UVALive-5013 Similarity(二分图最大权匹配)

题意：

给出一正确字符序列，然后再给出相同长度的字符序列，第二个序列需要确定给定字符与正确序列相对应。求最大相似比率(正确的个数/总个数)。

思路：

很容易直接想到二分图的最大权匹配，将两个相对应的字符建边，并以其重复个数为权值，然后跑一遍KM就可以了。

代码：

#include <algorithm>
#include <iostream>
#include <string.h>
#include <string>
#include <cstdio>
#include <map>
using namespace std;
const int maxn = 10005;
const int inf = 0x3f3f3f3f;
char s[maxn*2];
map<int, int> mp;
int T, N, K, M;
int num1[maxn], num2[maxn];
int cnt;
int weight[30][30];
int match[30];
int lx[30], ly[30];
int visx[30], visy[30];
int slack[30];
int dfs(int cur)
{
	visx[cur] = 1;
	for(int i = 1; i <= K; ++i)
	{
		if(visy[i]) continue;
		int gap = lx[cur]+ly[i]-weight[cur][i];
		if(gap == 0)
		{
			visy[i] = 1;
			if(match[i] == -1 || dfs(match[i]))
			{
				match[i] = cur;
				return 1;
			}
		}
		else slack[i] = min(slack[i], gap);
	}
	return 0;
}
int KM()
{
	memset(match, -1, sizeof match);
	memset(ly, 0, sizeof ly);
	for(int i = 1; i <= cnt; ++i)
	{
		lx[i] = weight[i][1];
		for(int j = 2; j <= K; ++j)
		lx[i] = max(lx[i], weight[i][j]);
	}
	for(int i = 1; i <= cnt; ++i)
	{
		memset(slack, 0x3f, sizeof slack);
		while(1)
		{
			memset(visx, 0, sizeof visx);
			memset(visy, 0, sizeof visy);
			if(dfs(i)) break;
			int mins = inf;
			for(int j = 1; j <= K; ++j)
			if(!visy[j]) mins = min(mins, slack[j]);
			for(int j = 1; j <= cnt; ++j)
			if(visx[j]) lx[j] -= mins;
			for(int j = 1; j <= K; ++j)
			if(visy[j]) ly[j] += mins;
		}
	}
	int ans = 0;
	for(int i = 1; i <= K; ++i)
	if(match[i] != -1) ans += weight[match[i]][i];
	return ans;
}
int main()
{
	scanf("%d", &T);
	while(T--)
	{
		scanf("%d %d %d", &N, &K, &M);
		getchar();
		gets(s); cnt = 0;
		mp.clear();
		for(int i = 0; i < N; ++i)
		if(mp.count(s[i*2])) num1[i+1] = mp[s[i*2]];
		else
		{
			mp[s[i*2]] = ++cnt;
			num1[i+1] = cnt;
		}
		for(int _ = 1; _ <= M; ++_)
		{
			memset(weight, 0, sizeof weight);
			cnt = 0; mp.clear();
			gets(s);
			for(int i = 0; i < N; ++i)
			if(mp.count(s[i*2])) num2[i+1] = mp[s[i*2]];
			else
			{
				mp[s[i*2]] = ++cnt;
				num2[i+1] = cnt;
			}
			for(int i = 1; i <= N; ++i)	//令cnt为A部个数，防止死循环
			++weight[num2[i]][num1[i]];
			printf("%.4f\n", KM()*1.0/N);
		}
	}
	return 0;
}

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