Description
Joe works in a maze.
Unfortunately, portions of the maze have caught on fire,and the owner of the maze neglected to create a fire escape plan.Help Joe escape the maze.
Given Joe‘s location in the maze and which squares of the maze are on fire,you must determine whether Joe can exit the maze before the fire reaches him,and how fast he can do it.
Joe and the fire each move one square per minute, vertically or horizontally (notdiagonally). The fire spreads all four directions from each square that is onfire. Joe may exit the maze from any square that borders the edge of the maze.Neither Joe nor the fire may enter a square that is occupied by a wall.
2 4 4 #### #JF# #..# #..# 3 3 ### #J. #.F
3 IMPOSSIBLE
图论基础题。。
AC代码：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
using namespace std;
const int maxn = 1010;
int n, m;
char g[maxn][maxn]; //用于存储整个图
queue<pair<int,int> > q; //用于bfs的队列
int a[maxn][maxn]; //用于存储每个格子开始起火的时间
int move[][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}}; //用于上下左右走
void bfs1() //第一次遍历，预处理每个格子起火的时间
{
memset(a, -1, sizeof(a)); //初始化
while(!q.empty()) q.pop(); //清空队列
for(int i=0; i<n; i++)
for(int j=0; j<n; j++)
if(g[i][j] == 'F') //找出所有起始着火点
{
a[i][j] = 0;
q.push(make_pair(i,j));
}
while(!q.empty()) //bfs来处理所有点的开始着火的时间
{
pair<int ,int> tmp = q.front();
q.pop();
int x = tmp.first, y = tmp.second;
for(int i=0; i<4; i++)
{
int t1 = x + move[i][0], t2 = y + move[i][1];
if(a[t1][t2] != -1) continue;
if(t1 < 0 || t2 < 0 || t1 >= n || t2 >= m) continue;
if(g[t1][t2] == '#') continue;
a[t1][t2] = a[x][y] + 1;
q.push(make_pair(t1, t2));
}
}
}
int b[maxn][maxn];
int bfs2() //第二次遍历
{
memset(b, -1, sizeof(b));
while(!q.empty()) q.pop();
for(int i=0; i<n; i++)
for(int j=0; j<n; j++)
if(g[i][j] == 'J') //找到Joe
{
q.push(make_pair(i, j));
b[i][j] = 0;
break;
}
while(!q.empty()) //从Joe这里开始bfs
{
pair<int, int> tmp = q.front();
q.pop();
int x = tmp.first, y = tmp.second;
if(x == 0 || y == 0 || x == n - 1 || y == m - 1)
return b[x][y] + 1;
for(int i=0; i<4; i++)
{
int t1 = x + move[i][0], t2 = y + move[i][1];
if(t1<0 || t2<0 || t1>=n || t2>=m)continue;
if(b[t1][t2]!=-1)continue;
if(g[t1][t2]=='#')continue;
if(a[t1][t2] != -1 && b[x][y] + 1 >= a[t1][t2]) continue;
b[t1][t2] = b[x][y] + 1;
q.push(make_pair(t1, t2));
}
}
return -1; //没有路可走出去
}
int main()
{
int T;
scanf("%d", &T);
while(T--)
{
scanf("%d %d", &n, &m);
for(int i=0; i<n; i++)
scanf("%s", g[i]);
bfs1();
int ans = bfs2();
if(ans == -1) printf("IMPOSSIBLE\n");
else printf("%d\n", ans);
}
return 0;
}
原文地址：http://blog.csdn.net/u014355480/article/details/43423141
