zoj 2071(Technology Trader) 最大权闭合图

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2071

题意简述：商人有m个订单，每个订单能挣一定的钱，但是要完成每个订单需要购买不同的机器，题目问该商人最多能够挣多少钱？

分析：建图 源点到每个机器连边，权值为花费，订单到汇点连边权值为权值为收益，之后机器和订单有联系的连一条有向边从机器到订单，最后所有订单对应的收益之和减去最大流即为所求！

代码如下：

#include <iostream>
#include <cstdio>
#include<cstring>
using namespace std;

const int N = 400;
const int E = 50000;
const int inf=0x3f3f3f3f;
int e,head
;
int dep
,que
,cur
;
struct node
{
int x,y;
int nxt;
int c;
}edge[E];
void addedge(int u,int v,int c)
{
edge[e].x=u;
edge[e].y=v;
edge[e].nxt=head[u];
edge[e].c=c;
head[u]=e++;

edge[e].x=v;
edge[e].y=u;
edge[e].nxt=head[v];
edge[e].c=0;
head[v]=e++;
}

int maxflow(int s,int t)
{
int i,j,k,front,rear,top,min,res=0;
while(1)
{
memset(dep,-1,sizeof(dep));
front=0;
rear=0;
que[rear++]=s;
dep[s]=0;
while(front!=rear)
{
i=que[front++];
for(j=head[i];j!=-1;j=edge[j].nxt)
if(edge[j].c&&dep[edge[j].y]==-1)
{
dep[edge[j].y]=dep[i]+1;
que[rear++]=edge[j].y;

}
}
if(dep[t]==-1)
break;
memcpy(cur,head,sizeof(head));
for(i=s,top=0;;)
{
if(i==t)
{
min=inf;
for(k=0;k<top;k++)
if(min>edge[que[k]].c)
{
min=edge[que[k]].c;
front=k;
}
for(k=0;k<top;k++)
{
edge[que[k]].c-=min;
edge[que[k]^1].c+=min;
}
res+=min;
i=edge[que[top=front]].x;

}
for(j=cur[i];cur[i]!=-1;j=cur[i]=edge[cur[i]].nxt)
if(dep[edge[j].y]==dep[i]+1&&edge[j].c)
break;
if(cur[i]!=-1)
{
que[top++]=cur[i];
i=edge[cur[i]].y;
}
else
{
if(top==0)
break;
dep[i]=-1;
i=edge[que[--top]].x;
}
}
}
return res;
}

int main ()
{
int t;
int n, i,m,j,num,k;
char name[260][35];
char s[35];
bool flag[110];
bool tag[260];
char order[110][35];
int c;
scanf("%d",&t);
int	h = 0;
while(t--)
{
h++;
scanf("%d",&n);
memset(head,-1,sizeof(head));
e = 0;
for(i=1;i<=n;i++)
{
scanf("%s%d",name[i],&c);
addedge(0,i,c);
}

scanf("%d",&m);
int sum = 0;
for(i=1;i<=m;i++)
{
scanf("%s%d",order[i],&c);
sum += c;
addedge(i+n,n+m+1,c);
scanf("%d",&num);
for(j=0;j<num;j++)
{
scanf("%s",s);
for(k=1;k<=n;k++)
if(strcmp(name[k],s)==0)
break;
if(k>n)
continue;
addedge(k,i+n,inf);
}
}

int flow = maxflow(0,n+m+1);
if(h>1)
printf("\n");
printf("%d\n",sum-flow);
memset(flag,0,sizeof(flag));
memset(tag,0,sizeof(tag));
num = 0;
for(i=head[n+m+1];i!=-1;i=edge[i].nxt)
{
if(edge[i^1].c>0)
{
flag[edge[i].y-n]=1;
num++;
}
}

for(i=n+1;i<n+m+1;i++)
if(flag[i-n])
for(j=head[i];j!=-1;j=edge[j].nxt)
if(edge[j].y!=n+m+1)
tag[edge[j].y]=1;

int num1 = 0;
for(i=1;i<=n;i++)
if(tag[i])
num1++;

printf("%d\n",num);
for(i=1;i<=m;i++)
if(flag[i])
printf("%s\n",order[i]);
printf("%d\n",num1);
for(i=1;i<=n;i++)
if(tag[i])
puts(name[i]);
//	system("Pause");
}
return 0;
}