hdu2813：One fihgt one（KM匹配）

http://acm.hdu.edu.cn/showproblem.php?pid=2813

Problem Description

Lv Bu and his soldiers are facing a cruel war——Cao Cao had his best generals just miles away.

There’s little time , but Lv Bu is unaware of how to arrange his warriors , what he know is that he have n brave generals while Cao Cao has m , and he has k fights to choose from , he’d like to make all his n warriors participate in the battle but get the least injuries . Lv Bu is happy because there is always a good solution . So , now is your task to tell Lv Bu the least injuries his troop would get.
No one could take part in two fights.

Input

Multiple cases. For each case ,there are three integers in the first line , namely n,m (1<=n<=m<=200)and k (n<=k<=m*n).
The next k lines are the information about k possible fights , for each line are two strings (no more than 20 characters ) and an integer. The first string indicates Lv Bu’s general and the second , of course , Cao Cao’s , and the integer is the injury Lv Bu’s general would get if this fight were chosen.

Output

One integer , the least injuries Lv Bu’s generals would get.

Sample Input

2 3 5
LvBu ZhangFei 6
LvBu GuanYu 5
LvBu XuChu 4
ZhangLiao ZhangFei 8
ZhangLiao XuChu 3

Sample Output

8

题意分析：

吕布和曹操对决，吕布有n个武将，曹操有m个武将，武将之间对决会有伤害，吕布的每个武将都要上场，求吕布方受到的最小伤害值。

解题思路：

map函数存图，KM求最大匹配。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <map>
#define N 220
using namespace std;
int e[N][N], lx[N], ly[N], slack[N], march[N], bookx[N], booky[N], n, m, nx, ny, inf=99999999;
map<string, int> a;
map<string, int> b;
int dfs(int u)
{
	bookx[u]=1;
	int v, d;
	for(v=1; v<=ny; v++)
	{
		if(!booky[v])
		{
			d=lx[u]+ly[v]-e[u][v];
			if(d==0)
			{
				booky[v]=1;
				if(march[v]==0 || dfs(march[v]))
				{
					march[v]=u;
					return 1;
				}	
			}
			else
				slack[v]=min(slack[v], d);
		}
	}
	return 0;
}
int KM()
{
	int i, j, x, d;
	memset(ly, 0, sizeof(ly));
	memset(march, 0, sizeof(march));
	for(i=1; i<=nx; i++)
	{
		lx[i]=-inf;
		for(j=1; j<=ny; j++)
			lx[i]=max(lx[i], e[i][j]);
	}
	for(x=1; x<=nx; x++)
	{
		for(i=1; i<=ny; i++)
			slack[i]=inf;
		while(1)
		{
			memset(bookx, 0, sizeof(bookx));
			memset(booky, 0, sizeof(booky));
			if(dfs(x))
				break;
			d=inf;
			for(i=1; i<=ny; i++)
				if(!booky[i])
					d=min(d, slack[i]);
			for(i=1; i<=nx; i++)
				if(bookx[i])
					lx[i]-=d;
			for(i=1; i<=ny; i++)
			{
				if(booky[i])
					ly[i]+=d;
				else
					slack[i]-=d;
			}
		}
	}
	int sum=0;
	for(i=1; i<=ny; i++)
		sum+=e[march[i]][i];
	return sum;
}
int main()
{
	int i, j, k, num;
	char s1[N], s2[N];
	while(scanf("%d%d%d", &n, &m, &k)!=EOF)
	{
		for(i=1; i<=n; i++)
			for(j=1; j<=m; j++)
				e[i][j]=-inf;
		a.clear();
		b.clear();
		nx=0;
		ny=0;
		for(i=1; i<=k; i++)
		{
			scanf("%s%s%d", s1, s2, &num);
			if(!a[s1])
			{
				nx++;
				a[s1]=nx;
			}
			if(!b[s2])
			{
				ny++;
				b[s2]=ny;	
			}
			e[a[s1]][b[s2]]=-num;
		}
		printf("%d
", -KM());
	}
	return 0;
}