poj 2396 zoj 1994 Budget（有源汇上下界的可行流）

本文主要是介绍poj 2396 zoj 1994 Budget（有源汇上下界的可行流），希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

思路：

无源汇 (附加源汇+最大解决)

有源汇 (附加(T,S)->无源汇)

Budget

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 6237		Accepted: 2367		Special Judge

Description

We are supposed to make a budget proposal for this multi-site competition. The budget proposal is a matrix where the rows represent different kinds of expenses and the columns represent different sites. We had a meeting about this, some time ago where we discussed the sums over different kinds of expenses and sums over different sites. There was also some talk about special constraints: someone mentioned that Computer Center would need at least 2000K Rials for food and someone from Sharif Authorities argued they wouldn't use more than 30000K Rials for T-shirts. Anyway, we are sure there was more; we will go and try to find some notes from that meeting.

And, by the way, no one really reads budget proposals anyway, so we'll just have to make sure that it sums up properly and meets all constraints.

Input

The first line of the input contains an integer N, giving the number of test cases. The next line is empty, then, test cases follow: The first line of each test case contains two integers, m and n, giving the number of rows and columns (m <= 200, n <= 20). The second line contains m integers, giving the row sums of the matrix. The third line contains n integers, giving the column sums of the matrix. The fourth line contains an integer c (c < 1000) giving the number of constraints. The next c lines contain the constraints. There is an empty line after each test case.

Each constraint consists of two integers r and q, specifying some entry (or entries) in the matrix (the upper left corner is 1 1 and 0 is interpreted as "ALL", i.e. 4 0 means all entries on the fourth row and 0 0 means the entire matrix), one element from the set {<, =, >} and one integer v, with the obvious interpretation. For instance, the constraint 1 2 > 5 means that the cell in the 1st row and 2nd column must have an entry strictly greater than 5, and the constraint 4 0 = 3 means that all elements in the fourth row should be equal to 3.

Output

For each case output a matrix of non-negative integers meeting the above constraints or the string "IMPOSSIBLE" if no legal solution exists. Put one empty line between matrices.

Sample Input

22 3 
8 10 
5 6 7 
4 
0 2 > 2 
2 1 = 3 
2 3 > 2 
2 3 < 5 2 2 
4 5 
6 7 
1 
1 1 > 10

Sample Output

2 3 3 
3 3 4 IMPOSSIBLE

一道好题

首先要根据题目要求构造出一个容量网络然后根据网络建图跑最大流得到可行流

容量网络的构造方法：

如图构造一个源点s（0）一个汇点t（n+m+1）源点与每行的节点建边权值为每行的总和

每列的节点与汇点之间建边权值为每列的总和

每行每列之间有上下界根据题目的要求求出low[i][j] high[i][j]

然后判断有木有不合理的情况不合理的直接输出IMPOSSIBLE

之后开始建图求最大流

超级源点ss（n+m+2）超级汇点tt（n+m+3）

每行的节点i与每列的节点j连接 i->j权值为high[i][j]-low[i][j]

容量网络里的每个点与ss或tt连接 out[i]为正与ss连接权值为out[i]

out[i]为负与tt连接权值为-out[i]

容量网络中s t之间也要建边 t->s权值INF

细节部门就是数据的读取处理出low[i][j] high[i][j] out[i]

<span style="font-family:KaiTi_GB2312;font-size:18px;">#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string.h>
#include <string>
#include <queue>
#include <vector>#define MEM(a,x) memset(a,x,sizeof a)
#define eps 1e-8
#define MOD 10009
#define MAXN 10010
#define INF 0x3fffffff
#define ll __int64
#define bug cout<<"here"<<endl;
#define fread freopen("ceshi.txt","r",stdin)
#define fwrite freopen("out.txt","w",stdout)using namespace std;void read(int &x)
{char ch;x=0;while(ch=getchar(),ch!=' '&&ch!='\n'){x=x*10+ch-'0';}
}struct Edge
{int from,to,cap,flow;bool operator <(const Edge e) const{if(e.from!=from)  return from<e.from;else return to<e.to;}Edge() {}Edge(int from,int to,int cap,int flow):from(from),to(to),cap(cap),flow(flow) {}
};struct Dinic
{vector<Edge> edges;vector<int> G[MAXN];bool vis[MAXN];//BFS使用int d[MAXN];   //从起点到i的距离int cur[MAXN]; //当前弧下标int n,m,s,t,maxflow;   //节点数 边数（包括反向弧） 源点编号和弧点编号void init(int n){this->n=n;for(int i=0;i<=n;i++)G[i].clear();edges.clear();}void addedge(int from,int to,int cap){edges.push_back(Edge(from,to,cap,0));edges.push_back(Edge(to,from,0,0));m=edges.size();G[from].push_back(m-2);G[to].push_back(m-1);}bool bfs(){MEM(vis,0);MEM(d,-1);queue<int> q;q.push(s);d[s]=maxflow=0;vis[s]=1;while(!q.empty()){int u=q.front(); q.pop();int sz=G[u].size();for(int i=0;i<sz;i++){Edge e=edges[G[u][i]];if(!vis[e.to]&&e.cap>e.flow){d[e.to]=d[u]+1;vis[e.to]=1;q.push(e.to);}}}return vis[t];}int dfs(int u,int a){if(u==t||a==0)  return a;int sz=G[u].size();int flow=0,f;for(int &i=cur[u];i<sz;i++){Edge &e=edges[G[u][i]];if(d[u]+1==d[e.to]&&(f=dfs(e.to,min(a,e.cap-e.flow)))>0){e.flow+=f;edges[G[u][i]^1].flow-=f;flow+=f;a-=f;if(a==0)  break;}}return flow;}int Maxflow(int s,int t){this->s=s; this->t=t;int flow=0;while(bfs()){MEM(cur,0);flow+=dfs(s,INF);}return flow;}
}Dic;int out[250];
int low[300][300];
int high[300][300];
int r[220],c[30];int main()
{
//    fread;int tc;scanf("%d",&tc);int cs=0;while(tc--){int n,m;scanf("%d%d",&n,&m);MEM(out,0);int line=0; int col=0;for(int i=1;i<=n;i++){int num;scanf("%d",&num);line+=num;out[0]-=num;out[i]+=num;}for(int j=1;j<=m;j++){int num;scanf("%d",&num);col+=num;out[n+m+1]+=num;out[j+n]-=num;}//out记录进入/出去 节点的流量  为正与源点建边  为负与汇点建边for(int i=0;i<300;i++)for(int j=0;j<300;j++){low[i][j]=0;high[i][j]=INF;}int ca;scanf("%d",&ca);int flag=0;while(ca--){int u,v,w;char ch;scanf("%d %d %c %d",&u,&v,&ch,&w);if(u==0&&v==0){for(int i=1;i<=n;i++)for(int j=1;j<=m;j++){if(ch=='>') low[i][j+n]=max(low[i][j+n],w+1);if(ch=='<') high[i][j+n]=min(high[i][j+n],w-1);if(ch=='=') low[i][j+n]=high[i][j+n]=w;if(low[i][j+n]>high[i][j+n])flag=1;}}else if(u==0){for(int i=1;i<=n;i++){if(ch=='>') low[i][v+n]=max(low[i][v+n],w+1);if(ch=='<') high[i][v+n]=min(high[i][v+n],w-1);if(ch=='=') low[i][v+n]=high[i][v+n]=w;if(low[i][v+n]>high[i][v+n])flag=1;}}else if(v==0){for(int j=1;j<=m;j++){if(ch=='>') low[u][j+n]=max(low[u][j+n],w+1);if(ch=='<') high[u][j+n]=min(high[u][j+n],w-1);if(ch=='=') low[u][j+n]=high[u][j+n]=w;if(low[u][j+n]>high[u][j+n])flag=1;}}else{if(ch=='>') low[u][v+n]=max(low[u][v+n],w+1);if(ch=='<') high[u][v+n]=min(high[u][v+n],w-1);if(ch=='=') low[u][v+n]=high[u][v+n]=w;if(low[u][v+n]>high[u][v+n])flag=1;}}if(cs) puts("");cs=1;if(flag||line!=col){puts("IMPOSSIBLE");continue;}int s=n+m+2,t=n+m+3;Dic.init(t);for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){out[i]-=low[i][j+n];out[j+n]+=low[i][j+n];Dic.addedge(i,j+n,high[i][j+n]-low[i][j+n]);}}Dic.addedge(n+m+1,0,INF);int sum=0;for(int i=0;i<=n+m;i++){if(out[i]>0){sum+=out[i];Dic.addedge(s,i,out[i]);}elseDic.addedge(i,t,-out[i]);}int mx=Dic.Maxflow(s,t);
//        cout<<mx<<endl;
//        cout<<"sum "<<sum<<endl;
//        int ffff=(Dic.Maxflow(s,t)==sum);
//        cout<<ffff<<endl;if(mx!=sum){
//            bug;puts("IMPOSSIBLE");continue;}
//        for(int i=0;i<n*m;i++)
//        {
//            int u=Dic.edges[i*2].from,v=Dic.edges[i*2].to;
//            cout<<"iii "<<i<<endl;
//            cout<<u<<" "<<v<<"  "<<Dic.edges[i*2].flow+low[u][v]<<endl;
//        }int id=0;for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){if(j==1){printf("%d",Dic.edges[id*2].flow+low[i][j+n]);}elseprintf(" %d",Dic.edges[id*2].flow+low[i][j+n]);id++;}printf("\n");}
//        if(tc)
//            puts("");}return 0;
}
</span>