hrbust 1621 迷宫问题II 广搜

题目链接：http://acm.hrbust.edu.cn/vj/index.php?/vj/index.php?c=&c=contest-contest&cid=134#problem/7

 很简单的广搜题。依然没有顺利的1A。没用优先队列。搞不清是不是还要回溯一下？【啊哈哈。我就是这么想的。】

// hrbust 1621 迷宫问题II
// 优先队列——广搜
// 什么时候需要回溯什么时候不需要？

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <queue>
using namespace std;
#define maxn 100000

char mp[210][210];
int dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};
int vis[210][210];
int r, c;
int step[210][210];
int stx, sty, edx, edy;

bool check(int x, int y) {
    if (x >= 0 && x < r && y >= 0 && y < c && !vis[x][y] && mp[x][y] != '#')
        return true;
    return false;
}

struct Node {
    friend bool operator < (Node n1, Node n2)
    {
        return step[n1.x][n1.y] > step[n2.x][n2.y];//"<"为从大到小排列，">"为从小打到排列
    }
    int x, y;
}node[40100];

priority_queue<Node> qn;

void bfs(int x, int y) {
    int head = 0, tail = 0;
    memset(vis, 0, sizeof(vis));
    for (int i=0; i<r; ++i) {
        for (int j=0; j<c; ++j) {
            step[i][j] = maxn;
        }
    }
    while(!qn.empty()) {
        qn.pop();
    }

    vis[x][y] = 1;
    step[x][y] = 0;
    Node temp;
    temp.x = x;
    temp.y = y;
   // node[tail++] = temp;
   qn.push(temp);
    //while(head < tail) {
    while(!qn.empty()) {
        //Node now = node[head++];
        Node now = qn.top();
        qn.pop();
        for (int i=0; i<4; ++i) {
            temp.x = now.x + dir[i][0];
            temp.y = now.y + dir[i][1];
            int tstep = 0;
            if (check(temp.x, temp.y)) {
                if (mp[temp.x][temp.y] >= '1' && mp[temp.x][temp.y] <= '9')
                    tstep = mp[temp.x][temp.y] - '0' + 1;
                else tstep = 1;
                step[temp.x][temp.y] = min(step[temp.x][temp.y], step[now.x][now.y] + tstep);
                vis[temp.x][temp.y] = 1;
                //node[tail++] = temp;
                qn.push(temp);
                if (temp.x == edx && temp.y == edy) break;
            }
        }
    }
    return;
}

int main() {
    int t;
    cin >> t;
    while(t--) {
        cin >> r >> c;

        for (int i=0; i<r; ++i) {
            for (int j=0; j<c; ++j) {
                cin >> mp[i][j];
                if (mp[i][j] == 'Z') {
                    stx = i, sty = j;
                }
                else if (mp[i][j] == 'W') {
                    edx = i, edy = j;
                }
            }
        }


        bfs(stx, sty);
        int ans = step[edx][edy];
        if (ans == maxn) {
            cout << "IMPOSSIBLE
";
        }
        else cout << ans << endl;
    }
    return 0;
}