http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=4744

Escape Time II

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Time Limit: 2 Seconds ???? Memory Limit: 65536 KB

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There is a fire in LTR ’ s home again. The fire can destroy all the things in t seconds, so LTR has to escape in t seconds. But there are some jewels in LTR ’ s rooms, LTR love jewels very much so he wants to take his jewels as many as possible before he goes to the exit. Assume that the ith room has j_i jewels. At the beginning LTR is in room s, and the exit is in room e.

Your job is to find a way that LTR can go to the exit in time and take his jewels as many as possible.

Input

There are multiple test cases.
For each test case:
The 1st line contains 3 integers n (2 ≤ n ≤ 10), m, t (1 ≤ t ≤ 1000000) indicating the number of rooms, the number of edges between rooms and the escape time.
The 2nd line contains 2 integers s and e, indicating the starting room and the exit.
The 3rd line contains n integers, the ith interger j_i (1 ≤ j_i ≤ 1000000) indicating the number of jewels in the ith room.
The next m lines, every line contains 3 integers a, b, c, indicating that there is a way between room a and room b and it will take c (1 ≤ c ≤ t) seconds.

Output

For each test cases, you should print one line contains one integer the maximum number of jewels that LTR can take. If LTR can not reach the exit in time then output 0 instead.

Sample Input

3 3 5 0 2 10 10 10 0 1 1 0 2 2 1 2 3 5 7 9 0 3 10 20 20 30 20 0 1 2 1 3 5 0 3 3 2 3 2 1 2 5 1 4 4 3 4 2

Sample Output

30 80 #include <iostream> #include <cstdio> #include <cstring> using namespace std; const int maxx = 20; int Edge[maxx][maxx]; int val[maxx]; bool vis[maxx]; int big = 0,e,n,t; void dfs(int s, int num,int ju) {if(num>t) return;if(s == e)if(ju > big && num<=t)big = ju;vis[s] = true;for(int i=0;i<n;i++){if(i!=s && Edge[s][i]<1e8 && !vis[i]){dfs(i,num + Edge[s][i],ju + val[i]);}}vis[s] = false;} void Floyd() {for(int i=0; i<n;i++)for(int j=0; j<n; j++)for(int k=0; k<n; k++)if(Edge[j][i] + Edge[i][k] < Edge[j][k])Edge[j][k] = Edge[j][i] + Edge[i][k]; } int main() {int m;int s;while(~scanf("%d %d %d",&n,&m,&t)){memset(Edge,0x6,sizeof(Edge));memset(val,0,sizeof(val));memset(vis,0,sizeof(vis));scanf("%d %d",&s,&e);for(int i=0;i<n;i++)scanf("%d",&val[i]);for(int i=1;i<=m;i++){int u,v,w;scanf("%d %d %d",&u,&v,&w);Edge[u][v] = Edge[v][u] = w;}Floyd();big = 0;vis[s] = true;dfs(s,0,val[s]);printf("%d\n",big);}return 0; }

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轉(zhuǎn)載于:https://www.cnblogs.com/cancangood/p/3946203.html

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