hdu 4284 Travel(压缩DP，4级)

Travel

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2641    Accepted Submission(s): 724

Problem Description

　　PP loves travel. Her dream is to travel around country A which consists of N cities and M roads connecting them. PP has measured the money each road costs. But she still has one more problem: she doesn't have enough money. So she must work during her travel. She has chosen some cities that she must visit and stay to work. In City_i she can do some work to earn Ci money, but before that she has to pay Di money to get the work license. She can't work in that city if she doesn't get the license but she can go through the city without license. In each chosen city, PP can only earn money and get license once. In other cities, she will not earn or pay money so that you can consider Ci=Di=0. Please help her make a plan to visit all chosen cities and get license in all of them under all rules above.
　　PP lives in city 1, and she will start her journey from city 1. and end her journey at city 1 too.

 

Input

　　The first line of input consists of one integer T which means T cases will follow.
　　Then follows T cases, each of which begins with three integers: the number of cities N (N <= 100) , number of roads M (M <= 5000) and her initiative money Money (Money <= 10^5) .
　　Then follows M lines. Each contains three integers u, v, w, which means there is a road between city u and city v and the cost is w. u and v are between 1 and N (inclusive), w <= 10^5.
　　Then follows a integer H (H <= 15) , which is the number of chosen cities.
　　Then follows H lines. Each contains three integers Num, Ci, Di, which means the i_th chosen city number and Ci, Di described above.(Ci, Di <= 10^5)

 

Output

　　If PP can visit all chosen cities and get all licenses, output "YES", otherwise output "NO".

 

Sample Input

2
4 5 10
1 2 1
2 3 2
1 3 2
1 4 1
3 4 2
3
1 8 5
2 5 2
3 10 1
2 1 100
1 2 10000
1
2 100000 1

 

Sample Output

YES
NO

 

Source

2012 ACM/ICPC Asia Regional Tianjin Online

 

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预处理两点距离，然后压缩DP。

dp[mask][k];最后在k 走过状态mask

注意重边

#include <iostream>
#include <cstring>
#include <cstdio>
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define clr(f,z) memset(f,z,sizeof(f))
#define ll(x) (1<<x)
using namespace std;
const int mp=109;
const int oo=0x3f3f3f3f;
int g[mp][mp],dis[mp][mp],C[mp],D[mp],id[mp];
int dp[ ll(16) ][16];
int n,m,money;

int main()
{
    int cas,a,b,c,H;
    while(~scanf("%d",&cas))
    {
      while(cas--)
      {
        clr(g,0x3f);
        scanf("%d%d%d",&n,&m,&money);
        FOR(i,1,m)
        {
          scanf("%d%d%d",&a,&b,&c);
          g[a][b]=g[b][a]=min(g[a][b],c);
        }

        scanf("%d",&H);
        FOR(i,0,H-1)
        {
          scanf("%d%d%d",&id[i],&C[i],&D[i]);
        }

        clr(dis,0x3f);
        FOR(i,1,n)
        g[i][i]=0;
        FOR(i,1,n)FOR(j,1,n)FOR(k,1,n)
        {
          g[j][k]=min(g[j][k],g[j][i]+g[i][k]);
        }

        clr(dp,-1);
        int zzz = ll(H)-1,then;
        FOR(i,0,H-1)
        if(money >= g[1][ id[i] ] + D[i])
        dp[ ll(i) ][i] = money - g[1][ id[i] ] - D[i] + C[i];

        FOR(i,1,zzz)FOR(j,0,H-1)FOR(k,0,H-1)
        if(j != k && (i&ll(k)) && (i&ll(j)) && dp[ i^ll(j) ][k] != -1)
        {
          then = i^ll(j);
          if(dp[then][k] >= g[ id[k] ][ id[j] ] + D[j])
            dp[i][j] = max(dp[i][j],dp[then][k] - g[ id[k] ][ id[j] ] - D[j] + C[j]);
        }

        bool flag=0;
        FOR(i,0,H-1)
        if(dp[zzz][i]>=g[ id[i] ][1])
          flag=1;
        if(flag)puts("YES");
        else puts("NO");
      }
    }
    return 0;
}