hdu 2813 One fihgt one (二分图最优匹配,KM算法)

Posted by 111qqz on Thursday, June 2, 2016

TOC

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
************************************************ */

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#define fst first
#define sec second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ms(a,x) memset(a,x,sizeof(a))
typedef long long LL;
#define pi pair < int ,int >
#define MP make_pair

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


bool find( int u)
{
    visx[u] = true;
    for ( int v = 1 ; v <= m ; v++)
    {
    if (visy[v]) continue;
    int tmp = lx[u] + ly[v] - w[u][v];
    if (tmp==0)
    {
        visy[v] = true;
        if (link[v]==-1||find(link[v]))
        {
        link[v] = u;
        return true;
        }
    }else if (tmp<slk[v]) slk[v] = tmp;
    }
    return false;
}
int KM()
{
    ms(lx,0xc0);
    ms(ly,0);
    ms(link,-1);

    for ( int  i = 1 ; i <= n ; i++)
    for ( int j = 1 ; j <= m ; j++)
        lx[i] = max(lx[i],w[i][j]);
    
  //  for ( int i = 1 ; i <= n ; i++) cout<<"lx[i]:"<<lx[i]<<endl;
    for ( int i = 1; i <= n ; i++)
    {
    ms(slk,0x3f);

    while (1)
    {
        ms(visx,false);
        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];
	    
        for ( int j = 1 ; j <= n ; j++) if (visx[j]) lx[j]-=d;
        for ( int j = 1 ; j <= m ; j++) if (visy[j]) ly[j]+=d; else slk[j]-=d;
    }
    }
    int res  =  0;
    for ( int i = 1 ; i <= m ; i++)
    if (link[i]>-1) res += w[link[i]][i];

    return -res;
}
int main()
{
    #ifndef  ONLINE_JUDGE 
    freopen("code/in.txt","r",stdin);
  #endif
//	ios::sync_with_stdio(false);
    while (scanf("%d%d%d",&n,&m,&k)!=EOF)
    {
        mp.clear();
        mp2.clear();
        tot1 = tot2 = 0;
        ms(w,0xc0);
        string u,v;
        char tmpu[25],tmpv[25];
        int cost;
        for ( int i = 1 ; i <= k ; i++)
        {	
        //cin>>u>>v>>cost;
        scanf("%s %s %d",tmpu,tmpv,&cost);
        u = string(tmpu);
        v = string(tmpv);
        if (!mp[u]) mp[u] = ++tot1;
        if (!mp2[v]) mp2[v] = ++tot2;
        w[mp[u]][mp2[v]] = -cost;
        }

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

  #ifndef ONLINE_JUDGE  
  fclose(stdin);
  #endif
    return 0;
}

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