UVa 11624，两次BFS

题目链接：http://vjudge.net/contest/132239#problem/A

题目链接：https://uva.onlinejudge.org/external/116/11624.pdf

《训练指南》P307

分析：只需要预处理每个格子起火的时间，在BFS扩展节点的时候加一个判断，到达该节点的时候，格子没有起火。

写法很巧妙，两次BFS类似，数据加一维kind，表示Joe到达该点和火到达该点。

#include <bits/stdc++.h>

using namespace std;

const int INF = 0x3f3f3f3f;
const int maxr = 1000+5;
const int maxc = 1000+5;

int R,C;
char maze[maxr][maxc];
struct Cell{
    int r,c;
    Cell (int r,int c) : r(r),c(c) {}
};
const int dr[] = {-1,1,0,0};
const int dc[] = {0,0,-1,1};
int d[maxr][maxc][2] ,vis[maxr][maxc][2];

queue<Cell> Q;
void bfs(int kind)
{
    while(!Q.empty())
    {
        Cell cell = Q.front();
        Q.pop();
        int r = cell.r, c = cell.c;
        for(int dir = 0; dir < 4; dir++)
        {
            int nr = r + dr[dir], nc = c + dc[dir];
            if(nr >= 0 && nr < R && nc >= 0 && nc < C && maze[nr][nc] == '.' && !vis[nr][nc][kind])
            {
                Q.push(Cell(nr, nc));
                vis[nr][nc][kind] = 1;
                d[nr][nc][kind] = d[r][c][kind] + 1;
            }
        }
    }
}


int ans;
void check(int r,int c)
{
    if(maze[r][c]!='.'||!vis[r][c][0]) return;      ///走不到
    if(!vis[r][c][1]||d[r][c][0]<d[r][c][1]) ans= min(ans,d[r][c][0]+1);    ///该点火到达不了，或者joe先于火
}

int main()
{
    //freopen("input.txt","r",stdin);
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&R,&C);
        int jr,jc;
        vector<Cell> fires;
        for(int i=0;i<R;i++)
        {
            scanf("%s",maze[i]);
            for(int j=0;j<C;j++)
            {
                if(maze[i][j]=='J')
                {
                    jr = i;
                    jc = j;
                    maze[i][j] = '.';
                }
                else if(maze[i][j]=='F')
                {
                    fires.push_back(Cell(i,j));
                    maze[i][j] = '.';
                }
            }
        }
        memset(vis,0,sizeof(vis));

        ///各个点到joe的距离
        ///Joe
        vis[jr][jc][0] = 1;
        d[jr][jc][0] = 0;
        Q.push(Cell(jr,jc));
        bfs(0);

        for(int i = 0;i<fires.size();i++)
        {
            vis[fires[i].r][fires[i].c][1] = 1;
            d[fires[i].r][fires[i].c][1] = 0;
            Q.push(fires[i]);
        }
        bfs(1);

        ans = INF;
        for(int i=0;i<R;i++)
        {
            check(i,0);
            check(i,C-1);
        }
        for(int i=0;i<C;i++)
        {
            check(0,i);
            check(R-1,i);
        }
        if(ans==INF) printf("IMPOSSIBLE
");
        else printf("%d
",ans);
    }
    return 0;
}