zoukankan      html  css  js  c++  java
  • 【POJ】[3620]Avoid The Lakes

    Avoid The Lakes

    Time Limit: 1000MS Memory Limit: 65536K

    Description

    Farmer John's farm was flooded in the most recent storm, a fact only aggravated by the information that his cows are deathly afraid of water. His insurance agency will only repay him, however, an amount depending on the size of the largest "lake" on his farm. The farm is represented as a rectangular grid with N (1 ≤ N ≤ 100) rows and M (1 ≤ M ≤ 100) columns. Each cell in the grid is either dry or submerged, and exactly K (1 ≤ KN × M) of the cells are submerged. As one would expect, a lake has a central cell to which other cells connect by sharing a long edge (not a corner). Any cell that shares a long edge with the central cell or shares a long edge with any connected cell becomes a connected cell and is part of the lake.

    Input

    * Line 1: Three space-separated integers: N, M, and K * Lines 2..K+1: Line i+1 describes one submerged location with two space separated integers that are its row and column: R and C

    Output

    * Line 1: The number of cells that the largest lake contains.

    Sample Input

    3 4 5
    3 2
    2 2
    3 1
    2 3
    1 1

    Sample Output

    4

    注意横纵坐标的对应关系 注意给的坐标是以1~n而不是0~n-1 每次查找后把当前位置填充防止重复查找 取查找的最大值为结果
    #include<stdio.h>
    #include<string.h>
    int map[120][120];
    int move[6]= {1,-1,0,0};
    int W,H,cnt;
    void dfs(int n,int m) {
        cnt++;
        map[n][m]=0;
        for(int i=0; i<4; i++) {
            int tn=n+move[i],tm=m+move[(i+2)%4];
            if(tn>0&&tn<=H&&tm>0&&tm<=W&&map[tn][tm]==1)
                dfs(tn,tm);
        }
        return ;
    }
    int main() {
        int T;
        while(scanf("%d %d %d",&H,&W,&T)!=EOF) {
            memset(map,0,sizeof(map));
            while(T--) {
                int n,m;
                scanf("%d %d",&n,&m);
                map[n][m]=1;
            }
            int res=0;
            for(int i=1; i<=H; i++)
                for(int j=1; j<=W; j++) {
                    if(map[i][j]==0)
                        continue;
                    cnt=0;
                    dfs(i,j);
                    if(res<cnt)
                        res=cnt;
                }
            printf("%d
    ",res);
        }
        return 0;
    }
    

    题目地址:【POJ】[3620]Avoid The Lakes

    查看原文:http://www.boiltask.com/blog/?p=1919
  • 相关阅读:
    for memory long term update
    Tomcat 部署 Jenkins (Linux系统)
    Linux新增硬盘
    jenkins常见问题
    Linux安装JDk
    Jenkins入门(一)
    主播必备工具Bongo Cat!
    Docker常用命令
    Docker二进制部署
    golang中for语句详解
  • 原文地址:https://www.cnblogs.com/BoilTask/p/12569457.html
Copyright © 2011-2022 走看看