uva11082 最大流和网络流建模

主要是建模想不到，下面两篇讲的不错。

http://www.cnblogs.com/xcw0754/p/4646140.html

http://www.cnblogs.com/AOQNRMGYXLMV/p/4280727.html

就是将矩阵的行看成一个水管，列看成一个水管，矩阵的值a[i][j]看成第i行流到第j列的流量。第i行的和就是该行流向所有列的流量之和。#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<string>
#include<map>
#include<set>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<sstream>
#define LL long long
#define OJ_PRINT 0
#define READ_FILE 0
using namespace std;
const int NN_MAX = 100;
const int INF = 0x1fffffff;
struct Edge{
int from,to,cap;
Edge (int a1=0,int b1=0,int c1=0):from(a1),to(b1),cap(c1) {}
};
/**********************************************************/
int r,c,s,t,sum;
int rr[NN_MAX],cc[NN_MAX];
int cap[NN_MAX][NN_MAX],flow[NN_MAX][NN_MAX],a[NN_MAX],p[NN_MAX];
/**********************************************************/
int min_2 (int x,int y) {return x<y?x:y;}
int max_2 (int x,int y) {return x>y?x:y;}
void InitCap ();
void EK ();
/**********************************************************/
int main()
{
if (READ_FILE) freopen ("in.txt","r",stdin);
int tt;
scanf ("%d",&tt);
for (int i=1;i<=tt;i++)
{
scanf ("%d%d",&r,&c);
rr[0]=cc[0]=0;
for (int j=1;j<=r;j++)
scanf ("%d",&rr[j]);
for (int j=r;j>=1;j--)
rr[j]-=rr[j-1];
for (int j=1;j<=c;j++)
scanf ("%d",&cc[j]);
for (int j=c;j>=1;j--)
cc[j]-=cc[j-1];
printf ("Matrix %d\n",i);
InitCap ();
EK ();
if (i!=tt) printf ("\n");
}
return 0;
}
void InitCap ()
{
memset (cap,0,sizeof (cap));
memset (flow,0,sizeof (flow));
s=0,t=r+c+1;
for (int i=1;i<=r;i++)
cap[s][i]=rr[i]-c;
for (int i=r+1;i<=r+c;i++)
cap[i][t]=cc[i-r]-r;
for (int i=1;i<=r;i++)
for (int j=r+1;j<=r+c;j++)
cap[i][j]=19;
if (OJ_PRINT)
for (int i=0;i<
4000
=r+c+1;i++)
for (int j=0;j<=r+c+1;j++)
if (cap[i][j])
printf ("%d %d %d\n",i,j,cap[i][j]);
}
void EK ()
{
queue<int>qee;
sum=0;
while (1)
{
memset (a,0,sizeof(a));a[s]=INF;
qee.push (s);
while (!qee.empty ())
{
int x=qee.front ();qee.pop ();
for (int y=0;y<=r+c+1;y++){
if (!a[y] && cap[x][y]>flow[x][y]){
p[y]=x;
qee.push (y);
a[y]=min_2 (a[x],cap[x][y]-flow[x][y]);
}
}
}
if (a[t]==0) break;
for (int i=t;i!=s;i=p[i]){
flow[p[i]][i]+=a[t];
flow[i][p[i]]-=a[t];
}
}
for (int i=1;i<=r;i++)
for (int j=r+1;j<=r+c;j++){
printf ("%d",flow[i][j]+1);
j==r+c?printf ("\n"):printf (" ");
}
}