zoj 2081 - Mission Impossible

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2081

刚看到最短路就想到了bfs搜索，但是看完题后发现bfs搜索解决完最短路问题后不好处理炸弹问题。我想过在state里边加上一个Boolean常量，来判断这个最短路是否过地雷，后来突然想到可以再dfs一下来判断过雷的最短路径的数目，所以那个方法就没有再尝试。dfs很容易想到要剪枝，有点像hdu1010的路径剪枝吧。。。悲剧的是我一直以为要求的是间谍被炸的概率，因此一直wa，更悲剧的时样例数据正好是50.00%，因此这个错误好难发现。。。。看错题的孩纸伤不起啊。。。。要吸取教训！！！

代码如下：

#include<iostream>
#include<queue>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
int N, M, minstep, sum, msum;
char maze[15][15];
int vis[15][15];
int dx[] = { 0, -1, 0, 1};
int dy[] = {-1,  0, 1, 0};
typedef struct 
{
    int x, y;
    int step;
}State;
State begin, end;
queue<State> q;
//char s[15];

int mayreach(int x, int y, int step)
{
    if((int)abs((double)(end.x - x)) + abs(double(end.y - y)) > minstep - step)
        return 0;
    return 1;
}

void dfs(int x, int y, int step, bool mines)
{
    if(x < 0 || x > N-1 || y < 0 || y > M-1)
        return;
    if(x == end.x && y == end.y && step == minstep)
    {
        if(mines == true)
            msum++;
        sum++;
    }
    if(maze[x][y] == '#')
        return;
    if(!mayreach(x, y, step))
        return;
    if(maze[x][y] == 'M')
        mines = true;
    dfs(x, y-1, step + 1, mines);
    dfs(x, y+1, step + 1, mines);
    dfs(x-1, y, step + 1, mines);
    dfs(x+1, y, step + 1, mines);
}

int main()
{
    int ncases, i, j, ok, c;
    scanf("%d", &ncases);
    getchar();
    for(c = 1; c <= ncases; c++)
    {
        scanf("%d%d", &N, &M);
        getchar();
        
        for(i = 0; i < N; i++)
        {
            for(j = 0; j < M; j++)
            {
                scanf("%c", &maze[i][j]);
                if(maze[i][j] == 'S')
                {
                    begin.x = i;
                    begin.y = j;
                }
                else if(maze[i][j] == 'T')
                {
                    end.x = i;
                    end.y = j;
                }
            }
            getchar();
        }
        memset(vis, 0, sizeof(vis));
        ok = 0;
        begin.step = 0;
        sum = msum = 0;
        vis[begin.x][begin.y] = 1;
        q.push(begin);
        while(!q.empty())
        {
            State u = q.front();
            q.pop();
            if(u.x == end.x && u.y == end.y)
            {
                ok = 1;
                minstep = u.step;
                break;
            }
            for(i = 0; i < 4; i++)
            {
                State v;
                v.step = u.step + 1;
                v.x = u.x + dx[i];
                v.y = u.y + dy[i];
                if(v.x >= 0 && v.x < N && v.y >=0 && v.y < M && maze[v.x][v.y] != '#' && !vis[v.x][v.y])
                {
                    q.push(v);
                    vis[v.x][v.y] = 1;
                }
            }
        }
        dfs(begin.x, begin.y, 0, false);
        //printf("%d %d %d\n", minstep, msum, sum);
        if(!ok || sum == msum)
            printf("Mission #%d:\nMission Impossible.\n\n", c);
        else printf("Mission #%d:\nThe probability for the spy to get to the telegraph transmitter is %.2lf%%.\n\n", c, (0.0+sum-msum)*100/sum);
        while(!q.empty())
            q.pop();
        //gets(s);
    }
    return 0;
}