2020-03-22（ccf最优配餐201409-4 ）

这道题只有60分，我对比了网上的博客，也没发现有什么不一样的地方，希望有大佬告知原因！谢谢！

#include <iostream>
#include <queue>
#include <cstring>
 using namespace std;
 
const int N = 1050;
bool g[N][N];  // true  代表不能走 
int n, m, k , d;
typedef pair<int, int> PII;
int min_cost[N];
typedef pair<int, pair<int, int> > PIII;
PII store[N];
PIII buyer[N];
queue<PII> q;
int dx[4] = {-1, 0, 1, 0},  dy[4] = {0, 1, 0, -1};
int mincost[N][N];
bool st[N][N];

//void bfs(int a, int b)
void bfs()
{
//  q.push({a, b});
//  mincost[a][b] = 0;
//  st[a][b] = true;
    while(q.size())
    {
        PII t = q.front();
        q.pop();
        int x = t.first;   int y = t.second;
//      cout<<x<<" "<<y<<endl;
        int cost = mincost[x][y];
        for(int i = 0; i<4; i++)
        {
            int xx = x + dx[i],  yy = y + dy[i];
//          cout<<"test:"<<xx<<" "<<yy<<" "<<st[xx][yy] <<" "<< g[xx][yy]<<endl;
            if(!st[xx][yy] && xx > 0 && xx <= n && yy >0 && yy <=n && g[xx][yy] == 0)
            {
//              cout<<"aaa"<<xx<<" "<<yy<<endl;
                mincost[xx][yy] = min( cost + 1, mincost[xx][yy] );
                q.push({xx, yy});
                st[xx][yy] = true;
            }
        }
    }
}
 int main() 
 {
    memset(mincost, 0x3f, sizeof (mincost));
    scanf("%d%d%d%d", &n, &m, &k, &d);
    
    for(int i = 0; i<m; i++) 
    {
        int a, b;
        scanf("%d%d", &a, &b);
        store[i].first = n+1 - b;   store[i].second = a; 
     }
     
     for(int i = 0; i<k; i++)
     {
        int a , b ,c;
        scanf("%d%d%d", &a, &b, &c);
        buyer[i].first =  n+1 - b;  buyer[i].second.first = a;  buyer[i].second.second = c;
      } 
      
      for(int i  = 0; i<d; i++)
      {
        int a;int b;
        scanf("%d%d", &a, &b);
        g[n+ 1 - b][a] = true;;
      }
      
      for(int i = 0; i < m ; i++)
      {
//          memset(st, 0, sizeof(st));
          int a = store[i].first;   int b= store[i].second;
          q.push({a, b});
          mincost[a][b] = 0;
          st[a][b] = true;
//        bfs(a, b);
      }
      bfs();
      
      long long sum = 0;
      for(int i = 0; i<k; i++)
      {
        int x = buyer[i].first;  int y = buyer[i].second.first;   int c = buyer[i].second.second;
//      cout<<x<<" "<<y<<" "<<endl;
        sum += mincost[x][y] * c;
      }
      cout<<sum<<endl;
 }