HDU-3718 Similarity

题目只有26个字母，所以我们新建一个二分图，v[i][j]表示字母i对应字母j时能成功匹配的个数，给这个边矩阵v求个最大匹配就是答案。

#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <fstream>
#include <iostream>
#include <vector>
#include <cctype>
#define rep(i, l, r) for(int i=l; i<=r; i++)
#define clr(x, c) memset(x, c, sizeof(x))
#define N 12345
#define MAX 1<<30
#define ll long long
using namespace std;
int read()
{
	int x=0, f=1; char ch=getchar();
	while (!isdigit(ch)) { if (ch=='-') f=-1; ch=getchar(); }
	while (isdigit(ch)) { x=x*10+ch-'0'; ch=getchar(); }
	return x*f;
}
int readch()
{
	char ch=getchar();
	while (!isupper(ch)) ch=getchar();
	return ch-'A'+1;
}

int n, m, l[27], st[27], lx[27], ly[27], v[27][27], k[N];
bool vx[27], vy[27];

bool Find(int x)
{
	vx[x]=1;
	rep(y, 1, 26)
	{
		if (vy[y]) continue;
		int a=lx[x]+ly[y]-v[x][y];
		if (!a)
		{
			vy[y]=1; if (!l[y] || Find(l[y])) { l[y]=x; return 1; }
		}
		else st[y]=min(st[y], a);
	}
	return false;
}

inline int km()
{
	clr(ly, 0); clr(l, 0); rep(i, 1, 26) lx[i]=-MAX;
	rep(i, 1, 26) rep(j, 1, 26) if (lx[i]<v[i][j]) lx[i]=v[i][j];
	rep(i, 1, 26)
	{
		rep(j, 1, 26) st[j]=MAX;
		while (1)
		{
			clr(vx, 0); clr(vy, 0);
			if (Find(i)) break; int a=MAX;
			rep(j, 1, 26) if (!vy[j] && st[j]<a) a=st[j];
			rep(j, 1, 26) if (vx[j]) lx[j]-=a;
			rep(j, 1, 26) if (vy[j]) ly[j]+=a; else st[j]-=a;
		}
	}
	int a=0; rep(i, 1, 26) a+=lx[i]+ly[i];
	return a;
}

int main()
{
	int t=read(); while (t--)
	{
		n=read(); m=read(); m=read();
		rep(i, 1, n) k[i]=readch();
		rep(o, 1, m)
		{
			rep(i, 1, 26) rep(j, 1, 26) v[i][j]=0;
			rep(i, 1, n) v[k[i]][readch()]++;
			printf("%.4lf\n", double(km())/n);
		}
	}
	return 0;
}