zoukankan      html  css  js  c++  java
  • POJ2226 Muddy Fields

    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 10149   Accepted: 3783

    Description

    Rain has pummeled the cows' field, a rectangular grid of R rows and C columns (1 <= R <= 50, 1 <= C <= 50). While good for the grass, the rain makes some patches of bare earth quite muddy. The cows, being meticulous grazers, don't want to get their hooves dirty while they eat. 

    To prevent those muddy hooves, Farmer John will place a number of wooden boards over the muddy parts of the cows' field. Each of the boards is 1 unit wide, and can be any length long. Each board must be aligned parallel to one of the sides of the field. 

    Farmer John wishes to minimize the number of boards needed to cover the muddy spots, some of which might require more than one board to cover. The boards may not cover any grass and deprive the cows of grazing area but they can overlap each other. 

    Compute the minimum number of boards FJ requires to cover all the mud in the field.

    Input

    * Line 1: Two space-separated integers: R and C 

    * Lines 2..R+1: Each line contains a string of C characters, with '*' representing a muddy patch, and '.' representing a grassy patch. No spaces are present.

    Output

    * Line 1: A single integer representing the number of boards FJ needs.

    Sample Input

    4 4
    *.*.
    .***
    ***.
    ..*.
    

    Sample Output

    4
    

    Hint

    OUTPUT DETAILS: 

    Boards 1, 2, 3 and 4 are placed as follows: 
    1.2. 
    .333 
    444. 
    ..2. 
    Board 2 overlaps boards 3 and 4.

    Source

    将连续的一段横/纵条算作同一个,若横纵条相交则连边,跑二分图匹配。

     1 /*by SilverN*/
     2 #include<algorithm>
     3 #include<iostream>
     4 #include<cstring>
     5 #include<cstdio>
     6 #include<cmath>
     7 #include<vector>
     8 using namespace std;
     9 const int mxn=1010;
    10 int m,n;
    11 char a[mxn][mxn];
    12 //
    13 int idx,idy;
    14 int hs[mxn][mxn];
    15 int link[mxn];
    16 int mp[mxn][mxn];
    17 bool vis[mxn];
    18 //
    19 void init(){
    20     int i,j;
    21     for(i=1;i<=n;i++){
    22         for(j=1;j<=m;j++){
    23             if(a[i][j]=='*'){
    24                 ++idx;
    25                 while(a[i][j]=='*')hs[i][j++]=idx;
    26             }
    27         }
    28     }
    29     idy=idx;
    30     for(j=1;j<=m;j++)
    31         for(i=1;i<=n;i++){
    32             if(a[i][j]=='*'){
    33                 ++idy;
    34                 while(a[i][j]=='*'){
    35                     mp[idy][hs[i][j]]=1;
    36                     mp[hs[i++][j]][idy]=1;
    37                 }
    38             }
    39         }
    40     return;
    41 }
    42 bool DFS(int u){
    43     for(int i=1;i<=idy;i++)
    44         if(!vis[i] && mp[u][i]){
    45             vis[i]=1;
    46             if(link[i]==-1 || DFS(link[i])){
    47                 link[i]=u;
    48                 return 1;
    49             }
    50         }
    51     return 0;
    52 }
    53 int ans=0;
    54 void solve(){
    55     memset(link,-1,sizeof link);
    56     for(int i=1;i<=idx;i++){
    57         memset(vis,0,sizeof vis);
    58         if(DFS(i))ans++;
    59     }
    60     return;
    61 }
    62 int main(){
    63     scanf("%d%d",&n,&m);
    64     int i,j;
    65     for(i=1;i<=n;i++)
    66         scanf("%s",a[i]+1);
    67     init();
    68     solve();
    69     printf("%d
    ",ans);
    70     return 0;
    71 }
  • 相关阅读:
    线段树专辑——hdu 1698 Just a Hook
    bool树的计算
    线段树专辑—— hdu 2795 Billboard
    数据库索引技术聚集索引和非聚集索引介绍(二)
    数据库索引技术—索引介绍 (一)
    AJAX核心对象 XMLHttpRequest 对象使用详解 (二)
    AJAX核心对象 XMLHttpRequest 对象使用详解 (一)
    输入框的提示信息(获得焦点和失去焦点时显示提示信息)
    不管你今天做了多少,加一点亮点总结今天的得分就是1.01。
    新情况——需要按照这个规定堆
  • 原文地址:https://www.cnblogs.com/SilverNebula/p/6047793.html
Copyright © 2011-2022 走看看