UVA 11624 Fire!(广度优先搜索)

题目大意:在一个N*M的迷宫内,J代表某人(只有一个),F代表火(可能不只一个),#代表墙,火每分钟会向四周除了墙以外的地方扩散一层,问人能否在没被火烧到

之前逃出迷宫,若能逃出输出最短时间。很明显的bfs。但由于火到达的地方人不能抵达,故需先对火进行bfs,标记后若人在火烧到之前抵达即可。最后逃出时间需要加一,

因为当时只是抵达边界,若逃出时间需加一。

#include <stdio.h>
#include <queue>
#include <string.h>
#include <algorithm>
using namespace std;
#define oo 0x3f3f3f3f
int dir[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
int m, n, time[1005][1005], vis[1005][1005];
char maps[1005][1005];
struct point
{
    int x, y, step;
};
point start;
queue<point>Fire;
void Fire_time()
{
    point now, next;
    while(Fire.size())
    {
        now = Fire.front();
        Fire.pop();
        time[now.x][now.y] = now.step;
        for(int i=0; i<4; i++)
        {
            next.x = now.x + dir[i][0];
            next.y = now.y + dir[i][1];
            next.step = now.step + 1;
            if(next.x>=0 && next.x<m && next.y>=0 && next.y<n && !vis[next.x][next.y]
               && maps[next.x][next.y]!='#')
            {
                vis[next.x][next.y] = 1;
                Fire.push(next);
            }
        }
    }
}
int bfs()
{
    queue<point>Q;
    Q.push(start);
    memset(time, oo, sizeof(time));
    Fire_time();
    memset(vis, 0, sizeof(vis));
    point now, next;
    while(Q.size())
    {
        now = Q.front();
        Q.pop();
        if(now.x==m-1 || now.y==n-1 || now.x==0 || now.y==0)return now.step+1;
        for(int i=0; i<4; i++)
        {
            next.x = now.x + dir[i][0];
            next.y = now.y + dir[i][1];
            next.step = now.step + 1;
            if(next.x>=0 && next.x<m && next.y>=0 && next.y<n && !vis[next.x][next.y]
               &&  maps[next.x][next.y]=='.' && next.step<time[next.x][next.y])
            {
                vis[next.x][next.y] = 1;
                    Q.push(next);
            }
        }
    }
    return -1;
}
int main()
{
    int Text;
    scanf("%d", &Text);
    while(Text--)
    {
        scanf("%d %d", &m, &n);
        memset(vis, 0, sizeof(vis));
        for(int i=0; i<m; i++)
            scanf("%s", maps[i]);
        for(int i=0; i<m; i++)
            for(int j=0; j<n; j++)
        {
            if(maps[i][j]=='J')
            {
                start.x = i;
                start.y = j;
                start.step = 0;
            }
            if(maps[i][j]=='F')
            {
                point s;
                s.x = i;
                s.y = j;
                s.step = 0;
                Fire.push(s);
                vis[i][j] = 1;
            }
        }
        int ans = bfs();
        if(ans==-1)printf("IMPOSSIBLE
");
        else printf("%d
", ans);
    }
    return 0;
}
原文地址:https://www.cnblogs.com/zznulw/p/5666188.html