题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2853
Last year a terrible earthquake attacked Sichuan province. About 300,000 PLA soldiers attended the rescue, also ALPCs. Our mission is to solve difficulty problems to optimization the assignment of troops. The assignment is measure by efficiency, which is an integer, and the larger the better.
We have N companies of troops and M missions, M>=N. One
company can get only one mission. One mission can be assigned to only one
company. If company i takes mission j, we can get efficiency Eij.
We have a
assignment plan already, and now we want to change some companies’ missions to
make the total efficiency larger. And also we want to change as less companies
as possible.
We have N companies of troops and M missions, M>=N. One
company can get only one mission. One mission can be assigned to only one
company. If company i takes mission j, we can get efficiency Eij.
We have a
assignment plan already, and now we want to change some companies’ missions to
make the total efficiency larger. And also we want to change as less companies
as possible.
题目描述：n个组和m个任务，Eij表示第i个组完成第j个任务的效率，每个组只能完成一个任务，每个任务只能由一个组完成，目前已经有了一个计划，但是现在我们想要让总效率达到最大，并且在此前提下还需要改变重新分配任务的组的个数最少。求出最大效率减去原先计划的效率和重新分配任务的组的个数。
算法分析：这道题的思维方式的确很独特，也很巧妙。首先解决第一个问题：最大效率减去原先计划的效率的差值。最大效率很好解决，用KM算法即可，原先计划的效率直接根据输入统计即可。那么第二个问题呢？重新分配任务的组的最小个数。
方法一：首先为了保证在最大效率情况下尽量选择原先已经分配了的任务，所以我们可以对原先已经分配了的任务在效率上加1，这样即使两个组对同一个任务效率相同也会选择原先的计划，然后我们标记一下有哪些边是原先计划里的。剩下的就是KM了。
说明：这种方法为什么会WA呢，还没有找到原因， 关键在于对每条边权都要乘以一个k(k>n)，下面的代码就没有乘以k，想想应该是这种方法下求得的不是最大效率吧，但为什么不是最大效率呢？ 每条边都乘以k最后的最大效率再除以k，和直接求得的最大效率不是一样的吗？
若有大牛明白其中奥妙，还望指点一二，在此感谢。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#define inf 0x7fffffff
using namespace std;
const int maxn=; int n,m,k,sum;
int lx[maxn],ly[maxn],visx[maxn],visy[maxn];
int link[maxn],slack[maxn],w[maxn][maxn];
int vis[maxn][maxn]; int dfs(int x)
{
visx[x]=;
for (int y= ;y<=m ;y++)
{
if (visy[y]) continue;
int t=lx[x]+ly[y]-w[x][y];
if (t==)
{
visy[y]=;
if (link[y]==- || dfs(link[y]))
{
link[y]=x;
return ;
}
}
else if (slack[y]>t) slack[y]=t;
}
return ;
} void KM()
{
memset(link,-,sizeof(link));
memset(ly,,sizeof(ly));
for (int i= ;i<=n ;i++)
{
lx[i]=-inf;
for (int j= ;j<=m ;j++)
lx[i]=max(lx[i],w[i][j]);
}
for (int x= ;x<=n ;x++)
{
for (int i= ;i<=m ;i++) slack[i]=inf;
while ()
{
memset(visx,,sizeof(visx));
memset(visy,,sizeof(visy));
if (dfs(x)) break;
int d=inf;
for (int i= ;i<=m ;i++)
{
if (!visy[i] && slack[i]<d) d=slack[i];
}
for (int i= ;i<=n ;i++)
if (visx[i]) lx[i] -= d;
for (int i= ;i<=m ;i++)
{
if (visy[i]) ly[i] += d;
else slack[i] -= d;
}
}
}
int ans=,cnt=;
for (int i= ;i<=m ;i++)
{
if (link[i]!=-)
{
ans += w[link[i] ][i];
if (vis[link[i] ][i]) cnt++;
}
}
printf("%d %d\n",n-cnt,ans-sum-cnt);
// for (int i=1 ;i<=m ;i++)
// {
// if (link[i]!=-1) ans += w[link[i] ][i];
// }
// printf("%d %d\n",n-ans%k,ans/k-sum);
} int main()
{
while (scanf("%d%d",&n,&m)!=EOF)
{
memset(w,,sizeof(w));
memset(vis,,sizeof(vis));
k=;
for (int i= ;i<=n ;i++)
{
for (int j= ;j<=m ;j++)
{
scanf("%d",&w[i][j]);
/// w[i][j] *= k;
}
}
int a;
sum=;
for (int i= ;i<=n ;i++)
{
scanf("%d",&a);
sum += w[i][a];
///sum += w[i][a]/k;
w[i][a] ++ ;
vis[i][a]=;
}
KM();
}
return ;
}
方法二：和方法一的区别就在于对每条边都乘以k（比如k=200），对于原有匹配w[x][y]++，最后的答案最大效率为ans。
那么差值=ans/k-sum；个数=n-ans%k。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#define inf 0x7fffffff
using namespace std;
const int maxn=; int n,m,k,sum;
int lx[maxn],ly[maxn],visx[maxn],visy[maxn];
int link[maxn],slack[maxn],w[maxn][maxn]; int dfs(int x)
{
visx[x]=;
for (int y= ;y<=m ;y++)
{
if (visy[y]) continue;
int t=lx[x]+ly[y]-w[x][y];
if (t==)
{
visy[y]=;
if (link[y]==- || dfs(link[y]))
{
link[y]=x;
return ;
}
}
else if (slack[y]>t) slack[y]=t;
}
return ;
} void KM()
{
memset(link,-,sizeof(link));
memset(ly,,sizeof(ly));
for (int i= ;i<=n ;i++)
{
lx[i]=-inf;
for (int j= ;j<=m ;j++)
lx[i]=max(lx[i],w[i][j]);
}
for (int x= ;x<=n ;x++)
{
for (int i= ;i<=m ;i++) slack[i]=inf;
while ()
{
memset(visx,,sizeof(visx));
memset(visy,,sizeof(visy));
if (dfs(x)) break;
int d=inf;
for (int i= ;i<=m ;i++)
{
if (!visy[i] && slack[i]<d) d=slack[i];
}
for (int i= ;i<=n ;i++)
if (visx[i]) lx[i] -= d;
for (int i= ;i<=m ;i++)
{
if (visy[i]) ly[i] += d;
else slack[i] -= d;
}
}
}
int ans=,cnt=;
for (int i= ;i<=m ;i++)
{
if (link[i]!=-) ans += w[link[i] ][i];
}
printf("%d %d\n",n-ans%k,ans/k-sum);
} int main()
{
while (scanf("%d%d",&n,&m)!=EOF)
{
memset(w,,sizeof(w));
k=;
for (int i= ;i<=n ;i++)
{
for (int j= ;j<=m ;j++)
{
scanf("%d",&w[i][j]);
w[i][j] *= k;
}
}
int a;
sum=;
for (int i= ;i<=n ;i++)
{
scanf("%d",&a);
sum += w[i][a]/k;
w[i][a] ++ ;
}
KM();
}
return ;
}