ZOJ 1649 Rescue （BFS）

来源：互联网发布：springboot配置端口号编辑：程序博客网时间：2024/04/29 18:32

题目大意：在一个矩阵里，有五种字符，一种是起点，一种是终点，一种代表墙不能走的，一种代表警卫需要花时间杀警卫的，最后一种代表路，每走一个格要花一个单位时间，杀一个警卫要花一个单位时间。求从起点到终点，花得最短时间！

分析：最短路径，其实很简单的，bfs搜索，然后判断一下是不是墙，是不是警卫和路，分别处理；在用一个数组d来记录每个点到起点的最短路径，同时也是用来标记已经遍历过的状态。

代码：

#include <cstdio>#include <cstring>#include <queue>#include <vector>using namespace std;const int N = 210;const int M = 210;const int INF = 1000000;int n, m, xa, ya, xr, yr, d[N][M], x[4] = { 0, 0, 1, -1 }, y[4] = { 1, -1, 0, 0 };bool vis[N][N];char g[N][M];struct P { int x, y; };bool bfs() {    queue<P> Q;    P e;     e.x = xr, e.y = yr;    Q.push( e );    for ( int i = 1; i <= n; ++i )        for ( int j = 1; j <= m; ++j ) d[i][j] = INF;    d[xr][yr] = 0;    while ( !Q.empty() ) {        P u = Q.front(); Q.pop();        int ux = u.x , uy = u.y;              //printf("%d %d\n", ux, uy);          for ( int i = 0; i < 4; ++i ){             int x1 = u.x + x[i], y1 = u.y + y[i];            if ( g[x1][y1] == '#' ) continue;            if ( g[x1][y1] == 'x' ) {                if ( d[x1][y1] > d[ux][uy] + 2 ) {                    d[x1][y1] = d[ux][uy] +2 ;                    e.x = x1, e.y = y1;                    Q.push(e);                }            }            else if ( g[x1][y1] == '.' || g[x1][y1] == 'a' ) {               if ( d[x1][y1] > d[ux][uy] + 1 ) {                    d[x1][y1] = d[ux][uy] + 1;                    e.x = x1; e.y = y1;                    Q.push(e);                }            }        }    }    if ( d[xa][ya] == INF ) return false;    else return true;}int main() {    while ( scanf("%d%d", &n, &m) != EOF ) {        getchar();        memset(vis, 0, sizeof(vis));        for ( int i = 0; i <= m+1; ++i ) g[i][0] = g[i][m+1] = '#';        for ( int i = 0; i <= n+1; ++i ) g[0][i] = g[n+1][i] = '#';        for ( int i = 1; i <= n; ++i, getchar() )             for ( int j = 1; j <= m; ++j ) {                scanf("%c", &g[i][j]);                if ( g[i][j] == 'a' ) xa = i, ya = j;                if ( g[i][j] == 'r' ) xr = i, yr = j;            }        //for ( int i = 0; i <= m + 1; ++i ) printf("%s\n", g[i]);        if ( bfs() ) printf("%d\n", d[xa][ya]);        else printf("Poor ANGEL has to stay in the prison all his life.\n");    }}