hdu 3435 A new Graph Game (有向环覆盖，拆点，二分图最优匹配，KM算法)

2016-06-02 · 2 min read · KM 二分图最佳匹配 拆点

hdu 3435题目链接

题意：给你一张图，图上可能有多个哈密顿回路。叫你求出形成多个哈密顿回路的总距离最小值

思路：题意杀啊。。。什么鬼了。。。然后时间。。1000的数据。。n3复杂度。。。还多组数据。。。。不是很懂这个时间是怎么算的。。为毛才2600MS啊。。。。

选区_073

做法同hdu 1853...

/* ***********************************************
Author :111qqz
Created Time :2016年06月02日 星期四 20时13分25秒
File Name :code/hdu/1853.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=1E3+4;
 7int n,m;
 8bool visx[N],visy[N];
 9int link[N];
10int lx[N],ly[N];
11int slk[N];
12int w[N][N];

 1bool find( int u)
 2{
 3    visx[u] = true;
 4    for ( int v = 1 ; v <= n ; v++)
 5    {
 6	if (visy[v]) continue;
 7	int tmp = lx[u] + ly[v] - w[u][v];
 8	if (tmp==0)
 9	{
10	    visy[v] = true;

 1	    if (link[v]==-1||find(link[v]))
 2	    {
 3		link[v] =  u;
 4		return true;
 5	    }
 6	}else if (tmp<slk[v]) slk[v] = tmp;
 7    }
 8    return false;
 9}
10int KM()
11{
12    ms(lx,0xc0);
13    ms(ly,0);
14    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]);
4    for ( int i = 1 ; i <= n ; i++) if (lx[i]==-inf-1) 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++)
2		if (visx[j]) lx[j]-=d;
3	    for ( int j = 1 ; j <= n ; j++)
4		if (visy[j]) ly[j]+=d ; else slk[j]-=d;
5	}
6    }
7    int res = 0 ;

    for ( int i  = 1 ; i <= n ; i++)
	if (link[i]>-1) res += w[link[i]][i];

    return -res;

 1}
 2int main()
 3{
 4	#ifndef  ONLINE_JUDGE 
 5	freopen("code/in.txt","r",stdin);
 6  #endif
 7	int T;
 8	int cas = 0 ;
 9	cin>>T;
10	while (T--)
11	{
12	    scanf("%d %d",&n,&m);
13	    ms(w,0xc0);
14	    for ( int i = 1 ; i <= m ; i++)
15	    {
16		int u,v,cost;
17		scanf("%d%d%d",&u,&v,&cost);
18		w[u][v] =max(w[u][v],-cost); //会不会有重边。..果然有。
19		w[v][u] =max(w[v][u],-cost);
20	    }

1	    int ans = KM();
2	    printf("Case %d: ",++cas);
3	    if (ans==-1) puts("NO");
4	    else printf("%d\n",ans);

	}

1  #ifndef ONLINE_JUDGE  
2  fclose(stdin);
3  #endif
4    return 0;
5}