Gym
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Gym - 100814D Frozen Rivers
这题就是给你一个树形的河,所有的河都结冰了,从根节点u开始,u的所有儿子边都同时融化,每融化一条边,它的兄弟节点融化速度减半。查询在某个时间内,融化后到达饿叶子节点的个数。这题就是一个搜索题,我是用bfs写的,因为查询次数比较多,可以用二分。
/*upper_bound(begin,end,key),start是查找的起点,end是终点,key是关键值,lower_bound()用法一样,upper_bound()函数,返回第一个大于要找的值得位置而Lower_bound是小于等于关键字的位置*/#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <stack>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>#include <bitset>#define INF 0x3f3f3f3f#define eps 1e-6#define PI 3.1415926#define mod 1000000009#define base 2333using namespace std;typedef long long LL;const int inf = 1e9;const int maxn = 1e5 + 10;const int maxx = 1e3 + 10;int t, n, v, q, cnt;LL c, x, ans[maxn], tmp[maxn], val[maxn];vector<int> G[maxn];void bfs() { queue<int> q; q.push(1); while(!q.empty()) { int cur = q.front(); q.pop(); if(G[cur].size() == 0) ans[++cnt] = tmp[cur]; else { LL minn = 1e18; for(int i = 0; i < G[cur].size(); i++) { int v = G[cur][i]; minn = min(minn, val[v]); q.push(v); } for(int i = 0; i < G[cur].size(); i++) { int v = G[cur][i]; tmp[v] = tmp[cur]+minn+2*(val[v]-minn); } } }}void solve() { scanf("%d", &t); while(t--) { scanf("%d", &n); memset(ans, 0, sizeof(ans)); memset(tmp, 0, sizeof(tmp)); cnt = 0; for(int i = 0; i <= n; i++) G[i].clear(); for(int i = 2; i <= n; i++) { scanf("%d%lld", &v, &c); val[i] = c, G[v].push_back(i); } bfs(); sort(ans+1, ans+cnt+1); scanf("%d", &q); while(q--) { scanf("%lld", &x); LL *xx = upper_bound(ans+1, ans+cnt+1, x); printf("%d\n", xx-(ans+1)); } }}int main() { //freopen("kingdom.in","r",stdin); //freopen("kingdom.out","w",stdout); solve();}
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