uva 10816 Travel in Desert（简单的好题~两种方法）
题意：
给出 一个图
点与点之间的路径上有两个权值 路径长度和温度
要求在所走路径中的温度的最大值最小的前提下 走最短路径
解题思路1：
首先用 最小生成树 的方法走出 最小瓶颈路 。把在这期间用到的全部温度小于 路径上最大温度 的边存下来，作为接下来求最短路径的图。
在新生成的图中求最短路径就可以；
code
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
const int maxm = 10005;
const int maxn = 105;
struct Edge{
int u,v;
double dist,tm;
void read(){
scanf("%d%d%lf%lf",&u,&v,&tm,&dist);
u--;v--;
}
bool operator<(const Edge et)const{
if(tm != et.tm) return tm < et.tm;
else return dist < et.dist;
}
}e[maxm];
int n,m,s,t;
int parent[maxn];
vector<Edge> g[maxn];
void init(){
scanf("%d%d",&s,&t);
s--;t--;
for(int i = 0; i < m; i++){
e[i].read();
}
for(int i = 0; i < n; i++){
g[i].clear();
}
}
int find(int x){
if(parent[x] == x) return x;
else return parent[x] = find(parent[x]);
}
const int INF = 0x3f3f3f3f;
const int MAXNODE = 105;
struct Edge2 {
int u, v;
double dist;
Edge2() {}
Edge2(int u, int v, double dist) {
this->u = u;
this->v = v;
this->dist = dist;
}
};
struct HeapNode {
double d;
int u;
HeapNode() {}
HeapNode(double d, int u) {
this->d = d;
this->u = u;
}
bool operator < (const HeapNode& c) const {
return d > c.d;
}
};
struct Dijkstra {
int n, m;
vector<Edge2> edges;
vector<int> g[MAXNODE];
bool done[MAXNODE];
double d[MAXNODE];
int p[MAXNODE];
void init(int tot) {
n = tot;
for (int i = 0; i < n; i++)
g[i].clear();
edges.clear();
}
void add_Edge(int u, int v, double dist) {
edges.push_back(Edge2(u, v, dist));
m = edges.size();
g[u].push_back(m - 1);
}
void print(int s, int e) {//shun xu
if (s == e) {
printf("%d", e + 1);
return;
}
print(s, edges[p[e]].u);
printf(" %d", e + 1);
}
void print2(int s, int e) {//ni xu
if (s == e) {
printf("%d", e + 1);
return;
}
printf("%d ", e + 1);
print2(s, edges[p[e]].u);
}
void dijkstra(int s) {
priority_queue<HeapNode> Q;
for (int i = 0; i < n; i++) d[i] = INF*1.0;
d[s] = 0.0;
memset(done, false, sizeof(done));
Q.push(HeapNode(0, s));
while (!Q.empty()) {
HeapNode x = Q.top(); Q.pop();
int u = x.u;
if (done[u]) continue;
done[u] = true;
for (int i = 0; i < g[u].size(); i++) {
Edge2& e = edges[g[u][i]];
if (d[e.v] > d[u] + e.dist) {
d[e.v] = d[u] + e.dist;
p[e.v] = g[u][i];
Q.push(HeapNode(d[e.v], e.v));
}
}
}
}
} graph;
void solve(){
// printf("...\n");
double ans = 0;
sort(e,e+m);
// for(int i = 0; i < m; i++){
// printf("%.1lf %.1lf %d %d\n",e[i].dist,e[i].tm,e[i].u,e[i].v);
// }
for(int i = 0; i < n; i++) parent[i] = i;
double max_tm = 500.0;
graph.init(n);
for(int i = 0; i < m; i++){
if(e[i].tm > max_tm) break;
graph.add_Edge(e[i].u,e[i].v,e[i].dist);
graph.add_Edge(e[i].v,e[i].u,e[i].dist);
int pu = find(e[i].u);
int pv = find(e[i].v);
if(pu == pv) continue;
parent[pu] = pv;
if(find(s) == find(t)){
max_tm = e[i].tm;
}
}
graph.dijkstra(s);
graph.print(s,t);
printf("\n");
printf("%.1lf %.1lf\n",graph.d[t],max_tm);
}
int main(){
while(scanf("%d%d",&n,&m) != EOF){
init();
solve();
}
return 0;
}
解题思路二：
把原图存下来，然后二分温度，再把全部小于温度mid的边拿出来构成一个新图，然后继续dijkstra求最短路，有成功和不成功两种结果，找到能成功的最小温度就可以
code
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
using namespace std;
const int MAXNODE = 105;
const int MAXEDGE = 20005;
typedef double Type;
const Type INF = 0x3f3f3f3f;
struct Edge {
int u, v;
Type dist, d;
Edge() {}
Edge(int u, int v, Type dist, Type d = 0) {
this->u = u;
this->v = v;
this->dist = dist;
this->d = d;
}
void read() {
scanf("%d%d%lf%lf", &u, &v, &d, &dist);
u--; v--;
}
};
struct HeapNode {
Type d;
int u;
HeapNode() {}
HeapNode(Type d, int u) {
this->d = d;
this->u = u;
}
bool operator < (const HeapNode& c) const {
return d > c.d;
}
};
int n, m, s, t;
struct Dijkstra {
int n, m;
Edge edges[MAXEDGE];
int first[MAXNODE];
int next[MAXEDGE];
bool done[MAXNODE];
Type d[MAXNODE];
int p[MAXNODE];
void init(int n) {
this->n = n;
memset(first, -1, sizeof(first));
m = 0;
}
void add_Edge(int u, int v, Type dist) {
edges[m] = Edge(u, v, dist);
next[m] = first[u];
first[u] = m++;
}
void add_Edge(Edge e) {
edges[m] = e;
next[m] = first[e.u];
first[e.u] = m++;
}
void print(int e) {//shun xu
if (p[e] == -1) {
printf("%d", e + 1);
return;
}
print(edges[p[e]].u);
printf(" %d", e + 1);
}
void print2(int e) {//ni xu
if (p[e] == -1) {
printf("%d\n", e + 1);
return;
}
printf("%d ", e + 1);
print2(edges[p[e]].u);
}
bool dijkstra(int s, int t) {
priority_queue<HeapNode> Q;
for (int i = 0; i < n; i++) d[i] = INF;
d[s] = 0;
p[s] = -1;
memset(done, false, sizeof(done));
Q.push(HeapNode(0, s));
while (!Q.empty()) {
HeapNode x = Q.top(); Q.pop();
int u = x.u;
if (u == t)
return true;
if (done[u]) continue;
done[u] = true;
for (int i = first[u]; i != -1; i = next[i]) {
Edge& e = edges[i];
if (d[e.v] > d[u] + e.dist) {
d[e.v] = d[u] + e.dist;
p[e.v] = i;
Q.push(HeapNode(d[e.v], e.v));
}
}
}
return false;
}
} gao;
Edge e[MAXEDGE];
bool judge(double mid) {
gao.init(n);
for (int i = 0; i < m; i++) {
if (e[i].d > mid) continue;
gao.add_Edge(e[i]);
gao.add_Edge(Edge(e[i].v, e[i].u, e[i].dist, e[i].d));
}
if (gao.dijkstra(s, t)) return true;
return false;
}
int main() {
while (~scanf("%d%d", &n, &m)) {
scanf("%d%d", &s, &t);
s--; t--;
for (int i = 0; i < m; i++)
e[i].read();
double l = 0, r = 50, mid;
while (r - l > 1e-8) {
mid = (l + r) / 2;
if (judge(mid)) r = mid;
else l = mid;
}
if (judge(r)) {
gao.print(t); printf("\n");
printf("%.1lf %.1lf\n", gao.d[t], mid);
}
}
return 0;
}
二分在非常多情况下都是非常好用的一种方法~
