题目:

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

2

2 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

Source

题解:

算是求有源汇上下界最大流的一道模板题···只不过建图有点麻烦····

先说有源汇上下界最大流最小流的基本求法(引用自农民伯伯-Coding):

有源汇网络的可行流

对于流量有上下界的有源汇网络,原图中存在源点S和汇点T,为了求可行流,先将其转换为无源汇网络。

从T-->S引入一条边,其流量上下界为[0, INF],此时原图变为了无源汇网络,然后按照无源汇网络求解可行流的方法来求解。 

有源汇网络的最大流

要求最大流,先求可行流,通过“有源汇网络的可行流”的求解方法来判断有源汇网络存在可行流。 
若存在可行流,记从S流出的流量sum1,然后将T-->S的边取消,再次从S到T求解网络的最大流,记从S流出的流量sum2. 那么该有源汇网络的最大流为 sum1 + sum2. 
其中,sum1是在网络满足流量下界的条件下,从源点S流出的流量;求出sum1之后,网络中可能还有余量可以继续增广,那么再次求解从S到T的最大流,得到sum2,sum1 + sum2即为最终的最大流。

有源汇网络的最小流

求解有源汇网络最小流分为以下几步: 
(1)对SS到TT求一次最大流,即为f1.(在有源汇的情况下,先把整个网络趋向必须边尽量满足的情况); 
(2)添加一条边T-->S,流量上限为INF,这条边即为P.(构造无源网络) 
(3)对SS到TT再次求最大流,即为f2。(判断此时的网络中存在可行流,同时求出SS-->TT最大流) 
    如果所有必须边都满流,证明存在可行解,原图的最小流为“流经边P的流量”(原图已经构造成无源汇网络,对于S同样满足 入流和==出流和,只有新添加的边流向S,而S的出流就是原图的最小流)。 
注: 最小流求法的正确性不知如何证明,待继续学习。

代码:

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<ctime>
#include<cctype>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
const int N=;
const int M=;
const int inf=0x3f3f3f3f;
int tot=,first[N],next[M],go[M],rest[M],lev[N],cur[N];
int tr[N],tc[N],n,m,low[N][N],up[N][N],T,src,des,ss,tt,sum1,sum2,k,ans,id[N][N],p;
bool jud[N][N];
inline void comb(int a,int b,int c)
{
next[++tot]=first[a],first[a]=tot,go[tot]=b,rest[tot]=c;
next[++tot]=first[b],first[b]=tot,go[tot]=a,rest[tot]=;
}
inline void comb1(int a,int b,int c)
{
next[tot]=first[a],first[a]=tot,go[tot]=b,rest[tot]=c;
next[++tot]=first[b],first[b]=tot,go[tot]=a,rest[tot]=;
}
inline int R()
{
char c;
int f=;
for(c=getchar();c<''||c>'';c=getchar());
for(;c<=''&&c>='';c=getchar())
f=(f<<)+(f<<)+c-'';
return f;
}
inline void pre()
{
memset(first,,sizeof(first));
memset(low,,sizeof(low));
memset(up,inf,sizeof(up));
memset(tr,,sizeof(tr));
memset(tc,,sizeof(tc));
tot=,src=,des=n+m+,ss=n+m+,tt=n+m+,sum1=,sum2=,ans=;
}
inline bool bfs()
{
for(int i=src;i<=tt;i++) cur[i]=first[i],lev[i]=-;
static int que[N],tail,u,v;
que[tail=]=ss;
lev[ss]=;
for(int head=;head<=tail;head++)
{
u=que[head];
for(int e=first[u];e;e=next[e])
{
if(lev[v=go[e]]==-&&rest[e])
{
lev[v]=lev[u]+;
que[++tail]=v;
if(v==tt) return true;
}
}
}
return false;
}
inline int dinic(int u,int flow)
{
if(u==tt)
return flow;
int res=,delta,v;
for(int &e=cur[u];e;e=next[e])
{
if(lev[v=go[e]]>lev[u]&&rest[e])
{
delta=dinic(v,min(flow-res,rest[e]));
if(delta)
{
rest[e]-=delta;
rest[e^]+=delta;
res+=delta;
if(res==flow) break;
}
}
}
if(flow!=res) lev[u]=-;
return res;
}
inline void maxflow()
{
while(bfs())
ans+=dinic(ss,inf);
}
int main()
{
//freopen("a.in","r",stdin);
T=R();
int a,b,c;
char s[];
while(T--)
{
scanf("\n");
n=R(),m=R();
pre();
for(int i=;i<=n;i++)
{
a=R();
tc[src]+=a,tr[i]+=a;
sum1+=a;
}
for(int i=;i<=m;i++)
{
a=R();
tr[des]+=a,tc[n+i]+=a;
sum2+=a;
}
if(sum1!=sum2)
{
cout<<"IMPOSSIBLE"<<endl;
cout<<endl;
continue;
}
k=R();
while(k--)
{
scanf("%d%d%s%d",&a,&b,s,&c);
if(a==&&b==)
{
for(int i=;i<=n;i++)
for(int j=;j<=m;j++)
{
if(s[]=='=')
{
low[i][j]=max(low[i][j],c);
up[i][j]=min(up[i][j],c);
}
if(s[]=='<')
up[i][j]=min(up[i][j],c-);
if(s[]=='>')
low[i][j]=max(low[i][j],c+);
}
}
else if(a==)
for(int i=;i<=n;i++)
{
if(s[]=='=')
{
low[i][b]=max(low[i][b],c);
up[i][b]=min(up[i][b],c);
}
if(s[]=='<')
up[i][b]=min(up[i][b],c-);
if(s[]=='>')
low[i][b]=max(low[i][b],c+);
}
else if(b==)
for(int i=;i<=m;i++)
{
if(s[]=='=')
{
low[a][i]=max(low[a][i],c);
up[a][i]=min(up[a][i],c);
}
if(s[]=='<')
up[a][i]=min(up[a][i],c-);
if(s[]=='>')
low[a][i]=max(low[a][i],c+);
}
else
{
if(s[]=='=')
{
low[a][b]=max(low[a][b],c);
up[a][b]=min(up[a][b],c);
}
if(s[]=='<')
up[a][b]=min(up[a][b],c-);
if(s[]=='>')
low[a][b]=max(low[a][b],c+);
}
}
bool flag=false;
for(int i=;i<=n;i++)
{
if(flag==true) break;
for(int j=;j<=m;j++)
{
if(low[i][j]>up[i][j])
{
flag=true;
break;
}
else
{
tr[n+j]+=low[i][j];
tc[i]+=low[i][j];
id[i][j]=++tot;
comb1(i,n+j,up[i][j]-low[i][j]);
}
}
}
if(flag==true)
{
cout<<"IMPOSSIBLE"<<endl;
cout<<endl;
continue;
}
int cnt=;
p=++tot;
comb1(des,src,inf);
for(int i=src;i<=des;i++)
{
if(tr[i]>tc[i])
{
comb(ss,i,tr[i]-tc[i]);
cnt+=tr[i]-tc[i];
}
if(tc[i]>tr[i])
comb(i,tt,tc[i]-tr[i]);
}
maxflow();
if(ans!=cnt)
{
cout<<"IMPOSSIBLE"<<endl;
cout<<endl;
continue;
}
rest[p]=rest[p^]=;
maxflow();
for(int i=;i<=n;i++)
{
for(int j=;j<=m;j++)
cout<<rest[id[i][j]^]+low[i][j]<<" ";
cout<<endl;
}
cout<<endl;
}
return ;
}

算法复习——有源汇上下界可行流(bzoj2396)的更多相关文章

  1. POJ2396 Budget [有源汇上下界可行流]

    POJ2396 Budget 题意:n*m的非负整数矩阵,给出每行每列的和,以及一些约束关系x,y,>=<,val,表示格子(x,y)的值与val的关系,0代表整行/列都有这个关系,求判断 ...

  2. 有源汇上下界可行流(POJ2396)

    题意:给出一个n*m的矩阵的每行和及每列和,还有一些格子的限制,求一组合法方案. 源点向行,汇点向列,连一条上下界均为和的边. 对于某格的限制,从它所在行向所在列连其上下界的边. 求有源汇上下界可行流 ...

  3. 计蒜客 31447 - Fantastic Graph - [有源汇上下界可行流][2018ICPC沈阳网络预赛F题]

    题目链接:https://nanti.jisuanke.com/t/31447 "Oh, There is a bipartite graph.""Make it Fan ...

  4. poj2396 Budget(有源汇上下界可行流)

    [题目链接] http://poj.org/problem?id=2396 [题意] 知道一个矩阵的行列和,且知道一些格子的限制条件,问一个可行的方案. [思路] 设行为X点,列为Y点,构图:连边(s ...

  5. poj2396有源汇上下界可行流

    题意:给一些约束条件,要求算能否有可行流,ps:刚开始输入的是每一列和,那么就建一条上下界相同的边,这样满流的时候就一定能保证流量相同了,还有0是该列(行)对另一行每个点都要满足约束条件 解法:先按无 ...

  6. ZOJ1994有源汇上下界可行流

    http://fastvj.rainng.com/contest/236779#problem/G Description: n 行 m 列 给你行和 与 列和 然后有Q个限制,表示特定单元格元素大小 ...

  7. bzoj 2406 矩阵 —— 有源汇上下界可行流

    题目:https://www.lydsy.com/JudgeOnline/problem.php?id=2406 这题,首先把题目那个式子的绝对值拆成两个限制,就成了网络流的上下界: 有上下界可行流原 ...

  8. bzoj千题计划158:bzoj2406: 矩阵(有源汇上下界可行流)

    http://www.lydsy.com/JudgeOnline/problem.php?id=2406 设矩阵C=A-B 最小化 C 一行或一列和的最大值 整体考虑一行或者一列的和 二分最大值 这样 ...

  9. bzoj 2406 矩阵——有源汇上下界可行流

    题目:https://www.lydsy.com/JudgeOnline/problem.php?id=2406 二分答案.把 b 的 n 个行作为一排, m 个列作为一排,每行和每列之间连上下界为 ...

随机推荐

  1. Chisel语言

    1 What is Chisel?      Chisel(Constructing Hardware In a Scala Embedded Language)是一种嵌入在高级编程语言Scala的硬 ...

  2. Java MVC 分页实例

    共4个文件 requestLogList.jsp RequestInfoController.java RequestInfoBean.java RequestInfoService.java 1.r ...

  3. ThreadLocal遇到线程池时, 各线程间的数据会互相干扰, 串来串去

    最近遇到一个比较隐蔽而又简单地问题,在使用ThreadLocal时发现出现多个线程中值串来串去,排查一番,确定问题为线程池的问题,线程池中的线程是会重复利用的,而ThreadLocal是用线程来做Ke ...

  4. HDinsight 系列-使用证书登陆中国区Azure

    使用azure explorer 插件的时候,登陆默认是globle的azure网站,中国区的azure不能直接使用 可以使用auth文件认证 auth 文件生成 az cloud show -o j ...

  5. springmvc+maven搭建web项目

    1.创建一个maven project 为spring1 2.进行项目的配置:默认的java 1.5 在properties中选择project facts项目进行配置,反选web之后修改java环境 ...

  6. 通过90行代码学会HTML5 WebSQL的4种基本操作

    Web SQL数据库API是一个独立的规范,在浏览器层面提供了本地对结构化数据的存储,已经被很多现代浏览器支持了. 我们通过一个简单的例子来了解下如何使用Web SQL API在浏览器端创建数据库表并 ...

  7. webpack 使用流程

    webpack loader 读文件的 脚手架  vue-cli: 自动化的小工具,帮咱们把项目的架子搭起来 -------------------------------------- 开发环境 n ...

  8. java程序-类的高级特性

    创建Employee类,在类中定义三个属性:编号,姓名,年龄,然后在构造方法里初始化这三个属性,最后在实现接口中的定义的CompareTo方法,将对象按编号升序排列. 代码如下:(程序可能有些错误,方 ...

  9. SVN中检出(check out) 跟导出(export) 的区别

    SVN中检出(check out) 和导出(export) 的区别?观点一:SVN是常用的一种常见的版本控制软件.SVN中检出(check   SVN中检出(check out) 和导出(export ...

  10. WORD与DWORD

    在看C/C++的书或者试题时,有时会见到利用word或dword定义的变量,第一次看到的时候并不知其是什么,更不用说word或dword占几个字节了.幸好在VC安装文件夹下有相关的定义.如C:\Pro ...