hdu 3722 Card Game （KM ）

Card Game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1165    Accepted Submission(s): 472

Problem Description

Jimmy invents an interesting card game. There are N cards, each of which contains a string Si. Jimmy wants to stick them into several circles, and each card belongs to one circle exactly. When sticking two cards, Jimmy will get a score. The score of sticking two cards is the longest common prefix of the second card and the reverse of the first card. For example, if Jimmy sticks the card S1 containing "abcd" in front of the card S2 containing "dcab", the score is 2. And if Jimmy sticks S2 in front of S1, the score is 0. The card can also stick to itself to form a self-circle, whose score is 0.

For example, there are 3 cards, whose strings are S1="ab", S2="bcc", S3="ccb". There are 6 possible sticking:
1.  S1->S2, S2->S3, S3->S1, the score is 1+3+0 = 4
2.  S1->S2, S2->S1, S3->S3, the score is 1+0+0 = 1
3.  S1->S3, S3->S1, S2->S2, the score is 0+0+0 = 0
4.  S1->S3, S3->S2, S2->S1, the score is 0+3+0 = 3
5.  S1->S1, S2->S2, S3->S3, the score is 0+0+0 = 0
6.  S1->S1, S2->S3, S3->S2, the score is 0+3+3 = 6
So the best score is 6.

Given the information of all the cards, please help Jimmy find the best possible score.

 

Input

There are several test cases. The first line of each test case contains an integer N (1 <= N <= 200). Each of the next N lines contains a string Si. You can assume the strings contain alphabets ('a'-'z', 'A'-'Z') only, and the length of every string is no more than 1000.

 

Output

Output one line for each test case, indicating the corresponding answer.

 

Sample Input

   
   
   
   

    
    
    
    3
ab
bcc
ccb
1
abcd
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    6
0
   
   
   
   

 

Source

2010 Asia Tianjin Regional Contest

 

思路：

裸的KM，建图后直接用模板就够了，注意是“longest common prefix”，题目不要读错了。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
//#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 205
#define MAXN 1000005
#define mod 1000000007
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
typedef long long ll;
using namespace std;

int n,nx,ny,ans;
int link[maxn],lx[maxn],ly[maxn],slack[maxn]; //lx,ly为顶标，nx,ny分别为x点集y点集的个数
int visx[maxn],visy[maxn],w[maxn][maxn];
int len[maxn];
char s[maxn][1005];

int DFS(int x)
{
    visx[x] = 1;
    for (int y = 1;y <= ny;y ++)
    {
        if (visy[y])
            continue;
        int t = lx[x] + ly[y] - w[x][y];
        if (t == 0)
        {
            visy[y] = 1;
            if (link[y] == -1||DFS(link[y]))
            {
                link[y] = x;
                return 1;
            }
        }
        else if (slack[y] > t)  //不在相等子图中slack 取最小的
            slack[y] = t;
    }
    return 0;
}
int KM()
{
    int i,j;
    memset (link,-1,sizeof(link));
    memset (ly,0,sizeof(ly));
    for (i = 1;i <= nx;i ++)            //lx初始化为与它关联边中最大的
        for (j = 1,lx[i] = -INF;j <= ny;j ++)
            if (w[i][j] > lx[i])
                lx[i] = w[i][j];

    for (int x = 1;x <= nx;x ++)
    {
        for (i = 1;i <= ny;i ++)
            slack[i] = INF;
        while (1)
        {
            memset (visx,0,sizeof(visx));
            memset (visy,0,sizeof(visy));
            if (DFS(x))  break;
            int d = INF;
            for (i = 1;i <= ny;i ++)
                if (!visy[i]&&d > slack[i])
                    d = slack[i];
            for (i = 1;i <= nx;i ++)
                if (visx[i])
                    lx[i] -= d;
            for (i = 1;i <= ny;i ++)  //修改顶标后，要把所有不在交错树中的Y顶点的slack值都减去d
                if (visy[i])
                    ly[i] += d;
                else
                    slack[i] -= d;
        }
    }
    int res = 0;
    for (i = 1;i <= ny;i ++)
        if (link[i] > -1)
            res += w[link[i]][i];
    return res;
}
void presolve()
{
    int i,j,k,p1,p2,cnt;
    for(i=1;i<=n;i++)
    {
        for(j=1;j<=n;j++)
        {
            if(i==j)
            {
                w[i][j]=0;
                continue ;
            }
            cnt=0;
            p1=len[i]-1;
            p2=0;
            while(1)
            {
                if(s[i][p1]==s[j][p2]) cnt++;
                else break ;
                p1--,p2++;
                if(p1<0||p2>=len[j]) break ;
            }
            w[i][j]=cnt;
        }
    }
}
int main()
{
    int i,j;
    while(~scanf("%d",&n))
    {
        nx=ny=n;
        for(i=1;i<=n;i++)
        {
            scanf("%s",s[i]);
            len[i]=strlen(s[i]);
        }
        presolve();
        ans=KM();
        printf("%d\n",ans);
    }
}