【KM匹配】 HDOJ 2853 Assignment

题意：求最大权匹配，要求改动最少。。。。做法很巧妙，现在原权值乘上一个很大的倍数，比如100，然后再加上一个小量，加上小量以后对匹配出来的结果没有影响，但可以求解出改动最小。。。

#include <iostream>
#include <queue> 
#include <stack> 
#include <map> 
#include <set> 
#include <bitset> 
#include <cstdio> 
#include <algorithm> 
#include <cstring> 
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 55
#define maxm 400005
#define eps 1e-10
#define mod 1000000007
#define INF 999999999
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid 
#define rson o<<1 | 1, mid+1, R
#define pii pair<int, int>
#pragma comment(linker, "/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
// head

int g[maxn][maxn];
int linker[maxn];
int slack[maxn];
bool visx[maxn];
bool visy[maxn];
int lx[maxn];
int ly[maxn];
int n, m, nx, ny, init;

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) {
			visy[y] = true;
			if(linker[y] == -1 || dfs(linker[y])) {
				linker[y] = x;
				return true;
			}
		}
		else slack[y] = min(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++)
			lx[i] = max(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, 0, sizeof visx);
			memset(visy, 0, 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++) res += g[linker[i]][i];
	return res;
}

void read()
{
	memset(g, 0, sizeof g);
	for(int i = 0; i < n; i++)
		for(int j = 0; j < m; j++)
			scanf("%d", &g[i][j]), g[i][j] *= 100;
	nx = ny = m;
	int x;
	init = 0;
	for(int i = 0; i < n; i++) {
		scanf("%d", &x), x--;
		init += g[i][x] / 100;
		g[i][x]++;
	}
}

void work()
{
	int ans = km();
	printf("%d %d\n", n - ans % 100, ans / 100 - init);
}

int main()
{
	while(scanf("%d%d", &n, &m)!=EOF) {
		read();
		work();
	}
	
	
	return 0;
}