2017 ACM/ICPC 广西邀请赛 题解
题目链接 Problems
HDOJ上的题目顺序可能和现场比赛的题目顺序不一样，
我这里的是按照HDOJ的题目顺序来写的。
Problem 1001
签到
#include <bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for (int i(a); i <= (b); ++i)
typedef long long LL;
LL n, f[31];
int ans;
int main(){
f[1] = 1LL;
for (LL i = 2; i <= 15; ++i){
f[i] = 1LL;
rep(j, 1, i) f[i] *= i;
}
while (~scanf("%lld", &n)){
ans = 0;
rep(i, 1, 15) if (f[i] <= n) ans = max(ans, i);
printf("%d\n", ans);
}
return 0;
}
Problem 1002
这道题的话，其实一个点上是可以有多种颜色的
因为最多只有51种颜色，所以我们维护51棵线段树即可。
我们对y坐标建立线段树，对于每个y，我们只需要知道最小的x在哪里即可。
因为他的询问的x坐标下界总是1，那么我们只要看看y1到y2的最小值是否小于等于给定的x即可。
询问的时候做51次子询问就可以了。
#include <bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for (int i(a); i <= (b); ++i)
#define dec(i, a, b) for (int i(a); i >= (b); --i)
typedef long long LL;
const int N = 2e6 + 10;
int X, c, d, x, y, op, ans, ret, cnt;
int root[53];
int l[N], r[N], v[N];
void update(int &i, int L, int R, int x, int val){
if (i == 0){
i = ++cnt;
v[i] = val;
}
v[i] = min(v[i], val);
if (L == R) return;
int mid = (L + R) >> 1;
if (x <= mid) update(l[i], L, mid, x, val);
else update(r[i], mid + 1, R, x, val);
}
void query(int i, int L, int R){
if (ret || i == 0) return;
if (c <= L && R <= d){
if (v[i] <= X) ret = 1;
return;
}
int mid = (L + R) >> 1;
if (c <= mid) query(l[i], L, mid);
if (d > mid) query(r[i], mid + 1, R);
}
int main(){
while (true){
scanf("%d", &op);
if (op == 3) break;
if (op == 0){
rep(i, 1, cnt) l[i] = r[i] = 0;
memset(root, 0, sizeof root);
cnt = 0;
}
if (op == 1){
scanf("%d%d%d", &x, &y, &c);
update(root[c], 1, 1000000, y, x);
}
if (op == 2){
scanf("%d%d%d", &X, &c, &d);
ans = 0;
rep(i, 0, 50){
ret = 0;
query(root[i], 1, 1000000);
ans += ret;
}
printf("%d\n", ans);
}
}
return 0;
}
Problem 1003
我们对于每一条边，找到所有包含这条边的三元环，个数计为x
然后这对答案的贡献就是$C_{x}^{2}$
判断两点之间是否有边的时候要手写哈希才能过
#include <bits/stdc++.h>
const int N = 2e5 + 10;
#define rep(i, a, b) for (int i(a); i <= (b); ++i)
#define dec(i, a, b) for (int i(a); i >= (b); --i)
namespace Hashmap{
const int P = 1000007, seed = 2333;
int u[N << 2], v[N << 2], nt[N << 2];
int head[P], inum;
inline void init(){
inum = 0;
memset(u, 0, sizeof u);
memset(v, 0, sizeof v);
memset(nt, 0, sizeof nt);
memset(head, 0, sizeof head);
}
inline void add(int _u, int _v){
int t = (_u * seed + _v) % P;
u[++inum] = _u, v[inum] = _v, nt[inum] = head[t], head[t] = inum;
}
inline bool query(int _u, int _v){
int t = (_u * seed + _v) % P;
for (int p = head[t]; p; p = nt[p])
if (u[p] == _u && v[p] == _v) return 1;
return 0;
}
}
using namespace std;
using namespace Hashmap;
typedef long long LL;
vector<int> a[N];
struct ss{ int x, y; } e[N];
int n, m;
int main(){
while (~scanf("%d%d",&n, &m)){
init();
rep(i, 1, n) a[i].clear();
rep(i, 1, m){
int x, y;
scanf("%d%d", &x, &y);
e[i].x = x;
e[i].y = y;
a[x].push_back(y);
a[y].push_back(x);
add(x, y);
add(y, x);
}
LL ans = 0;
rep(i, 1, m){
int x = e[i].x, y = e[i].y;
if (a[e[i].y].size() < a[e[i].x].size()) swap(x, y);
LL tot = 0;
for (auto u: a[x]) if (query(u, y)) tot++;
ans += tot * (tot - 1) / 2;
}
printf("%lld\n", ans);
}
return 0;
}
Problem 1004
这题规律找了半天……
$f[n] = f[n - 1] + 5f[n - 2] + f[n - 3] - f[n - 4]$
然后矩阵加速下就可以了
#include <bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for (int i(a); i <= (b); ++i)
#define dec(i, a, b) for (int i(a); i >= (b); --i)
#define MP make_pair
#define fi first
#define se second
typedef long long LL;
const LL mod = 1e9 + 7;
struct Matrix{
LL arr[6][6];
} f, unit, a;
LL m, k;
int n;
Matrix Add(Matrix a, Matrix b){
Matrix c;
rep(i, 1, n) rep(j, 1, n){
c.arr[i][j] = (a.arr[i][j] + b.arr[i][j]) % mod;
}
return c;
}
Matrix Mul(Matrix a, Matrix b){
Matrix c;
rep(i, 1, n) rep(j, 1, n){
c.arr[i][j] = 0;
rep(k, 1, n) (c.arr[i][j] += (a.arr[i][k] * b.arr[k][j] % mod)) %= mod;
}
return c;
}
Matrix Pow(Matrix a, LL k){
Matrix ret(unit); for (; k; k >>= 1, a = Mul(a, a)) if (k & 1) ret = Mul(ret, a); return ret;
}
int main(){
n = 4;
memset(f.arr, 0, sizeof f.arr);
memset(a.arr, 0, sizeof a.arr);
memset(unit.arr, 0, sizeof unit.arr);
rep(i, 1, n) unit.arr[i][i] = 1;
f.arr[1][1] = 1;
f.arr[1][2] = 5;
f.arr[1][3] = 1;
f.arr[1][4] = 1000000006;
f.arr[2][1] = 1;
f.arr[3][2] = 1;
f.arr[4][3] = 1;
a.arr[1][1] = 36;
a.arr[2][1] = 11;
a.arr[3][1] = 5;
a.arr[4][1] = 1;
while (~scanf("%lld", &m)){
if (m == 1LL){ puts("1"); continue;}
if (m == 2LL){ puts("5"); continue;}
if (m == 3LL){ puts("11"); continue;}
if (m == 4LL){ puts("36"); continue;}
if (m == 5LL){ puts("95"); continue;}
k = m - 4;
Matrix b = Pow(f, k);
Matrix c = Mul(b, a);
printf("%lld\n", c.arr[1][1]);
}
return 0;
}
Problem 1010
这道题的话先转DFS序，然后求出以每个点为根的子树的对应区间，
于是就转化成了求区间里选一个数和x异或后的最大值。
可持久化trie即可。
#include <bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for (int i(a); i <= (b); ++i)
#define dec(i, a, b) for (int i(a); i >= (b); --i)
typedef long long LL;
const int N = 1e5 + 10;
const int M = 1e7 + 10;
int n, m, tot = 0;
int rt[N], id[M], t[M][2];
int l, r, x;
int a[N], c[N], L[N], R[N], g[N];
int ti;
int u;
vector <int> v[N];
inline void insert(int pre, int x, int k){
int now = rt[k] = ++tot; id[tot] = k;
dec(i, 30, 0){
int j = (x >> i) & 1;
t[now][j ^ 1] = t[pre][j ^ 1];
t[now][j] = ++tot;
id[tot] = k;
now = t[now][j];
pre = t[pre][j];
}
}
inline int query(int l, int r, int x){
int ans = 0, tmp = rt[r];
dec(i, 30, 0){
if (id[tmp] < l) break;
int j = ((x >> i) & 1) ^ 1;
if (id[t[tmp][j]] >= l) ans |= (1 << i);
else j ^= 1;
tmp = t[tmp][j];
}
return ans;
}
void dfs(int x){
L[x] = ++ti;
g[ti] = a[x];
for (auto u : v[x]){
dfs(u);
}
R[x] = ti;
}
int main(){
while (~scanf("%d%d", &n, &m)){
memset(id, 0, sizeof id);
memset(rt, 0, sizeof rt);
memset(t, 0, sizeof t);
memset(L, 0, sizeof L);
memset(R, 0, sizeof R);
tot = 0;
ti = 0;
rep(i, 0, n + 1) v[i].clear();
rep(i, 1, n) scanf("%d", a + i);
rep(i, 2, n){
scanf("%d", &x);
v[x].push_back(i);
}
dfs(1);
id[0] = -1;
insert(0, 0, 0);
rep(i, 1, n) insert(rt[i - 1], g[i], i);
rep(i, 1, m){
scanf("%d%d", &u, &x);
printf("%d\n", query(L[u], R[u], x));
}
}
return 0;
}
