题目地址:HDU 5045
终于在比赛中用网络流A了一道题。。。刷了那么多网络流,终于用到一次了。。虽然题目很简单,但是还是要纪念一下下。。。
我这题的思路就是求m/n次费用流,每n个算作同一轮,对这同一轮的求最大费用流。建图就很简单了,最简单的二分图模型。
代码如下:
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <stdlib.h>
#include <math.h>
#include <ctype.h>
#include <queue>
#include <map>
#include <set>
#include <algorithm>
using namespace std;
#define LL __int64
using namespace std;
double mp[20][2000], d[200], cost;
const int INF=0x3f3f3f3f;
int head[200], source, sink, cnt, flow;
int f[200], cur[200], vis[200];
struct node
{
int u, v, cap, next;
double cost;
} edge[100000];
void add(int u, int v, int cap, double cost)
{
edge[cnt].v=v;
edge[cnt].cap=cap;
edge[cnt].cost=cost;
edge[cnt].next=head[u];
head[u]=cnt++;
edge[cnt].v=u;
edge[cnt].cap=0;
edge[cnt].cost=-cost;
edge[cnt].next=head[v];
head[v]=cnt++;
}
int spfa()
{
int i;
for(i=0; i<200; i++)
d[i]=INF;
memset(vis,0,sizeof(vis));
cur[source]=-1;
f[source]=INF;
d[source]=0;
queue<int>q;
q.push(source);
while(!q.empty())
{
int u=q.front();
q.pop();
vis[u]=0;
for(i=head[u]; i!=-1; i=edge[i].next)
{
int v=edge[i].v;
if(d[v]>d[u]+edge[i].cost&&edge[i].cap)
{
d[v]=d[u]+edge[i].cost;
f[v]=min(f[u],edge[i].cap);
cur[v]=i;
if(!vis[v])
{
vis[v]=1;
q.push(v);
}
}
}
}
if(d[sink]==INF) return 0;
flow+=f[sink];
cost-=f[sink]*d[sink];
for(i=cur[sink]; i!=-1; i=cur[edge[i^1].v])
{
edge[i].cap-=f[sink];
edge[i^1].cap+=f[sink];
}
return 1;
}
void mcmf()
{
cost=flow=0;
while(spfa()) ;
}
int main()
{
int t, num=0, i, j, k, n, m;
double sum;
scanf("%d",&t);
while(t--)
{
sum=0;
num++;
scanf("%d%d",&n,&m);
for(i=1; i<=n; i++)
{
for(j=1; j<=m; j++)
{
scanf("%lf",&mp[i][j]);
}
}
k=1;
sum=0;
while(k<=m)
{
source=0;
sink=2*n+1;
memset(head,-1,sizeof(head));
cnt=0;
for(i=k; i<k+n&&i<=m; i++)
{
add(i-k+1+n,sink,1,0);
for(j=1; j<=n; j++)
{
add(j,i-k+1+n,1,-mp[j][i]);
}
}
for(i=1; i<=n; i++)
add(source,i,1,0);
mcmf();
sum+=cost;
//printf("%.2lf %d
",sum, k);
k+=n;
}
printf("Case #%d: %.5lf
",num,sum);
}
return 0;
}