hdu 4185 Oil Skimming (二分图最大匹配,匈牙利算法)
hdu 4185题目链接 题意:给出一个nn的字符maze,‘.’代表水,‘#’代表油田。 挖油的机器一次会挖两个相邻方块。要求是必须两块必须都是油,不然会有杂质。问最多能挖多少次。 思路:和那道用12的小矩形块填充是一个思路。根据奇偶性对点标号,然后建图,匈牙利,2A. 第一遍是dfs写错了一个变量QAQ.a
/* ***********************************************
Author :111qqz
Created Time :2016年05月30日 星期一 12时53分34秒
File Name :code/hdu/4185.cpp
************************************************ */
1#include <cstdio>
2#include <cstring>
3#include <iostream>
4#include <algorithm>
5#include <vector>
6#include <queue>
7#include <set>
8#include <map>
9#include <string>
10#include <cmath>
11#include <cstdlib>
12#include <ctime>
13#define fst first
14#define sec second
15#define lson l,m,rt<<1
16#define rson m+1,r,rt<<1|1
17#define ms(a,x) memset(a,x,sizeof(a))
18typedef long long LL;
19#define pi pair < int ,int >
20#define MP make_pair
1using namespace std;
2const double eps = 1E-8;
3const int dx4[4]={1,0,0,-1};
4const int dy4[4]={0,-1,1,0};
5const int inf = 0x3f3f3f3f;
6const int N=1000;
7int n;
8char maze[N][N];
9int f1[N][N],f2[N][N];
10int tot1;
11int tot2;
12int cnt;
13int link[N*N];
14bool vis[N*N];
15int head[N*N];
16struct Edge
17{
18 int v;
19 int nxt;
20}edge[N*N*2];
1void addedge( int u,int v)
2{
3 edge[cnt].v = v;
4 edge[cnt].nxt = head[u];
5 head[u] = cnt;
6 cnt++;
7}
1bool find( int u)
2{
3 for ( int i = head[u] ; i !=-1 ; i = edge[i].nxt)
4 {
5 int v = edge[i].v;
6 //cout<<"u:"<<u<<" v:"<<v<<endl;
7 if (vis[v]) continue;
8 vis[v] = true;
9 if (link[v]==-1||find(link[v]))
10 {
11 link[v] = u;
12 return true;
13 }
14 }
1 return false;
2}
3int hungary()
4{
5 ms(link,-1);
6 int res = 0;
7 for ( int i = 1 ; i <= tot1 ; i++)
8 {
9 ms(vis,false);
10 if (find(i)) res++;
11 }
12 return res;
13}
14int main()
15{
16 #ifndef ONLINE_JUDGE
17 freopen("code/in.txt","r",stdin);
18 #endif
1 int T;
2 int cas = 0 ;
3 scanf("%d",&T);
4 while (T--)
5 {
6 ms(f1,0);
7 ms(f2,0);
8 tot1 = 0 ;
9 tot2 = 0 ;
10 cnt = 0 ;
11 ms(head,-1);
1 scanf("%d",&n);
2 for ( int i = 0 ; i < n ; i++)
3 scanf("%s",maze[i]);
1 for ( int i = 0 ; i < n ; i++)
2 for ( int j = 0 ; j < n ; j++)
3 {
4 if (maze[i][j]=='.') continue;
5 if ((i+j)%2==0)
6 {
7 f1[i][j] = ++tot1;
8 }
9 else
10 {
11 f2[i][j] = ++tot2;
12 }
}
// cout<<"tot1:"<<tot1<<endl;
// cout<<"tot2:"<<tot2<<endl;
1 for ( int i = 0 ; i < n ; i++)
2 for ( int j = 0 ; j < n; j++)
3 {
4 if (j+1<n&&maze[i][j]=='#'&&maze[i][j+1]=='#')
5 {
6 int u = (i+j)%2==0?f1[i][j]:f2[i][j];
7 int v = (i+j+1)%2==0?f1[i][j+1]:f2[i][j+1];
8 if ((i+j)%2==1) swap(u,v);
9 v+=tot1;
10 addedge(u,v);
11 }
12 if (i+1<n&&maze[i][j]=='#'&&maze[i+1][j]=='#')
13 {
14 int u = (i+j)%2==0?f1[i][j]:f2[i][j];
15 int v = (i+1+j)%2==0?f1[i+1][j]:f2[i+1][j];
16 if ((i+j)%2==1) swap(u,v);
17 v+=tot1;
18 addedge(u,v);
19 }
20 }
int ans = hungary();
printf("Case %d: %d\n",++cas,ans);
}
1 #ifndef ONLINE_JUDGE
2 fclose(stdin);
3 #endif
4 return 0;
5}