How far away ?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5766 Accepted Submission(s): 2166
Problem Description
There are n houses in the village and some bidirectional roads connecting them. Every day peole always like to ask like this "How far is it if I want to go from house A to house B"? Usually it hard to
answer. But luckily int this village the answer is always unique, since the roads are built in the way that there is a unique simple path("simple" means you can‘t visit a place twice) between every two houses. Yout task is to answer all these curious people.
Input
First line is a single integer T(T<=10), indicating the number of test cases.
For each test case,in the first line there are two numbers n(2<=n<=40000) and m (1<=m<=200),the number of houses and the number of queries. The following n-1 lines each consisting three numbers i,j,k, separated bu a single space, meaning that there is a road
connecting house i and house j,with length k(0<k<=40000).The houses are labeled from 1 to n.
Next m lines each has distinct integers i and j, you areato answer the distance between house i and house j.
Output
For each test case,output m lines. Each line represents the answer of the query. Output a bland line after each test case.
Sample Input
2 3 2 1 2 10 3 1 15 1 2 2 3 2 2 1 2 100 1 2 2 1
Sample Output
10 25 100 100
Source
AC代码:
#include <iostream> #include <cstring> #include <vector> #include <cstdio> using namespace std; const int M = 4 * 1e4 + 100; typedef long long ll; typedef pair<int,int> P; vector<P>G[M]; int vis[M],isGO,e; void dfs(int x, int cost) { if(isGO || vis[x]) return ; if(x == e)//数据太大,要改成全局变量 { isGO = 1; printf("%d\n",cost); return ; } vis[x] = 1; for(int i = 0; i < G[x].size(); i++) { P a = G[x][i]; dfs(a.first,cost + a.second); } vis[x] = 0; } void solve() { int x,y,c,n,m; scanf("%d %d",&n,&m); for(int i = 1; i < n; i++) { scanf("%d %d %d",&x,&y,&c); G[x].push_back(make_pair(y,c)); G[y].push_back(make_pair(x,c)); } while(m--) { isGO = 0; scanf("%d %d",&x,&e); dfs(x,0); } for(int i = 1; i <= n; i++) G[i].clear(); } int main() { int T,cnt = 0; scanf("%d",&T); while(T--) { solve(); } return 0; }