POJ 2312 Battle City (BFS状态搜索)

题意：

。。。

思路：

R和S是不能走的，遇到B时发射子弹和移动可以合并成一个代价为2的移动。

我们的状态是 [ 当前的位置， 已经消耗的时间 ]

状态用优先队列保存， 每次取出， 剪枝，  扩展

因为优先队列中取出的始终是 time 最小的状态，所以判断到达了终点就可以退出。

//#include<bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <cassert>
#include <algorithm>
#include <cmath>
#include <climits>
#include <set>
#include <map>
using namespace std;

#define SPEED_UP iostream::sync_with_stdio(false);
#define FIXED_FLOAT cout.setf(ios::fixed, ios::floatfield);
#define rep(i, s, t) for(int (i)=(s);(i)<=(t);++(i))
#define urep(i, s, t) for(int (i)=(s);(i)>=(t);--(i))

typedef long long LL;

const int Maxn = 305;

char mm[Maxn][Maxn];
int n, m, t[Maxn][Maxn], sx, sy, ex, ey;

const int dx[] = {-1, 1, 0, 0};
const int dy[] = {0, 0, -1, 1};

struct st{
    int x;
    int y;
    int t;
    st(int x, int y, int t):x(x), y(y), t(t){};
    inline bool operator < (const st& rhs) const {
        return t > rhs.t;
    }
};

inline int check(int x, int y) {
    return 0 <= x && x < n && 0 <= y && y < m;
}

void init() {
    rep(i, 0, n-1) {
        scanf("%s", mm[i]);
        rep(j, 0, m-1) {
            if (mm[i][j] == 'Y') sx = i, sy = j;
            else if (mm[i][j] == 'T') ex = i, ey = j;
        }
    }
}

void solve() {
    priority_queue<st> q;
    //queue<st> q;
    q.push( st(sx, sy, 0) );
    rep(i, 0, n-1)
        rep(j, 0, m-1) t[i][j] = INT_MAX;

    int x, y, old_t;
    t[sx][sy] = 0;
    while (!q.empty()) {
        st fr = q.top();q.pop();
        x = fr.x, y = fr.y, old_t = fr.t;

        if (x == ex && y == ey) {
            break;
            //continue;
        }
        int nx, ny, nt;
        rep(i, 0, 3) {
            nx = x + dx[i];
            ny = y + dy[i];
            nt = old_t + 1;
            // skip ?
            if (!check(nx, ny) || mm[nx][ny] == 'S' || mm[nx][ny] == 'R') continue;
            if (mm[nx][ny] == 'B') ++nt;
            if (nt >= t[ex][ey] || nt >= t[nx][ny]) continue;
            t[nx][ny] = nt;
            //cout << "going to " << nx << ", " << ny << endl;
            //cout << "time = " << nt << endl;
            q.push(st(nx, ny, nt));
        }
    }

    if (t[ex][ey] == INT_MAX) {
        printf("-1\n");
    }
    else {
        printf("%d\n", t[ex][ey]);
    }
}
int main() {
#ifndef ONLINE_JUDGE
    freopen("input.in", "r", stdin);
#endif
    //SPEED_UP
    while (scanf("%d%d",&n, &m) != EOF && (m && n)) {
        init();
        if (sx == ex && sy == ey) cout << '0' << endl;
        solve();
    }

    return 0;
}