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

Posted by 111qqz on Thursday, June 2, 2016

TOC

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
************************************************ */

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#define fst first
#define sec second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ms(a,x) memset(a,x,sizeof(a))
typedef long long LL;
#define pi pair < int ,int >
#define MP make_pair

using namespace std;
const double eps = 1E-8;
const int dx4[4]={1,0,0,-1};
const int dy4[4]={0,-1,1,0};
const int inf = 0x3f3f3f3f;
const int N=205;
int n;
char st[N][1005];
int w[N][N];
int link[N];
int lx[N],ly[N];
bool visx[N],visy[N];
int slk[N];
int solve(string a,string b)
{
    int la = a.length();
    int lb = b.length();
    int len = min(la,lb);
    int res = 0 ;
    for ( int i = 0 ; i < len ; i++)
    {
    if (b[i]==a[la-1-i]) res++;
    else break;
    }
    return res;
}

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

    int tmp = lx[u] + ly[v] - w[u][v];
    if (tmp==0)
    {
        visy[v] = true;
        if (link[v]==-1||find(link[v]))
        {
        link[v] = u ;
        return true;
        }	
    }else if (tmp<slk[v]) slk[v] = tmp;
    }
    return false;
}
int KM()
{
    ms(lx,0);
    ms(ly,0);
    ms(link,-1);

    for ( int i = 1 ; i <= n ; i++)
    for  ( int j = 1 ; j <= n ; j++ )
        lx[i] = max(lx[i],w[i][j]);

    for ( int i = 1; i <= n ; i++)
    {
    ms(slk,0x3f);

    while (1)
    {
        ms(visx,false);
        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;

        for ( int j = 1 ; j <= n ; j++)
        if (visy[j]) ly[j]+=d ; else slk[j]-=d;
    }
    }
    int res = 0 ;
    for ( int i = 1 ; i <= n ; i++)
    if (link[i]>-1) res += w[link[i]][i];

    return res;
}
int main()
{
    #ifndef  ONLINE_JUDGE 
    freopen("code/in.txt","r",stdin);
  #endif

    while (~scanf("%d",&n))
    {
        for ( int i = 1 ; i <= n ; i++) scanf("%s",st[i]);
	    
        ms(w,0);

        for ( int i = 1 ; i <= n ; i++)
        for ( int j = 1 ; j <= n ; j++)
            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;
		
        int ans = KM();
        printf("%d\n",ans);
    }

  #ifndef ONLINE_JUDGE  
  fclose(stdin);
  #endif
    return 0;
}

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