hdu 3395 Special Fish (二分图最佳匹配,KM算法)
题意:鱼,一些鱼认为自己是汉子,然后他会去和他认为是妹子的鱼啪啪啪,然后被啪啪啪的妹子就会产卵? 卵的val是它parent的val的异或。给出n,为鱼的数量,然后给出一个n*n的 mat,a[i][j]==1表示第i条鱼认为第j条鱼是妹子。问卵的最大val之和是多少。需要注意的是:每条鱼最多可以去和一个妹子啪,并且可以作为妹子被啪一次(这两个是独立的。。。) (Each fish can attack one other fish and can only be attacked once)
思路:二分图最佳匹配,KM算法,2A.
/* ***********************************************
Author :111qqz
Created Time :2016年06月01日 星期三 20时38分58秒
File Name :code/hdu/3395.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=105;
7int n;
8int w[N][N];
9char a[N][N];
10int val[N];
11bool visx[N],visy[N];
12int link[N];
13int lx[N],ly[N];
14int slk[N];
1bool find( int u)
2{
3 visx[u] = true;
1 for ( int v = 1 ; v <= n ; v++)
2 {
3 if (visy[v]) continue;
int tmp = lx[u] + ly[v] - w[u][v];
1 if (tmp==0)
2 {
3 visy[v] = true;
4 if (link[v]==-1||find(link[v]))
5 {
6 link[v] = u;
7 return true;
8 }
9 }
10 else if (tmp<slk[v]) slk[v] = tmp;
11 }
12 return false;
13}
14int KM()
15{
16 ms(lx,0);
17 ms(ly,0);
18 ms(link,-1);
1 for ( int i = 1 ; i <= n ; i++)
2 for ( int j = 1 ; j <= n ; j++)
3 lx[i] = max(lx[i],w[i][j]);
// for ( int i = 1 ; i <= n ; i++) if (lx[i]==0) return -1;
1 for ( int i = 1 ; i <= n ; i++)
2 {
3 ms(slk,0x3f);
1 while (1)
2 {
3 ms(visx,false);
4 ms(visy,false);
if (find(i)) break;
int d = inf;
for ( int j = 1 ; j <= n ; j++) if (!visy[j]&&slk[j]<d) d = slk[j];
1 for ( int j = 1 ; j <= n ; j++) if (visx[j]) lx[j]-=d;
2 for ( int j = 1 ; j <= n ; j++) if (visy[j]) ly[j]+=d; else slk[j]-=d;
3 }
}
1 int res = 0 ;
2 for ( int i = 1 ; i <= n ; i++)
3 if (link[i]>-1) res += w[link[i]][i];
1 return res;
2}
3int main()
4{
5 #ifndef ONLINE_JUDGE
6 freopen("code/in.txt","r",stdin);
7 #endif
1 while (scanf("%d",&n)!=EOF)
2 {
3 if (n==0) break;
4 ms(w,0);
5 for ( int i = 1 ; i <= n ; i++) scanf("%d",&val[i]);
for ( int i = 1 ; i <= n ; i++)
for ( int j = 1 ; j <= n ;j++) w[i][j] = val[i] ^ val[j];
1 for ( int i = 1 ; i <= n ; i++) scanf("%s",a[i]+1);
2 for ( int i = 1 ; i <= n ; i++)
3 for ( int j = 1; j <= n ; j++)
4 {
5 if (a[i][j]=='0') w[i][j]=0;
6 }
int ans = KM();
printf("%d\n",ans);
}
1 #ifndef ONLINE_JUDGE
2 fclose(stdin);
3 #endif
4 return 0;
5}