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BNU25359Escape Time II（状态压缩DP）

2024-07-20 06:05:24   218人阅读

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 theith room has ji jewels. At the beginning LTR is in rooms, 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 ji (1 ≤ji ≤ 1000000) indicating the number of jewels in theith 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 output0 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

Source

ZOJ Monthly, June 2012

Author

FU, Yujun 

#include<stdio.h>
#include<string.h>
#define mulit(i) (1<<i)
#define inf 999999999
#define ll long long
ll map[15][15],v[15],n,s,e,maxtim;
ll dp[mulit(11)+5][15];//dp[sta][i]表示状态sta以点i为结尾点
void dfs(int sta,int i)
{
    for(int j=0;j<n;j++)
    if(map[i][j]!=-1&&dp[sta|mulit(j)][j]>dp[sta][i]+map[i][j]&&dp[sta][i]+map[i][j]<=maxtim)
    {
        dp[sta|mulit(j)][j]=dp[sta][i]+map[i][j];
        dfs(sta|mulit(j),j);
    }
}
ll findpath()
{
   ll maxv=0,tv;
   for(int i=0;i<mulit(n);i++)
   for(int j=0;j<n;j++)
   dp[i][j]=inf;

   dp[mulit(s)][s]=0;
   dfs(mulit(s),s);
   for(int sta=1;sta<mulit(n);sta++)
   {
       tv=0;
       for(int i=0;mulit(i)<=sta;i++)
        if(mulit(i)&sta)
        tv+=v[i];
        if(dp[sta][e]<=maxtim&&tv>maxv)maxv=tv;
   }
    return maxv;
}
int main()
{
    ll m,a,b,c;
    while(scanf("%lld%lld%lld",&n,&m,&maxtim)>0)
    {
        scanf("%lld%lld",&s,&e);
        for(int i=0;i<n;i++)scanf("%lld",&v[i]);
        memset(map,-1,sizeof(map));
        while(m--)
        {
            scanf("%lld%lld%lld",&a,&b,&c);
            if(map[a][b]>c||map[a][b]<0)
            map[a][b]=map[b][a]=c;
        }
        printf("%lld\n",findpath());
    }
}


 

BNU25359Escape Time II（状态压缩DP）


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