hdu4284 dfs+floyd

题意:
      给你n个城市,m条边,要有h个必须旅游和打工的城市,问你能不能从1把所有必须的h个城市全部旅游并且打工完...

思路:

      先一遍floyd跑出全局最短路,然后暴力枚举出打工的顺序,当打工的个数达到h的时候判断下是否能从第h个打工的点用当前剩余的前回到1如果能就ok..

#include<stdio.h>
#include<string.h>

#define inf 1000000000

int map[100+50][100+50];
int H[20] ,C[20] ,D[20];
int HH[20];

int minn(int a ,int b)
{
   return a < b ? a : b;
}

void floyd(int n)
{
   for(int k = 1 ;k <= n ;k ++)
   for(int i = 1 ;i <= n ;i ++)
   for(int j = 1 ;j <= n ;j ++)
   map[i][j] = minn(map[i][k] + map[k][j] ,map[i][j]);
}

int mark[20] ,ok ,H_n;

void DFS(int s ,int mony ,int sum)
{
   if(sum == H_n && mony >= map[s][1])
   {
      ok = 1;
      return;
   }
   for(int i = 1 ;i <= H_n ;i ++)
   {
      if(mark[i]) continue;
      if(mony < map[s][H[i]]) continue;
      if(mony  - map[s][H[i]] >= D[i] && !ok)
      {
         mark[i] = 1;
         DFS(H[i] ,mony - map[s][H[i]] - D[i] + C[i] ,sum + 1);
         mark[i] = 0;
      }
   }
}

int main ()
{
   int i ,j ,m ,n ,mon ,t ,a ,b ,c;
   scanf("%d" ,&t);
   while(t--)
   {
      scanf("%d %d %d" ,&n ,&m ,&mon);
      for(i = 1 ;i <= n ;i ++)
      {
         for(j = 1 ;j <= n ;j ++)
         map[i][j] = inf;
         map[i][i] = 0;
      }
      for(i = 1 ;i <= m ;i ++)
      {
         scanf("%d %d %d" ,&a ,&b ,&c);
         map[a][b] = map[b][a] = minn(c ,map[a][b]);
      }
      floyd(n);
      scanf("%d" ,&H_n);
      memset(mark ,0 ,sizeof(mark));
      for(i = 1 ;i <= H_n ;i ++)
      {
         scanf("%d %d %d" ,&H[i] ,&C[i] ,&D[i]);
      }
      ok = 0; 
      DFS(1 ,mon ,0);
      if(ok) printf("YES
");
      else printf("NO
");
   }
   return 0;
}