hdu 3722 Card Game (有向环覆盖，拆点，二分图最佳匹配，KM算法)

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

hdu 3722题目链接

题意：n个串，a串放在b串前面的val值是“The score of sticking two cards is the longest common prefix of the second card and the reverse of the first card”.问如何放使得总的val最大。

思路：先暴力处理出每两个的权值。。2002001000的复杂度。。还是可以接受的。。

然后把每个串看成了一个点，由于一个串最多可以被放在前面一次，被放在后面一次，所以可以类比图论中的环的入度和出度为1.

然后跑一遍KM. 1A,开心。

/* ***********************************************
Author :111qqz
Created Time :2016年06月02日 星期四 23时37分55秒
File Name :code/hdu/3722.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=205;
 7int n;
 8char st[N][1005];
 9int w[N][N];
10int link[N];
11int lx[N],ly[N];
12bool visx[N],visy[N];
13int slk[N];
14int solve(string a,string b)
15{
16    int la = a.length();
17    int lb = b.length();
18    int len = min(la,lb);
19    int res = 0 ;
20    for ( int i = 0 ; i < len ; i++)
21    {
22	if (b[i]==a[la-1-i]) res++;
23	else break;
24    }
25    return res;
26}

1bool find( int u)
2{
3   // cout<<"u:"<<u<<endl;
4    visx[u] = true;
5    for ( int v  = 1 ; v <= n ; v++)
6    {
7	if (visy[v]) continue;

 1	int tmp = lx[u] + ly[v] - w[u][v];
 2	if (tmp==0)
 3	{
 4	    visy[v] = true;
 5	    if (link[v]==-1||find(link[v]))
 6	    {
 7		link[v] = u ;
 8		return true;
 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]);

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];

	    for ( int j = 1 ; j <= n ; j++)
		if (visx[j]) lx[j]-=d;

1	    for ( int j = 1 ; j <= n ; j++)
2		if (visy[j]) ly[j]+=d ; else slk[j]-=d;
3	}
4    }
5    int res = 0 ;
6    for ( int i = 1 ; i <= n ; i++)
7	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))
2	{
3	    for ( int i = 1 ; i <= n ; i++) scanf("%s",st[i]);

	    ms(w,0);

1	    for ( int i = 1 ; i <= n ; i++)
2		for ( int j = 1 ; j <= n ; j++)
3		    if (i!=j) w[i][j] = solve(string(st[i]),string(st[j]));

	 //   for ( int i = 1 ; i <= n ; i++)
	//	for ( int j = 1 ; j <= n ; j++) if (i!=j) cout<<"w[i][j]:"<<w[i][j]<<endl;

1	    int ans = KM();
2	    printf("%d\n",ans);
3	}

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