hdu 1242 用bfs求最短路

最新推荐文章于 2020-02-07 17:42:41 发布

原创最新推荐文章于 2020-02-07 17:42:41 发布 · 721 阅读

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本文介绍了一个经典的最短路径问题——救援Angel，并通过广度优先搜索(BFS)算法来解决这一问题。该问题涉及如何在包含墙壁、道路及守卫的监狱地图中找到从起点到终点的最短时间路径。

Rescue

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21907 Accepted Submission(s): 7801

Problem Description
Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the prison.

Angel’s friends want to save Angel. Their task is: approach Angel. We assume that “approach Angel” is to get to the position where Angel stays. When there’s a guard in the grid, we must kill him (or her?) to move into the grid. We assume that we moving up, down, right, left takes us 1 unit time, and killing a guard takes 1 unit time, too. And we are strong enough to kill all the guards.

You have to calculate the minimal time to approach Angel. (We can move only UP, DOWN, LEFT and RIGHT, to the neighbor grid within bound, of course.)

Input
First line contains two integers stand for N and M.

Then N lines follows, every line has M characters. “.” stands for road, “a” stands for Angel, and “r” stands for each of Angel’s friend.

Process to the end of the file.

Output
For each test case, your program should output a single integer, standing for the minimal time needed. If such a number does no exist, you should output a line containing “Poor ANGEL has to stay in the prison all his life.”

思路：这是最短路问题，当然要用bfs来做，可是如果用最基础的bfs肯定是会出问题的，因为有‘x’的存在你需要多花1个单位的时间，所以你需要把有’x’的位置放在下一级花费同样时间的队列当中。

#include <algorithm>
#include <cstring>
#include <iostream>
#include <queue>
using namespace std;
typedef pair<int, int> P;
char a[210][210];
int b[210][210];
int vis[210][210];
int n,m;
int sx,sy;
int prove;
int dx[4] = {-1,1,0,0};
int dy[4] = {0,0,1,-1};
//这个顺序很重要，如果换一种方式很可能就会出错。
queue<pair<int, int>>que;
void bfs(int x,int y)
{
//常规的bfs再加一个if判断即可。
    memset(vis, 0, sizeof(vis));
    vis[x][y] = 1;
    b[x][y] = 0;
    que.push(P(x,y));
    while (que.size()) {
        P p = que.front();
        que.pop();
        int x = p.first;
        int y = p.second;
        if (a[x][y] == 'a') {
            break;
        }
        if (a[x][y] == 'x') {
            b[x][y] += 1;
            a[x][y] = '.';
            que.push(P(x,y));
            continue;
            //continue的存在确保了‘x’这个点放到了下一级队列中。
        }
        for (int i = 0; i<4; i++) {
            int x1 = x + dx[i];
            int y1 = y + dy[i];
            if (0<=x1&&x1<n&&0<=y1&&y1<m&&a[x1][y1]!='#'&&vis[x1][y1]==0) {
                b[x1][y1] = b[x][y]+1;
                vis[x1][y1] = 1;
                que.push(P(x1,y1));
            }
        }
    }
}
int main()
{
    while (cin>>n>>m) {
        int x,y,x1,y1;
        for (int i = 0; i<n; i++) {
            for (int j = 0; j<m; j++) {
                cin>>a[i][j];
                if (a[i][j] == 'r') {
                    x = i;y = j;
                }
                if (a[i][j] == 'a') {
                    x1 = i;y1 = j;
                }
            }
        }
        bfs(x,y);
        if (vis[x1][y1]) {
            cout<<b[x1][y1]<<endl;
        }
        else
            cout << "Poor ANGEL has to stay in the prison all his life." << endl;
    }
    return 0;
}