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。

  1/* ***********************************************
  2Author :111qqz
  3Created Time :2016年06月02日 星期四 19时17分19秒
  4File Name :code/hdu/2813.cpp
  5************************************************ */
  6
  7#include <cstdio>
  8#include <cstring>
  9#include <iostream>
 10#include <algorithm>
 11#include <vector>
 12#include <queue>
 13#include <set>
 14#include <map>
 15#include <string>
 16#include <cmath>
 17#include <cstdlib>
 18#include <ctime>
 19#define fst first
 20#define sec second
 21#define lson l,m,rt<<1
 22#define rson m+1,r,rt<<1|1
 23#define ms(a,x) memset(a,x,sizeof(a))
 24typedef long long LL;
 25#define pi pair < int ,int >
 26#define MP make_pair
 27
 28using namespace std;
 29const double eps = 1E-8;
 30const int dx4[4]={1,0,0,-1};
 31const int dy4[4]={0,-1,1,0};
 32const int inf = 0x3f3f3f3f;
 33const int N=205;
 34int n,m,k;
 35map<string,int>mp,mp2;
 36int tot1,tot2;
 37int w[N][N];
 38int lx[N],ly[N];
 39int link[N];
 40bool visx[N],visy[N];
 41int slk[N];
 42
 43
 44bool find( int u)
 45{
 46    visx[u] = true;
 47    for ( int v = 1 ; v <= m ; v++)
 48    {
 49	if (visy[v]) continue;
 50	int tmp = lx[u] + ly[v] - w[u][v];
 51	if (tmp==0)
 52	{
 53	    visy[v] = true;
 54	    if (link[v]==-1||find(link[v]))
 55	    {
 56		link[v] = u;
 57		return true;
 58	    }
 59	}else if (tmp<slk[v]) slk[v] = tmp;
 60    }
 61    return false;
 62}
 63int KM()
 64{
 65    ms(lx,0xc0);
 66    ms(ly,0);
 67    ms(link,-1);
 68
 69    for ( int  i = 1 ; i <= n ; i++)
 70	for ( int j = 1 ; j <= m ; j++)
 71	    lx[i] = max(lx[i],w[i][j]);
 72
 73  //  for ( int i = 1 ; i <= n ; i++) cout<<"lx[i]:"<<lx[i]<<endl;
 74    for ( int i = 1; i <= n ; i++)
 75    {
 76	ms(slk,0x3f);
 77
 78	while (1)
 79	{
 80	    ms(visx,false);
 81	    ms(visy,false);
 82
 83	    if (find(i)) break;
 84
 85	    int d = inf;
 86
 87	    for ( int j = 1 ; j <= m ; j++)
 88		if (!visy[j]&&slk[j]<d) d = slk[j];
 89
 90	    for ( int j = 1 ; j <= n ; j++) if (visx[j]) lx[j]-=d;
 91	    for ( int j = 1 ; j <= m ; j++) if (visy[j]) ly[j]+=d; else slk[j]-=d;
 92	}
 93    }
 94    int res  =  0;
 95    for ( int i = 1 ; i <= m ; i++)
 96	if (link[i]>-1) res += w[link[i]][i];
 97
 98    return -res;
 99}
100int main()
101{
102	#ifndef  ONLINE_JUDGE 
103	freopen("code/in.txt","r",stdin);
104  #endif
105//	ios::sync_with_stdio(false);
106	while (scanf("%d%d%d",&n,&m,&k)!=EOF)
107	{
108	    mp.clear();
109	    mp2.clear();
110	    tot1 = tot2 = 0;
111	    ms(w,0xc0);
112	    string u,v;
113	    char tmpu[25],tmpv[25];
114	    int cost;
115	    for ( int i = 1 ; i <= k ; i++)
116	    {	
117		//cin>>u>>v>>cost;
118		scanf("%s %s %d",tmpu,tmpv,&cost);
119		u = string(tmpu);
120		v = string(tmpv);
121		if (!mp[u]) mp[u] = ++tot1;
122		if (!mp2[v]) mp2[v] = ++tot2;
123		w[mp[u]][mp2[v]] = -cost;
124	    }
125
126	    int ans = KM();
127	    //cout<<ans<<endl;
128	    printf("%d\n",ans);
129	}
130
131  #ifndef ONLINE_JUDGE  
132  fclose(stdin);
133  #endif
134    return 0;
135}