ACM/ICPC 2018亚洲区预选赛北京赛站网络赛 A.Saving Tang Monk II（优先队列广搜）

#include<bits/stdc++.h>
using namespace std;
const int maxN = 123;
const int inf = 1e9 + 7;
char G[maxN][maxN];
int times[maxN][maxN][6];
int n, m, sx, sy, ex, ey, ans;
int dir[4][2] = {{0,1},{1,0},{0,-1},{-1,0}};

struct node {
    int x, y, t, o;
    bool operator < (const node& p) const {
        return t > p.t;
    }
};
void bfs() {
    node t = (node) {
        sx, sy, 0, 0
    };
    times[sx][sy][0] = 0;
    priority_queue<node> q;
    q.push(t);
    while(!q.empty()) {
        node u = q.top();
        q.pop();
        int x = u.x, y = u.y, o = u.o, t = u.t;

        if(x == ex && y == ey) {
            ans = min(ans, t);
            return;
        }
        for(int d = 0; d < 4; d++) {
            int tx = x + dir[d][0];
            int ty = y + dir[d][1];
            if(tx < 0 || tx >= n || ty < 0 || ty >= m) //越界
                continue;

            if(G[tx][ty] == ‘#‘) {
                if(o == 0) {
                    continue;
                } else if(times[tx][ty][o-1] > t + 2) {

                    q.push((node) {
                        tx,ty,t+2,o-1
                    });

                    times[tx][ty][o-1] = t + 2;
                }
            } else if(G[tx][ty] == ‘P‘ && times[tx][ty][o] > t) {

                q.push((node) {
                    tx,ty,t,o
                });

                times[tx][ty][o] = t;
            } else if(G[tx][ty] == ‘B‘ && o < 5 && times[tx][ty][o + 1] > t + 1) {

                q.push((node) {
                    tx,ty,t+1,o+1
                });

                times[tx][ty][o + 1] = t + 1;
            } else if(times[tx][ty][o] > t + 1) {

                q.push((node) {
                    tx,ty,t+1,o
                });

                times[tx][ty][o] = t + 1;
            }
        }
    }
}
int main() {

    while(~scanf("%d %d", &n, &m)) {
        if(n == 0)
            break;
        for(int i = 0; i < maxN; i++)
            for(int j = 0; j < maxN; j++)
                for(int k = 0 ; k <= 5; k++)
                    times[i][j][k] = inf;
        for(int i = 0; i < n; i++) {
            scanf("%s", G[i]);
        }
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                if(G[i][j] == ‘S‘)
                    sx = i, sy = j;
                if(G[i][j] == ‘T‘)
                    ex = i, ey = j;
            }
        }

        ans = inf;
        bfs();
        if(ans == inf) {
            printf("-1\n");
        } else {
            printf("%d\n", ans);
        }
    }
    return 0;
}

原文地址：https://www.cnblogs.com/Jadon97/p/9691594.html