给出一个地图,地图有四种路面,经过每种路面花费的时间不同,问从起点到终点所花费的最少时间是多少
把到各个点的花费存入队列中,然后弹出,即可得到最小
Sample Input
4 6
1 2 10
T...TT
TTT###
TT.@#T
..###@
0 1 3 0
1 2 10
T...TT
TTT###
TT.@#T
..###@
0 1 3 0
4 6
1 2 2
T...TT
TTT###
TT.@#T
..###@
0 1 3 0
2 2
5 1 3
T@
@.
0 0 1 1
Sample Output
Case 1: 14
Case 2: 8
Case 3: -1
Case 2: 8
Case 3: -1
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
int x,y,v1,v2,v3,b,n,m;
int x1,x2,y1,y2;
int p[25][25];
int vis[25][25];
int v[5];
int dir[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};
struct node
{
int x,y,step;
friend bool operator < (node a,node b)
{
return a.step > b.step;
}
}; bool judge(int x,int y)
{
if(x < 0 || y < 0 || x >= n || y >= m || p[x][y] < 0 || vis[x][y])
return false;
return true;
} int work()
{
priority_queue<node>que;
node a,b;
a.x=x1;
a.y=y1;
a.step = 0;
que.push(a);
memset(vis,0,sizeof(vis));
vis[x1][y1] = 1;
while(!que.empty())
{
a = que.top();
que.pop();
if(a.x == x2 && a.y == y2)
return a.step;
for(int i = 0;i < 4;i++)
{
b = a;
b.x += dir[i][0];
b.y += dir[i][1];
if(!judge(b.x,b.y))
continue;
b.step += v[p[b.x][b.y]];
vis[b.x][b.y] = 1;
que.push(b);
}
}
return -1;
} int main()
{
int i,j,cas = 1;
char s[1000];
while(~scanf("%d%d",&n,&m))
{
scanf("%d%d%d",&v[3],&v[2],&v[1]);
for(i = 0; i < n; i++)
{
scanf("%s",s);
for(j = 0;s[j]; j++)
{
if(s[j]=='T') p[i][j] = 1;
else if(s[j] == '.') p[i][j] = 2;
else if(s[j] == '#') p[i][j] = 3;
else if(s[j] == '@') p[i][j] = -1;
}
}
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
int ans = work();
printf("Case %d: %d\n",cas++,ans);
}
return 0;
}