题目链接：hdu2612
思路：题意是求两个人到某一个KFC花费时间和最小，其实就是求最短距离和，用两个BFS，分别以两个人为起点，分别记录下两人到每个KFC的距离，然后求出最小的和
#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
#define N 205
using namespace std;
char map[N][N];
int v[N][N],ans1[N][N],ans2[N][N],d[4][2] = { {-1,0},{1,0},{0,-1},{0,1} };
int x1,x2,y1,y2,n,m,num;
struct node
{
int x,y,step;
friend bool operator < (node a,node b)
{
return a.step > b.step;
}
};
void bfs1(int x,int y,int total)//total表示KFC的总数目
{
memset(ans1,1,sizeof(ans1));
memset(v,0,sizeof(v));
priority_queue <node> q;
node s,temp;
s.x = x;
s.y = y;
s.step = 0;
v[x][y] = 1;
q.push(s);
while(!q.empty())
{
temp = q.top();
q.pop();
if(map[temp.x][temp.y] == '@')
{
ans1[temp.x][temp.y] = temp.step;
total -- ;
}
if(total == 0) return;//所有的KFC都已经计算过
for(int i= 0 ; i < 4 ; i ++)
{
s = temp;
s.x += d[i][0];
s.y += d[i][1];
if(s.x < 0 || s.x >= n || s.y < 0 || s.y >= m || v[s.x][s.y] || map[s.x][s.y] == '#')
continue;
v[s.x][s.y] = 1;
s.step ++;
q.push(s);
}
}
}
void bfs2(int x,int y,int total)
{
memset(ans2,1,sizeof(ans2));
memset(v,0,sizeof(v));
priority_queue <node> q;
node s,temp;
s.x = x;
s.y = y;
s.step = 0;
v[x][y] = 1;
q.push(s);
while(!q.empty())
{
temp = q.top();
q.pop();
if(map[temp.x][temp.y] == '@')
{
ans2[temp.x][temp.y] = temp.step;
total -- ;
}
if(total == 0) return;
for(int i= 0 ; i < 4 ; i ++)
{
s = temp;
s.x += d[i][0];
s.y += d[i][1];
if(s.x < 0 || s.x >= n || s.y < 0 || s.y >= m || v[s.x][s.y] || map[s.x][s.y] == '#')
continue;
v[s.x][s.y] = 1;
s.step ++;
q.push(s);
}
}
}
int main()
{
int i,j;
while(~scanf("%d%d",&n,&m))
{
num = 0;
for(i = 0 ; i < n ; i ++)
{
scanf("%s",map[i]);
for(j = 0 ; j < m ; j ++)
{
if(map[i][j] == 'Y')
{
x1 = i;
y1 = j;
}
else if(map[i][j] == 'M')
{
x2 = i;
y2 = j;
}
else if(map[i][j] == '@')
num++;
}
}
bfs1(x1,y1,num);
bfs2(x2,y2,num);
int sum = 10000000;
for(i = 0 ; i < n ; i ++)
for(j = 0 ; j < m ; j ++)
if(map[i][j] == '@')
{
sum = min(sum,ans1[i][j] + ans2[i][j]);
}
printf("%d\n",sum * 11);
}
return 0;
}