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  • HDU 4771

    http://acm.hdu.edu.cn/showproblem.php?pid=4771

    给一个地图,@是起点,给一些物品坐标,问取完所有物品的最小步数,不能取完输出-1

    物品数最多只有四个,状态压缩一下bfs即可

    #include <iostream> 
    #include <cstdio>
    #include <algorithm>
    #include <queue>
    #include <cstring>
    
    using namespace std;
    
    int n, m, k;
    
    char G[105][105];
    int vis[105][105][1<<4];
    
    struct p {
        int x, y, key, step;
    };
    
    int dx[] = {1, -1, 0, 0};
    int dy[] = {0, 0, 1, -1};
    
    p now;
    
    int bfs() {
        queue <p> q;
        now.step = 0;
        q.push(now);
        vis[now.x][now.y][0] = 1;
        while(!q.empty()) {
            p u = q.front();
            q.pop();
            for(int i = 0; i < 4; i++) {
                int xx = u.x + dx[i];
                int yy = u.y + dy[i];
                if(xx < 0 || xx >= n || yy < 0 || yy >= m) continue;
                if(G[xx][yy] == '#') continue;
                p next;
                if(G[xx][yy] >= '0' && G[xx][yy] <= '3') {
                    if(!vis[xx][yy][u.key]) {
                        vis[xx][yy][u.key|(1<<(G[xx][yy]-'0'))] = 1;
                        next.x = xx, next.y = yy, next.key = u.key|(1<<(G[xx][yy]-'0')), next.step = u.step + 1;
                        if(next.key == (1<<k)-1) return next.step;
                        q.push(next);
                    }
                }
                else {
                    if(!vis[xx][yy][u.key]) {
                        vis[xx][yy][u.key] = 1;
                        next.x = xx, next.y = yy, next.key = u.key, next.step = u.step + 1;
                        q.push(next);
                    }
                }
            }
        }
        return -1;
    }
    
    int main() {
        while(~scanf("%d%d", &n, &m)) {
            if(!n && !m) break;
            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] == '@')
                        now.x = i, now.y = j;
            scanf("%d", &k);
            int flag = 1;
            now.key = 0;
            memset(vis, 0, sizeof(vis));
            for(int i = 0; i < k; i++) {
                int x, y;
                scanf("%d%d", &x, &y);
                if(G[x-1][y-1] == '#') flag = 0;
                else if(G[x-1][y-1] == '@') {
                    G[x-1][y-1] = i + '0';
                    vis[x-1][y-1][1<<i] = 1;    
                }
                else G[x-1][y-1] = i + '0';    
            }
            if(!flag) {
                puts("-1");
                continue;
            }
            printf("%d
    ", bfs());
        }
        return 0; 
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/xiaohongmao/p/4587328.html
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