis[v] = dis[u] + w[i];
q.push({dis[v], v});
}
}
}
}
void Main(){
cin >> n >> m >> s;
for (int i = 1; i <= m;i++){
int u, v, w;
cin >> u >> v >> w;
add(u, v, w);
}
dij();
for (int i = 1; i <= n;i++)
cout << dis[i] << " ";
}
}
signed main(){
// freopen(".in","r",stdin);
// freopen(".out","w",stdout);
ClockA;
int T=1;
// T=rd;
while(T--) Star_F::Main();
// ClockB;
return 0;
}
P5905 【模板】全源最短路（Johnson）
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ll long long
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define DEBUG(x) cerr << #x << '=' << x << endl
inline int rd() {
int x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-')
f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
return x * f;
}
void print(ll x) {
if (x < 0)
putchar('-'), x = -x;
if (x > 9)
print(x / 10);
putchar(x % 10 + '0');
return;
}
namespace Star_F {
struct edge {
int v, w, next;
} e[10005];
struct node {
int dis, id;
bool operator<(const node& a) const { return dis > a.dis; }
node(int d, int x) { dis = d, id = x; }
};
const int INF = 1e9;
int head[5005], vis[5005], t[5005];
int cnt, n, m;
long long h[5005], dis[5005];
void addedge(int u, int v, int w) {
e[++cnt].v = v;
e[cnt].w = w;
e[cnt].next = head[u];
head[u] = cnt;
}
bool spfa(int s) {
queue<int> q;
memset(h, 63, sizeof(h));
h[s] = 0, vis[s] = 1;
q.push(s);
while (!q.empty()) {
int u = q.front();
q.pop();
vis[u] = 0;
for (int i = head[u]; i; i = e[i].next) {
int v = e[i].v;
if (h[v] > h[u] + e[i].w) {
h[v] = h[u] + e[i].w;
if (!vis[v]) {
vis[v] = 1;
q.push(v);
t[v]++;
if (t[v] == n + 1) return false;
}
}
}
}
return true;
}
void dijkstra(int s) {
priority_queue<node> q;
for (int i = 1; i <= n; i++) dis[i] = INF;
memset(vis, 0, sizeof(vis));
dis[s] = 0;
q.push(node(0, s));
while (!q.empty()) {
int u = q.top().id;
q.pop();
if (vis[u]) continue;
vis[u] = 1;
for (int i = head[u]; i; i = e[i].next) {
int v = e[i].v;
if (dis[v] > dis[u] + e[i].w) {
dis[v] = dis[u] + e[i].w;
if (!vis[v]) q.push(node(dis[v], v));
}
}
}
return;
}
void Main() {
n = rd(); m = rd();
for (int i = 1; i <= m; i++) {
int u = rd(), v = rd(), w = rd();
addedge(u, v, w);
}
for (int i = 1; i <= n; i++) addedge(0, i, 0);
if (!spfa(0)) {
cout << -1 << endl;
return;
}
for (int u = 1; u <= n; u++)
for (int i = head[u]; i; i = e[i].next) e[i].w += h[u] - h[e[i].v];
for (int i = 1; i <= n; i++) {
dijkstra(i);
long long ans = 0;
for (int j = 1; j <= n; j++) {
if (dis[j] == INF)
ans += j * INF;
else
ans += j * (dis[j] + h[j] - h[i]);
}
cout << ans << endl;
}
}
}
signed main() {
// freopen(".in", "r", stdin);
// freopen(".out", "w", stdout);
return Star_F::Main(), 0;
}
P3385 【模板】负环
#include <bits/stdc++.h>
using namespace std;
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define DEBUG(x) cerr << #x << " = " << x << endl
#define ll long long
typedef pair <int, int> PII;
typedef unsigned int uint;
typedef unsigned long long ull;
#define i128 __int128
#define fi first
#define se second
mt19937 rnd(chrono::system_clock::now().time_since_epoch().count());
#define ClockA clock_t start, end; start = clock()
#define ClockB end = clock(); cerr << "time = " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;
//#define int long long
inline int rd(){
int x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9'){
if (ch == '-')
f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
return x * f;
}
#define rd rd()
void wt(int x){
if (x < 0)
putchar('-'), x = -x;
if (x > 9)
wt(x / 10);
putchar(x % 10 + '0');
return;
}
void wt(char x){
putchar(x);
}
void wt(int x, char k){
wt(x),putchar(k);
}
namespace Star_F{
const int N = 2005;
int dis[N], cnt[N];
vector<PII> G[N];
bool vis[N];
int n, m;
bool spfa(){
queue<int> q;
vis[1] = 1, dis[1] = 0, cnt[1] = 1;
q.push(1);
while(!q.empty()){
int u = q.front();
q.pop();
vis[u] = 0;
for (int i = 0; i < G[u].size();i++){
int v = G[u][i].fi, w = G[u][i].se;
if(dis[v]>dis[u]+w){
dis[v] = dis[u] + w;
if(!vis[v]){
cnt[v]++;
q.push(v);
if(cnt[v]>=n)
return true;
}
}
}
}
return false;
}
void Main(){
memset(dis, 0x3f, sizeof(dis));
memset(vis, 0, sizeof(vis));
memset(cnt, 0, sizeof(cnt));
cin >> n >> m;
for (int i = 1; i <= n;i++)
G[i].clear();
for (int i = 1; i <= m;i++){
int u, v, w;
cin >> u >> v >> w;
G[u].push_back({v, w});
if(w>=0)
G[v].push_back({u, w});
}
cout << (spfa() ? "YES" : "NO") << endl;
}
}
signed main(){
// freopen(".in","r",stdin);
// freopen(".out","w",stdout);
ClockA;
int T=1;
T=rd;
while(T--) Star_F::Main();
// ClockB;
return 0;
}
P1119 灾后重建
动态加边，动态求全源最短路。
由于输入的时间 \(t\) 按升序排序，不用再排序。直接暴力加边，用 \(\text{Floyd}\) 维护全源最短路即可。
注意编号是 \(0 \sim N-1\) ，所以数组从 \(0\) 开始。
#include <bits/stdc++.h>
using namespace std;
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define DEBUG(x) cerr << #x << " = " << x << endl
#define ll long long
typedef pair <int, int> PII;
typedef unsigned int uint;
typedef unsigned long long ull;
#define i128 __int128
#define fi first
#define se second
mt19937 rnd(chrono::system_clock::now().time_since_epoch().count());
#define ClockA clock_t start, end; start = clock()
#define ClockB end = clock(); cerr << "time = " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;
//#define int long long
inline int rd(){
int x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9'){
if (ch == '-')
f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
return x * f;
}
#define rd rd()
void wt(int x){
if (x < 0)
putchar('-'), x = -x;
if (x > 9)
wt(x / 10);
putchar(x % 10 + '0');
return;
}
void wt(char x){
putchar(x);
}
void wt(int x, char k){
wt(x),putchar(k);
}
namespace Star_F{
const int N = 205;
int T, n, m, a[N], f[N][N];
void update(int k){
for (int i = 0; i < n;i++){
for (int j = 0; j < n;j++)
f[i][j] = f[j][i] = min(f[i][j], f[i][k] + f[k][j]);
}
}
void Main(){
cin >> n >> m;
for (int i = 0; i < n;i++)
cin >> a[i];
memset(f, 0x3f, sizeof(f));
for (int i = 0; i <= n;i++)
f[i][i] = 0;
for (int i = 1; i <= m;i++){
int u, v, w;
cin >> u >> v >> w;
f[u][v] = f[v][u] = w;
}
cin >> T;
int x = 0;
while(T--){
int u, v, t;
cin >> u >> v >> t;
while(a[x]<=t&&x<n)
update(x++);
if(a[u]>t||a[v]>t)
cout << -1 << endl;
else{
if(f[u][v]==0x3f3f3f3f)
cout << -1 << endl;
else
cout << f[u][v] << endl;
}
}
}
}
signed main(){
// freopen(".in","r",stdin);
// freopen(".out","w",stdout);
ClockA;
int T=1;
// T=rd;
while(T--) Star_F::Main();
// ClockB;
return 0;
}
P1462 通往奥格瑞玛的道路
看到最大值最小，果断考虑二分。
我们二分一个 \(mid\)，每次只经过边权小于 \(mid\) 的边，判断是否能到达终点。
如何判断？和最短路类似，我们用 dis[v]=dis[u]-w[i] 更新 dis 数组即可，所以用大根堆维护。
#include <bits/stdc++.h>
using namespace std;
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define DEBUG(x) cerr << #x << " = " << x << endl
#define ll long long
typedef pair <int, int> PII;
typedef unsigned int uint;
typedef unsigned long long ull;
#define i128 __int128
#define fi first
#define se second
mt19937 rnd(chrono::system_clock::now().time_since_epoch().count());
#define ClockA clock_t start, end; start = clock()
#define ClockB end = clock(); cerr << "time = " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;
//#define int long long
inline int rd(){
int x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9'){
if (ch == '-')
f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
return x * f;
}
#define rd rd()
void wt(int x){
if (x < 0)
putchar('-'), x = -x;
if (x > 9)
wt(x / 10);
putchar(x % 10 + '0');
return;
}
void wt(char x){
putchar(x);
}
void wt(int x, char k){
wt(x),putchar(k);
}
namespace Star_F{
const int N = 10005, M = 100005, INF = 0x3f3f3f3f;
int h[N], ne[M], e[M], w[M], idx;
int n, m, k;
int a[N], l = INF, r, flag;
void add(int u,int v,int W){
e[++idx] = v, ne[idx] = h[u], w[idx] = W, h[u] = idx;
}
int dis[N];
priority_queue<PII> q;
bool check(int mid){
memset(dis, -1, sizeof(dis));
dis[1] = k;
q.push({dis[1], 1});
while(!q.empty()){
int u = q.top().se, d = q.top().fi;
q.pop();
if(dis[u]!=d)
continue;
for (int i = h[u]; i;i=ne[i]){
int v = e[i];
if(a[v]>mid)
continue;
if(dis[v]<dis[u]-w[i]&&dis[u]-w[i]>=0){
dis[v] = dis[u] - w[i];
q.push({dis[v], v});
}
}
}
return dis[n] != -1;
}
void Main(){
cin >> n >> m >> k;
for (int i = 1; i <= n;i++){
cin >> a[i];
l = min(l, a[i]), r = max(r, a[i]);
}
flag = r;
for (int i = 1; i <= m;i++){
int u, v, w;
cin >> u >> v >> w;
add(u, v, w), add(v, u, w);
}
l = a[1];
while(l<=r){
int mid = l + r >> 1;
if(check(mid))
r = mid - 1;
else
l = mid + 1;
}
if(l==flag+1)
cout << "AFK" << endl;
else
cout << l << endl;
}
}
signed main(){
// freopen(".in","r",stdin);
// freopen(".out","w",stdout);
ClockA;
int T=1;
// T=rd;
while(T--) Star_F::Main();
// ClockB;
return 0;
}
UVA11090 Going in Cycle!!
经典套路二分题
二分答案 \(w\)，原式 \(\sum c_i\ge w \times *cnt\)。化简得 \(\sum^{cnt}(c_i-w)\ge 0\)。转换为找负环。
#include <bits/stdc++.h>
using namespace std;
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define DEBUG(x) cerr << #x << " = " << x << endl
#define ll long long
typedef pair <int, int> PII;
typedef unsigned int uint;
typedef unsigned long long ull;
#define i128 __int128
#define fi first
#define se second
mt19937 rnd(chrono::system_clock::now().time_since_epoch().count());
#define ClockA clock_t start, end; start = clock()
#define ClockB end = clock(); cerr << "time = " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;
//#define int long long
inline int rd(){
int x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9'){
if (ch == '-')
f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
return x * f;
}
#define rd rd()
void wt(int x){
if (x < 0)
putchar('-'), x = -x;
if (x > 9)
wt(x / 10);
putchar(x % 10 + '0');
return;
}
void wt(char x){
putchar(x);
}
void wt(int x, char k){
wt(x),putchar(k);
}
namespace Star_F{
const int INF = 0x3f3f3f3f;
const int N = 305, M = 15000;
int m, n;
int h[N], e[M], ne[M], idx;
double w[M];
int tmp = 1;
void add(int u,int v,double W){
e[++idx] = v, ne[idx] = h[u], w[idx] = W, h[u] = idx;
}
int nm[N], vis[N];
double dis[N];
bool spfa(double x){
queue<int> q;
memset(vis, 1, sizeof(vis));
for (int i = 1; i <= n;i++)
dis[i] = (double)INF;
memset(nm, 0, sizeof(nm));
for (int i = 1; i <= m;i++)
w[i] -= x;
for (int i = 1; i <= n;i++)
q.push(i);
while(!q.empty()){
int now = q.front();
q.pop();
vis[now] = 0;
for (int i = h[now]; i;i=ne[i]){
int v=e[i];
if(dis[v]>=dis[now]+w[i]){
dis[v] = dis[now] + w[i];
nm[v] = nm[now] + 1;
if(vis[v]==0){
q.push(v);
}
if(nm[v]>=n+1){
for (int i = 1; i <= m;i++)
w[i] += x;
return 1;
}
}
}
}
for (int i = 1; i <= m;i++) w[i]+=x;
return 0;
}
void Main(){
idx = 0;
memset(h, 0, sizeof(h));
cin >> n >> m;
for (int i = 1; i <= m;i++){
int u, v, w;
cin >> u >> v >> w;
add(u, v, (double)w);
}
double l = 0, r = 100000001;
while(r-l>0.0000001){
double mid = (l + r) / 2;
if(spfa(mid))
r = mid;
else l = mid;
}
cout << "Case #" << tmp++ << ": ";
if(r==100000001)
cout << "No cycle found." << endl;
else
printf("%.2lf\n", l);
}
}
signed main(){
// freopen(".in","r",stdin);
// freopen(".out","w",stdout);
ClockA;
int T=1;
T=rd;
while(T--) Star_F::Main();
// ClockB;
return 0;
}
P2865 [USACO06NOV] Roadblocks G
单源次短路模板题目
#include <bits/stdc++.h>
using namespace std;
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define DEBUG(x) cerr << #x << " = " << x << endl
#define ll long long
typedef pair <int, int> PII;
typedef unsigned int uint;
typedef unsigned long long ull;
#define i128 __int128
#define fi first
#define se second
mt19937 rnd(chrono::system_clock::now().time_since_epoch().count());
#define ClockA clock_t start, end; start = clock()
#define ClockB end = clock(); cerr << "time = " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;
//#define int long long
inline int rd(){
int x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9'){
if (ch == '-')
f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
return x * f;
}
#define rd rd()
void wt(int x){
if (x < 0)
putchar('-'), x = -x;
if (x > 9)
wt(x / 10);
putchar(x % 10 + '0');
return;
}
void wt(char x){
putchar(x);
}
void wt(int x, char k){
wt(x),putchar(k);
}
namespace Star_F{
const int N=200005;
int n, m;
int h[N], e[N], ne[N], w[N], idx;
int dis[2][N];
void add(int u,int v,int W){
e[++idx] = v, ne[idx] = h[u], w[idx] = W, h[u] = idx;
}
struct node{
int id, d;
bool friend operator<(node a,node b){
return a.d > b.d;
}
};
priority_queue<node> q;
void dij(){
for (int i = 1; i <= n;i++)
dis[0][i] = dis[1][i] = 2147483647;
dis[0][1] = 0;
q.push({1, 0});
while(!q.empty()){
int u = q.top().id, d = q.top().d;
q.pop();
if(d>dis[1][u])
continue;
for (int i = h[u]; i;i=ne[i]){
int v = e[i];
if(dis[0][v]>d+w[i]){
dis[1][v] = dis[0][v];
dis[0][v] = d + w[i];
q.push({v, dis[0][v]});
}
if(dis[1][v]>d+w[i]&&dis[0][v]<d+w[i]){
dis[1][v] = d + w[i];
q.push({v, dis[1][v]});
}
}
}
}
void Main(){
cin >> n >> m;
for (int i = 1; i <= m;i++){
int u, v, w;
cin >> u >> v >> w;
add(u, v, w), add(v, u, w);
}
dij();
cout << dis[1][n] << endl;
}
}
signed main(){
// freopen(".in","r",stdin);
// freopen(".out","w",stdout);
ClockA;
int T=1;
// T=rd;
while(T--) Star_F::Main();
// ClockB;
return 0;
}
P1875 佳佳的魔法药水
最短路计数问题。
#include <bits/stdc++.h>
using namespace std;
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define DEBUG(x) cerr << #x << " = " << x << endl
#define ll long long
typedef pair <int, int> PII;
typedef unsigned int uint;
typedef unsigned long long ull;
#define i128 __int128
#define fi first
#define se second
mt19937 rnd(chrono::system_clock::now().time_since_epoch().count());
#define ClockA clock_t start, end; start = clock()
#define ClockB end = clock(); cerr << "time = " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;
//#define int long long
inline int rd(){
int x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9'){
if (ch == '-')
f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
return x * f;
}
#define rd rd()
void wt(int x){
if (x < 0)
putchar('-'), x = -x;
if (x > 9)
wt(x / 10);
putchar(x % 10 + '0');
return;
}
void wt(char x){
putchar(x);
}
void wt(int x, char k){
wt(x),putchar(k);
}
namespace Star_F{
const int N = 3005;
int c[N], ans[N], t[N][N];
bool f[N];
void Main(){
int n;
cin >> n;
for (int i = 1; i <= n;i++)
cin >> c[i], ans[i] = 1;
int u, v, w;
while(scanf("%d%d%d",&u,&v,&w)!=EOF)
t[u + 1][v + 1] = t[v + 1][u + 1] = w + 1;
for (int i = 1; i < n;i++){
int maxn = 0x3f3f3f3f;
int b;
for (int j = 1; j <= n;j++)
if(!f[j]&&c[j]<maxn)
b = j, maxn = c[j];
f[b] = 1;
for (int j = 1; j <= n;j++)
if(f[j]&&t[b][j]){
if(c[b]+c[j]==c[t[b][j]])
ans[t[b][j]] += ans[b] * ans[j];
if(c[b]+c[j]<c[t[b][j]])
c[t[b][j]] = c[b] + c[j], ans[t[b][j]] = ans[b] * ans[j];
}
}
cout << c[1] << " " << ans[1] << endl;
}
}
signed main(){
// freopen(".in","r",stdin);
// freopen(".out","w",stdout);
ClockA;
int T=1;
// T=rd;
while(T--) Star_F::Main();
// ClockB;
return 0;
}
P1948 [USACO08JAN] Telephone Lines S
最大值最小，还是考虑二分。
二分 \(mid\) ，把大于 \(mid\) 的边看为 \(1\)，小于等于 \(mid\) 得便看为 \(0\)，跑最短路即可
当然更优秀的做法是 01BFS。
#include <bits/stdc++.h>
using namespace std;
const int N = 1010, M = 20010;
int n, m, k;
int h[N], e[M], w[M], ne[M], idx;
int dist[N];
deque<int> q;
bool st[N];
void add(int a, int b, int c){
e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx ++ ;
}
bool check(int bound){
memset(dist, 0x3f, sizeof dist);
memset(st, 0, sizeof st);
q.push_back(1);
dist[1] = 0;
while (q.size()){
int t = q.front();
q.pop_front();
if (st[t]) continue;
st[t] = true;
for (int i = h[t]; ~i; i = ne[i]){
int j = e[i], x = w[i] > bound;
if (dist[j] > dist[t] + x){
dist[j] = dist[t] + x;
if (!x) q.push_front(j);
else q.push_back(j);
}
}
}
return dist[n] <= k;
}
int main(){
cin >> n >> m >> k;
memset(h, -1, sizeof h);
while (m -- ){
int a, b, c;
cin >> a >> b >> c;
add(a, b, c), add(b, a, c);
}
int l = 0, r = 1e6 + 1;
while (l < r){
int mid = l + r >> 1;
if (check(mid)) r = mid;
else l = mid + 1;
}
if (r == 1e6 + 1) cout << -1 << endl;
else cout << r << endl;
return 0;
}
P2371 [国家集训队] 墨墨的等式
奇妙好题！！
同余最短路：与差分约束有异曲同工之妙，都将约束条件转化为边，每种状态转化为点。把本来与图论毫不相干的问题抽象到具体的图上，通过拓扑排序，最短路等基础算法获得最小状态，从而解决问题。
在本题中，以 \(0\) 到 \(a_1-1\) 为节点，对于检点 \(v\)，遍历数组 \(a\)，将 \(v\) 和 \((v+a_j) \bmod a_1\) 连一条权值为 \(a_j\) 的边。把图建完之后，就形成了一个有向图，跑最短路后 \(dis_v\) 的值即为用题目给出的数能获得的最小的，对 \(a_1\) 取模为 \(v\) 的值。
$ \forall v$, \(1\) 到 n 中有 \(\frac{n-d i s_{v}}{a_{1}}+1\) 数是可以用题目给出的数到达，所以答案为 $\sum_{i}^{a_{1}-1} \frac{r-d i s_{v}}{a_{1}}- \frac{l-1-d i s_{v}}{a_{1}} $
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define DEBUG(x) cerr << #x << '=' << x << endl
inline int rd()
{
int x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9')
{
if (ch == '-')
f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
return x * f;
}
void print(int x)
{
if (x < 0)
putchar('-'), x = -x;
if (x > 9)
print(x / 10);
putchar(x % 10 + '0');
return;
}
namespace Star_F
{
#define int long long
const int maxn = 13, N = 5e5 + 10, M = 11 * 5e5 + 10;
int n, l, r;
int a[maxn];
struct edge
{
int u, v, w, nxt;
} e[M];
int tot, head[N];
void add(int u, int v, int w) { e[++tot] = {u, v, w, head[u]}, head[u] = tot; }
struct node
{
int u, dis;
bool operator<(const node &a) const { return dis > a.dis; }
};
int vis[N], dis[N];
void init() {
FOR(i, 1, N - 1)
dis[i] = 1e18 + 10,
vis[i] = 0;
}
void dj(int s)
{
init();
dis[s] = 0;
priority_queue<node> q;
q.push({s, dis[s]});
while (!q.empty())
{
int u = q.top().u, k = q.top().dis;
q.pop();
if (vis[u] && k != dis[u])
continue;
vis[u] = 1;
for (int i = head[u]; i; i = e[i].nxt)
{
int v = e[i].v, w = e[i].w;
if (dis[v] > dis[u] + w)
{
dis[v] = dis[u] + w;
q.push({v, dis[v]});
}
}
}
}
int minn = 1e18 + 10, id = 0;
int sov(int x)
{
int ans = 0;
FOR(i, 0, minn - 1)
{
if (dis[i] <= x)
{
ans += (x - dis[i]) / minn + 1;
}
}
return ans;
}
void Main()
{
cin >> n >> l >> r;
FOR(i, 1, n)
{
cin >> a[i];
if (a[i] == 0)
{
i--;
n--;
continue;
}
if (minn > a[i])
minn = a[i], id = i;
}
FOR(i, 0, minn - 1)
{
FOR(j, 1, n)
{
if (a[j] == minn)
continue;
add(i, (i + a[j]) % minn, a[j]);
}
}
dj(0);
int ans = sov(r) - sov(l - 1);
cout << ans;
}
}
signed main(){
//freopen("inq.in","r",stdin);
//freopen("inq.out","w",stdout);
return Star_F::Main(), 0;
return 0;
}
P2446 [SDOI2010] 大陆争霸
由题可知，有的城市被保护。设 \(u\) 被 \(v\) 保护，我们从 \(v\) 建一条有向边到 \(u\) ，并记录 \(u\) 的入度 \(in_u\) 。广度遍历图，在摧毁 \(v\) 时，删去 \(v\) 到 \(u\) 的边，并更新入度。当 \(in_u=0\) 时，方可进入城市 \(u\) 。
设 \(arrive_i\) 表示到达 \(i\) 的时间（可能要在门口等待）。设 \(into_i\) 为进入 \(i\) 的时间（即什么时候所有保护 \(i\) 的城市被摧毁了）。设 \(dis_i\)表示摧毁 \(i\) 的时间，可得 \(dis_i=\max (arrive_i, into_i)\) 。设 \(\left\{j_{n}\right\}\) 为 \(i\) 的所有前驱，则 \(into_i=\max \{into_{j_n} \}\)。
#include <bits/stdc++.h>
using namespace std;
const int N=3001,M=200001;
struct edge{
int u,nxt,v;
}e[M];
int he[N],n,m,tote;
struct Graph{
int u,nxt;
}G[M];
int hg[N],totg;
struct node{
int key,len;
friend bool operator < (node x,node y){
return x.len<y.len;
}
};
int dis[N],into[N],in[N],arr[N];
bool vis[N];
priority_queue<node> q;
void dijkstra(int s){
for(int i=1;i<=n;i++)
dis[i]=arr[i]=1e9;
dis[s]=into[s]=arr[s]=0;
in[s]=0;
q.push({s,0});
int u,v;
while(!q.empty()){
u=q.top().key;
q.pop();
if(vis[u]) continue;
vis[u]=1;
for(int k=he[u];k;k=e[k].nxt){
v=e[k].u;
if(dis[u]+e[k].v<arr[v]){
arr[v]=dis[u]+e[k].v;
if(!in[v]){
dis[v]=max(into[v],arr[v]);
q.push({v,-dis[v]});
}
}
}
for(int k=hg[u];k;k=G[k].nxt){
v=G[k].u;
into[v]=max(into[v],dis[u]);
in[v]--;
if(!in[v]){
dis[v]=max(into[v],arr[v]);
q.push({v,-dis[v]});
}
}
}
}
int main(){
cin >> n >> m;
while(m--){
int x,y,l;
cin >> x >> y >> l;
e[++tote].u = y, e[tote].nxt = he[x], e[tote].v = l, he[x] = tote;
}
for(int i=1;i<=n;i++){
int x, y;
cin >> x;
while(x--){
cin >> y;
++in[i], G[++totg].u = i, G[totg].nxt = hg[y], hg[y] = totg;
}
}
dijkstra(1);
cout << dis[n] << endl;
return 0;
}
P8817 [CSP-S 2022] 假期计划
-
\(u\) 和 \(v\) 之间可达意为 \(u\) 和 \(v\) 之间可以不多于 \(k\) 次转车到达，及 \(u\)，\(v\) 的距离不多于 \(k+1\)。
-
一个点 \(u\) 在家附近，意为 \(u\) 和 1 之间可达。
注意，为方便，特殊地，我们认为自己和自己不可达。
观察数据范围，\(n^2\) 可过，已经足够处理出每个点对之间的距离了。因此首先应该采用 BFS，\(n^2\) 计算任意点对之间是否可达。
注意：对于满足边权全为 1 的图，单源最短路可以做到 \(\mathcal{O}(n)\) 的复杂度（采用 BFS）；
对于满足边权只有 0，1 两种的图，单源最短路也可以做到 \(\mathcal{O}(n)\) 的复杂度（采用 0-1 BFS）。
我们考虑一条 \(1 - a - b - c - d - 1\) 的路径，\(n^4\) 的枚举是不可行的。
很直接的想法是，依次考虑 \(a\)，\(b\)，\(c\)，\(d\)，发现贪心是不对的，从 1 选择了更大的权值景点 \(a\)，但是可能接下来能到达的 \(b\) 就小的可怜；而选一个权值稍小一点的 \(a\)，可能可达的 \(b\) 会很大。
但有一种贪心是对的，那就是确定了 \(a\)，\(b\)，\(c\)，选择 \(d\) 的时候。如果我们确定了 \(c\)，那么直接选择 \(c\) 可达的，在家附近的，不为 \(a\) 也不为 \(b\) 的权值最大的景点作为 \(d\) 即可。反正 \(d\) 之后没有景点了，所以可以直接贪心，没有后顾之忧。
由于环的对称性，可以发现确定了 \(b\)，\(c\)，\(d\) 之后，也可以贪心地选择 \(a\)。
有了大体思路。
我们定义 \(f(u)\) 表示 \(u\) 可达，且在家附近的权值最大的景点。
第一步：我们预处理出 \(f(u)\)。
第二步：直接 \(n^2\) 枚举，循环确定景点 \(b\) 和 \(c\)（\(b \ne c\)），然后记 \(a = f(b)\)，\(d = f(c)\)，然后试图用 \(w(a) + w(b) + w(c) + w(d)\) 更新答案，其中 \(w(u)\) 表示景点 \(u\) 的权值。
发现重复性的细节是存在问题的，比如：会出现 \(f(b) = c\) 的情况，此时让 \(a = f(b)\) 会导致 \(a = c\)，这是不允许的。
不过，贪心思想还是不变的：\(a\) 应该尝试更换为 \(u\) 可达，且在家附近的权值第二大的景点。
更换定义，\(f(u, k)\) 表示 \(u\) 可达，且在家附近的权值第 \(k\) 大的景点。
当发现 \(f(b, 1) = c\) 时，将 \(a\) 设置为 \(f(b, 2)\) 即可。
当发现 \(f(c, 1) = b\) 时，将 \(d\) 设置为 \(f(c, 2)\) 即可。
发现并没有完全解决：如果这样处理后的 \(a\) 和 \(d\) 仍然重复怎么办？
还是贪心思想，考虑把 \(a\) 或者 \(d\) 换成更小的那个比较一下就行，具体来说，比如原先 \(a = f(b, 2)\)，就把 \(a\) 下调为 \(f(b, 3)\)；或者原先 \(d = f(c, 2)\)，就把 \(d\) 下调为 \(f(c, 3)\)，然后比较两种方案哪种更好就行了。
重复性解决了，但是还有个细节：如果原先 \(a = f(b, 1)\)，发现 \(a = d\)，我们想下调 \(a\)。是直接把 \(a\) 设置成 \(f(b, 2)\) 吗？错误，我们还需要检查一下 \(f(b, 2)\) 是否等于 \(c\)，如果等于 \(c\) 还需要继续下调到 \(f(b, 3)\)。下调 \(d\) 同理。
至于原先 \(a = f(b, 2)\) 为啥下调 \(a\) 不需检查？因为如果原先 \(a = f(b, 2)\) 了说明 \(f(b, 1)\) 已经等于 \(c\) 了。所以 \(f(b, 3)\) 肯定什么问题都没有。
如果直接这么写是没有问题的，但是麻烦了。下面是一种更好写的处理方法：
直接分别枚举 \(a\) 分别作为 \(f(b, 1)\)，\(f(b, 2)\)，\(f(b, 3)\) 和 \(d\) 分别作为 \(f(c, 1)\)，\(f(c, 2)\)，\(f(c, 3)\) 组成的共九种情况，检查互异性然后更新答案即可。
需要提前预处理出 \(f(u, k \le 3)\)，其实只要在最开始 BFS 预处理的同时维护一下就行了。
注意某些点可达，还在家附近的点数可能没有 3 个，因此注意类似访问 \(f(b, 3)\) 时产生的越界问题。
如果枚举到某个 \(b\)，\(c\)，发现 \(b\) 或者 \(c\) 有对应点数不超过 3 的情况时，这组 \(b\)，\(c\) 不一定可以生成一组合法的解，属于正常现象。
题目保证全局上是有解的。
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
const int N = 2510, M = 20010;
int n, m, k;
LL w[N];
int h[N], e[M], ne[M], idx;
int dist[N], f[N][4], q[N];
bool st[N][N];
void add(int a, int b)
{
e[idx] = b, ne[idx] = h[a], h[a] = idx ++ ;
}
void bfs(int start, int f[])
{
int hh = 0, tt = 0;
memset(dist, 0x3f, sizeof dist);
dist[start] = 0;
q[0] = start;
while (hh <= tt)
{
int t = q[hh ++ ];
if (dist[t] == k + 1) continue;
for (int i = h[t]; ~i; i = ne[i])
{
int j = e[i];
if (dist[j] > dist[t] + 1)
{
dist[j] = dist[t] + 1;
st[start][j] = true;
q[ ++ tt] = j;
if (st[1][j])
{
f[3] = j;
for (int u = 3; u; u -- )
if (w[f[u]] > w[f[u - 1]])
swap(f[u], f[u - 1]);
else break;
}
}
}
}
}
int main()
{
scanf("%d%d%d", &n, &m, &k);
for (int i = 2; i <= n; i ++ ) scanf("%lld", &w[i]);
memset(h, -1, sizeof h);
while (m -- )
{
int a, b;
scanf("%d%d", &a, &b);
add(a, b), add(b, a);
}
for (int i = 1; i <= n; i ++ )
bfs(i, f[i]);
LL res = 0;
for (int b = 2; b <= n; b ++ )
for (int c = 2; c <= n; c ++ )
if (st[b][c])
for (int x = 0; x < 3; x ++ )
for (int y = 0; y < 3; y ++ )
{
int a = f[b][x], d = f[c][y];
if (a && d && a != d && a != c && b != d)
res = max(res, w[a] + w[b] + w[c] + w[d]);
}
printf("%lld\n", res);
return 0;
}
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