hdu 2813 One fihgt one (二分图最优匹配，KM算法)

2016-06-02 · 2 min read · KM 二分图最佳匹配

hdu 2813 题目链接

题意：吕布有n个武将，曹操有m（m>=n）个武将。给出k个关系，为吕布的某个武将和曹操的某个武将pk后会受到的伤害。吕布要求他所有n的武将都要上场，每个武将只能战斗一次，问如何安排，使得所有武将受到的伤害总和最小。

思路：裸的KM。 用map把武将名字变成点的编号。

/* ***********************************************
Author :111qqz
Created Time :2016年06月02日 星期四 19时17分19秒
File Name :code/hdu/2813.cpp
************************************************ */

 1#include <cstdio>
 2#include <cstring>
 3#include <iostream>
 4#include <algorithm>
 5#include <vector>
 6#include <queue>
 7#include <set>
 8#include <map>
 9#include <string>
10#include <cmath>
11#include <cstdlib>
12#include <ctime>
13#define fst first
14#define sec second
15#define lson l,m,rt<<1
16#define rson m+1,r,rt<<1|1
17#define ms(a,x) memset(a,x,sizeof(a))
18typedef long long LL;
19#define pi pair < int ,int >
20#define MP make_pair

 1using namespace std;
 2const double eps = 1E-8;
 3const int dx4[4]={1,0,0,-1};
 4const int dy4[4]={0,-1,1,0};
 5const int inf = 0x3f3f3f3f;
 6const int N=205;
 7int n,m,k;
 8map<string,int>mp,mp2;
 9int tot1,tot2;
10int w[N][N];
11int lx[N],ly[N];
12int link[N];
13bool visx[N],visy[N];
14int slk[N];

 1bool find( int u)
 2{
 3    visx[u] = true;
 4    for ( int v = 1 ; v <= m ; v++)
 5    {
 6	if (visy[v]) continue;
 7	int tmp = lx[u] + ly[v] - w[u][v];
 8	if (tmp==0)
 9	{
10	    visy[v] = true;
11	    if (link[v]==-1||find(link[v]))
12	    {
13		link[v] = u;
14		return true;
15	    }
16	}else if (tmp<slk[v]) slk[v] = tmp;
17    }
18    return false;
19}
20int KM()
21{
22    ms(lx,0xc0);
23    ms(ly,0);
24    ms(link,-1);

1    for ( int  i = 1 ; i <= n ; i++)
2	for ( int j = 1 ; j <= m ; j++)
3	    lx[i] = max(lx[i],w[i][j]);

1  //  for ( int i = 1 ; i <= n ; i++) cout<<"lx[i]:"<<lx[i]<<endl;
2    for ( int i = 1; i <= n ; i++)
3    {
4	ms(slk,0x3f);

1	while (1)
2	{
3	    ms(visx,false);
4	    ms(visy,false);

	    if (find(i)) break;

	    int d = inf;
	    
	    for ( int j = 1 ; j <= m ; j++)
		if (!visy[j]&&slk[j]<d) d = slk[j];

1	    for ( int j = 1 ; j <= n ; j++) if (visx[j]) lx[j]-=d;
2	    for ( int j = 1 ; j <= m ; j++) if (visy[j]) ly[j]+=d; else slk[j]-=d;
3	}
4    }
5    int res  =  0;
6    for ( int i = 1 ; i <= m ; i++)
7	if (link[i]>-1) res += w[link[i]][i];

 1    return -res;
 2}
 3int main()
 4{
 5	#ifndef  ONLINE_JUDGE 
 6	freopen("code/in.txt","r",stdin);
 7  #endif
 8//	ios::sync_with_stdio(false);
 9	while (scanf("%d%d%d",&n,&m,&k)!=EOF)
10	{
11	    mp.clear();
12	    mp2.clear();
13	    tot1 = tot2 = 0;
14	    ms(w,0xc0);
15	    string u,v;
16	    char tmpu[25],tmpv[25];
17	    int cost;
18	    for ( int i = 1 ; i <= k ; i++)
19	    {	
20		//cin>>u>>v>>cost;
21		scanf("%s %s %d",tmpu,tmpv,&cost);
22		u = string(tmpu);
23		v = string(tmpv);
24		if (!mp[u]) mp[u] = ++tot1;
25		if (!mp2[v]) mp2[v] = ++tot2;
26		w[mp[u]][mp2[v]] = -cost;
27	    }

1	    int ans = KM();
2	    //cout<<ans<<endl;
3	    printf("%d\n",ans);
4	}

1  #ifndef ONLINE_JUDGE  
2  fclose(stdin);
3  #endif
4    return 0;
5}