hdu 2813 One fihgt one（二分图最小权匹配KM）

题目链接：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
  
  

 

Author

shǎ崽

 

Source

HDU 1st “Old-Vegetable-Birds Cup” Programming Open Contest


题意是吕布和曹操对战，给出两队的对战情况，要求吕布伤害最小，即求最小权匹配，建图用到了map，在读入的时候我用了cin，一直超时，改为scanf就AC了。。
代码：

#include<stdio.h>
#include<string.h>
#include<map>
#include<string>
#include<iostream>
using namespace std;
map<string,int>mp1;
map<string,int>mp2;
const int maxn=210;
const int inf=1<<31-1;
int n,m;
int w[maxn][maxn];
int lx[maxn],ly[maxn];
int sx[maxn],sy[maxn];
int match[maxn];
int nx,ny;
int find(int u)
{
    sx[u]=1;
    for(int v=1;v<=ny;v++)
    {
        if(!sy[v]&&lx[u]+ly[v]==w[u][v])
        {
            sy[v]=1;
            if(match[v]==-1||find(match[v]))
            {
                match[v]=u;
                return 1;
            }
        }
    }
    return 0;
}
int km()
{
    for(int i=1;i<=nx;i++)
    {
        lx[i]=-inf;
        for(int j=1;j<=ny;j++)
            if(lx[i]<w[i][j])
            lx[i]=w[i][j];
    }
    memset(ly,0,sizeof(ly));
    memset(match,-1,sizeof(match));

    for(int u=1;u<=nx;u++)
    {
        while(1)
        {
            memset(sx,0,sizeof(sx));
            memset(sy,0,sizeof(sy));
            if(find(u))
                break;
            int d=inf;
            for(int i=1;i<=nx;i++)
                if(sx[i])
                for(int j=1;j<=ny;j++)
                if(!sy[j]&&d>lx[i]+ly[j]-w[i][j])
                   d=lx[i]+ly[j]-w[i][j];
            if(d==inf)
                return -1;
            for(int i=1;i<=nx;i++)
            {
                if(sx[i])
                    lx[i]-=d;
            }
            for(int i=1;i<=ny;i++)
            {
                if(sy[i])
                    ly[i]+=d;
            }
        }
    }
    int res=0;
    for(int i=1;i<=ny;i++)
        if(match[i]>=0)
        res+=w[match[i]][i];
    return -res;
}
int main()
{
    //freopen("in.txt","r",stdin);
    int p;
    while(scanf("%d %d %d",&n,&m,&p)==3)
    {
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
                w[i][j]=-inf;
        mp1.clear();
        mp2.clear();
        nx=n;
        ny=m;
        int u=1,v=1;
        while(p--)
        {
            char a[30],b[30];
            int c;
            scanf("%s%s%d",a,b,&c);
            if(mp1.find(a)==mp1.end())
                mp1[a]=u++;
            if(mp2.find(b)==mp2.end())
                mp2[b]=v++;
            w[mp1[a]][mp2[b]]=-c;
        }
        int ans=km();
        printf("%d\n",ans);
    }
    return 0;
}