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HDU 3722 Card Game (KM)

题目链接：点我~~

题意：给n个字符串，然后对于任意两个字符串进行匹配，第一个翻转后与第二个匹配，最长公共前缀为得分，自己到自己为0。

思路：将字符串预处理建图，然后跑KM。

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PI;
typedef pair< PI, int> PII;
const double eps=1e-5;
const double pi=acos(-1.0);
const int mod=1e9+7;
const int INF=0x3f3f3f3f;
#define Key_value ch[ch[root][1]][0]
const int MAXN = 800000;
const int MAXM = 1250;
int nx,ny;
int g[MAXM][MAXM];
int linker[MAXM],lx[MAXM],ly[MAXM];
int slack[MAXM];
bool visx[MAXM],visy[MAXM];
bool DFS(int x)
{
    visx[x] = true;
    for(int y = 0; y < ny; y++)
    {
        if(visy[y])continue;
        int tmp = lx[x] + ly[y] - g[x][y];
        if(tmp == 0)
        {
            visy[y] = true;
            if(linker[y] == -1 || DFS(linker[y]))
            {
                linker[y] = x;
                return true;
            }
        }
        else if(slack[y] > tmp)
            slack[y] = tmp;
    }
    return false;
}
int KM()
{
    memset(linker,-1,sizeof(linker));
    memset(ly,0,sizeof(ly));
    for(int i = 0; i < nx; i++)
    {
        lx[i] = -INF;
        for(int j = 0; j < ny; j++)
            if(g[i][j] > lx[i])
                lx[i] = g[i][j];
    }
    for(int x = 0; x < nx; x++)
    {
        for(int i = 0; i < ny; i++)
            slack[i] = INF;
        while(true)
        {
            memset(visx,false,sizeof(visx));
            memset(visy,false,sizeof(visy));
            if(DFS(x))break;
            int d = INF;
            for(int i = 0; i < ny; i++)
                if(!visy[i] && d > slack[i])
                    d = slack[i];
            for(int i = 0; i < nx; i++)
                if(visx[i])
                    lx[i] -= d;
            for(int i = 0; i < ny; i++)
            {
                if(visy[i])ly[i] += d;
                else slack[i] -= d;
            }
        }
    }
    int res = 0;
    for(int i = 0; i < ny; i++)
        if(linker[i] != -1)
            res += g[linker[i]][i];
    return res;
}
//HDU 2255
int main()
{
    int n;
    string str[205];
    while(~scanf("%d",&n))
    {
        for(int i = 0; i < n; i++)
            cin>>str[i];
        for(int i=0; i<n; ++i)
            for(int j=0; j<n; ++j)
                g[i][j]=0;
        for(int i=0; i<n; ++i)
        {
            int len=str[i].size();
            string tmp=str[i];
            reverse(tmp.begin(),tmp.end());
            for(int j=0; j<n; ++j)
            {
                if(i==j)
                continue;
                int t=min(len,(int)str[j].size());
                int cnt=0;
                for(int k=0; k<t; ++k)
                {
                    if(tmp[k]==str[j][k])
                    {
                        cnt++;
                    }
                    else
                        break;
                }
                g[i][j]=cnt;
            }
        }
        nx = ny = n;
        printf("%d\n",KM());
    }
    return 0;
}