标签：include clu fine can splay 匈牙利算法 square src 建图
InputThe input starts with an integer K (1 <= K <= 100) indicating the number of cases. Each case starts with an integer N (1 <= N <= 600) indicating the size of the square grid. Each of the following N lines contains N characters that represent the cells of a row in the grid. A character of ‘#‘ represents an oily cell, and a character of ‘.‘ represents a pure water cell.OutputFor each case, one line should be produced, formatted exactly as follows: "Case X: M" where X is the case number (starting from 1) and M is the maximum number of scoops of oil that may be extracted.Sample Input
1 6 ...... .##... .##... ....#. ....## ......
Sample Output
Case 1: 3
题意：给一张图，问最大的连续两个格子数目有多少
题解：建图太麻烦，实在没想到，看网上思路才a的，把整个图分成i+j%2是否=1两个；然后匹配匈牙利算法就行了
坑点：刚开始以为由于是从上到下，从左到右遍历的，不用算i-1，j-1的情况，才发现这样会少算几种情况。
#include<map> #include<set> #include<cmath> #include<queue> #include<stack> #include<vector> #include<cstdio> #include<cstdlib> #include<cstring> #include<iostream> #include<algorithm> #define pi acos(-1) #define ll long long #define mod 1000000007 using namespace std; const int N=600+5,maxn=100000+5,inf=0x3f3f3f3f; int n,color[N],col,row; int u[N][N]; char ma[N][N]; bool used[N],ok[N][N]; bool match(int x) { for(int i=1;i<=row;i++) { if(!used[i]&&ok[x][i]) { used[i]=1; if(color[i]==0||match(color[i])) { color[i]=x; return 1; } } } return 0; } int solve() { int ans=0; memset(color,0,sizeof color); for(int i=1;i<=col;i++) { memset(used,0,sizeof used); ans+=match(i); } return ans; } int main() { ios::sync_with_stdio(false); cin.tie(0); int t; cin>>t; for(int k=1;k<=t;k++) { cin>>n; for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) cin>>ma[i][j]; col=row=0; memset(u,0,sizeof u); memset(ok,0,sizeof ok); for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) { if(ma[i][j]==‘#‘) { if((i+j)%2==0)u[i][j]=++col; if((i+j)%2==1)u[i][j]=++row; } } for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) { if((i+j)%2==0&&ma[i][j]==‘#‘) { if(i+1<=n&&ma[i+1][j]==‘#‘)ok[u[i][j]][u[i+1][j]]=1; if(j+1<=n&&ma[i][j+1]==‘#‘)ok[u[i][j]][u[i][j+1]]=1; if(i-1>=1&&ma[i-1][j]==‘#‘)ok[u[i][j]][u[i-1][j]]=1; if(j-1>=1&&ma[i][j-1]==‘#‘)ok[u[i][j]][u[i][j-1]]=1; } if((i+j)%2==1&&ma[i][j]==‘#‘) { if(i+1<=n&&ma[i+1][j]==‘#‘)ok[u[i+1][j]][u[i][j]]=1; if(j+1<=n&&ma[i][j+1]==‘#‘)ok[u[i][j+1]][u[i][j]]=1; if(i-1>=1&&ma[i-1][j]==‘#‘)ok[u[i-1][j]][u[i][j]]=1; if(j-1>=1&&ma[i][j-1]==‘#‘)ok[u[i][j-1]][u[i][j]]=1; } } /* for(int i=1;i<=n;i++) { for(int j=1;j<=n;j++) cout<<u[i][j]; cout<<endl; } for(int i=1;i<=col;i++) { for(int j=1;j<=row;j++) cout<<ok[i][j]; cout<<endl; }*/ cout<<"Case "<<k<<": "<<solve()<<endl; } return 0; }
标签：include clu fine can splay 匈牙利算法 square src 建图
原文地址：http://www.cnblogs.com/acjiumeng/p/6741545.html
