(2312)POJ

题意：问你能否从R走到T，如果能，就算最小的步数

解法：BFS+记忆化搜索，设置一个数组存下当前位置的最小步数，最后返dp[x][y],x y为T的坐标即可

#include<iostream>
#include<cstdio>
#include<string.h>
#include<string>
#include<stack>
#include<set>
#include<algorithm>
#include<cmath>
#include<vector>
#include<map>
#include<queue>

#define ll __int64
#define lll unsigned long long
#define MAX 1000009
#define eps 1e-8
#define INF 0xfffffff
#define mod 1000000007

using namespace std;

int dir[4][2] = {-1,0,0,1,1,0,0,-1};

struct point
{
    int x;
    int y;
    int step;
};

queue<point>Q;

int N,M;
char ma[309][309];
int vis[309][309];
int mintime[309][309];
int sx,sy,dx,dy;
int flag;

int BFS(point x)
{
    while(!Q.empty()) Q.pop();
    Q.push(x);
    point hd;
    while(!Q.empty())
    {
        hd = Q.front();
        Q.pop();
        for(int i = 0; i<4; i++)
        {
            int xx = hd.x + dir[i][0];
            int yy = hd.y + dir[i][1];
            if(xx<0||xx>=N||yy<0||yy>=M) continue;
            if(ma[xx][yy]!='S'&&ma[xx][yy]!='R')
            {
                point t;
                t.x = xx;
                t.y = yy;
                t.step = hd.step + 1;
                if(ma[xx][yy] == 'B')
                    t.step++;
                if(t.step < mintime[xx][yy])
                {
                    mintime[xx][yy] = t.step;
                    Q.push(t);
                }
            }
        }
    }
    return mintime[dx][dy];
}

int main()
{
    while(~scanf("%d%d",&N,&M))
    {
        if(N==0&&M==0)break;
        memset(vis,0,sizeof(vis));
        for(int i = 0; i<N; i++)
            scanf("%s",ma[i]);
        for(int i = 0; i<N; i++)
        {
            for(int j = 0; j<M; j++)
            {
                mintime[i][j] = INF;
                if(ma[i][j] == 'Y')
                {
                    sx = i;
                    sy = j;
                }
                if(ma[i][j] == 'T')
                {
                    dx = i;
                    dy = j;
                }
            }
        }
        point start;
        start.x = sx;
        start.y = sy;
        start.step = 0;
        flag = 0;
        mintime[sx][sy] = 0;
        int ans = BFS(start);
        if(ans<INF)
        {
            printf("%d\n",ans);
        }
        else
            printf("-1\n");

    }
    return 0;
}