• ZOJ 3652 Maze 模拟,bfs,读题 难度:2


    http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=4842

    要注意题目中两点:

    1.在踏入妖怪控制的区域那一刹那,先减行动力,然后才能杀妖怪

    2.在妖怪控制区域行动力也会恢复

    3.妖怪也许不在自己的控制区域

    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <queue>
    
    using namespace std;
    const int maxn=100;
    const int dx[4] = {0,0,-1,1};
    const int dy[4] = {1,-1,0,0};
    int n,m,l,k,sx,sy,tx,ty;
    int maz[maxn][maxn];
    int mon[6][2];
    
    int vis[maxn][maxn][maxn];
    int ind[maxn][maxn];
    struct P{
        int x,y,sta,r;
        P(){x = y = sta = r = 0;}
        P(int _x,int _y,int _sta,int _r){
            x = _x;
            y = _y;
            sta = _sta;
            r = _r;
        }
        bool operator <(P p2)const {
            if(vis[x][y][sta] != vis[p2.x][p2.y][p2.sta])
                return vis[x][y][sta] > vis[p2.x][p2.y][p2.sta];
            return r < p2.r;
        }
    };
    int ok(int x,int y,int sta){
        if(x < 1 || y < 1 || x > n || y > m)return -1;
        int ki = maz[x][y];
        if(ki == -1)return -1;
        if(ki > 0){
            if(sta & (1 << (ki-1)))return 1;
            return 0;
        }
        return 1;
    }
    priority_queue <P> que;
    int bfs(){
        memset(vis ,-1, sizeof vis);
        vis[sx][sy][0] = 0;
        que.push(P(sx,sy,0,0));
        while(!que.empty()){
            P tp=que.top();que.pop();
           // printf("pop  x: %d y: %d sta: %d step %d turn %d
    ",tp.x,tp.y,tp.sta,tp.r,vis[tp.x][tp.y][tp.sta]);
            if(tx == tp.x && ty == tp.y)return vis[tx][ty][tp.sta];
            for(int i = 0;i < 4;i++){
                int tmpx = tp.x + dx[i];
                int tmpy = tp.y + dy[i];
                int sta = tp.sta;
                if(ind[tmpx][tmpy] != 0)sta |= (1 << (ind[tmpx][tmpy] - 1));
                if(ok(tmpx,tmpy,tp.sta) >= 0 && vis[tmpx][tmpy][sta] == -1){
                    if(tp.r == 0){
                        vis[tmpx][tmpy][sta] = vis[tp.x][tp.y][tp.sta] + 1;
                       // printf("push x: %d y: %d sta: %d step %d turn %d
    ",
                        //       tmpx,tmpy,sta,(ok(tmpx,tmpy,sta)==0?0:l-1),vis[tp.x][tp.y][tp.sta]+1);
                        que.push(P(tmpx,tmpy,sta,(ok(tmpx,tmpy,tp.sta) == 0?0:l-1)));
                    }
                    else {
                        vis[tmpx][tmpy][sta] = vis[tp.x][tp.y][tp.sta];
                        //printf("push x: %d y: %d sta: %d step %d turn %d
    ",
                          //     tmpx,tmpy,sta,(ok(tmpx,tmpy,sta)==0?0:tp.r-1),vis[tp.x][tp.y][tp.sta]);
                        que.push(P(tmpx,tmpy,sta,(ok(tmpx,tmpy,tp.sta) == 0?0:tp.r-1)));
                    }
                }
            }
        }
        return -1;
    }
    
    int main(){
        while(scanf("%d%d%d",&n,&m,&l) == 3){
            while(!que.empty())que.pop();
            memset(ind,0,sizeof ind);
            for(int i = 1;i <= n;i++){
                for(int j = 1;j <= m;j++){
                    scanf("%d",maz[i]+j);
                }
            }
            scanf("%d",&k);
            for(int i = 1;i <= k;i++){
                scanf("%d%d",mon[i],mon[i] + 1);
                ind[mon[i][0]][mon[i][1]]=i;
            }
            scanf("%d%d%d%d",&sx,&sy,&tx,&ty);
            int ans = bfs();
            if(ans < 0){
                puts("We need God's help!");
            }
            else {
                printf("%d
    ",ans);
            }
        }
        return 0;
    }
    
  • 相关阅读:
    upload1
    web2
    自动生成代码,简化开发
    rabbitmq简易安装
    jenkin安装
    mysql 数据插入为问号 ?
    git 基础复习
    git 使用,强制推远程仓库
    Spring 源码解析(持续集成,哈哈)
    ContainerBase.addChild: start: 错误
  • 原文地址:https://www.cnblogs.com/xuesu/p/4509951.html
Copyright © 2020-2023  润新知