题意：给一个矩阵，给出每行每列之和，附加一些条件，如第i行第j列数必需大于（小于）多少。
思路题解：矩阵模型，模拟网络流，行、列标号为结点，构图，附加s,t，s连行标（容量上下限每行之和（必需以这个
值全部送过去），每个列标连向t（容量上下限每列之和）），其他每个行到列都有边（有限制的按限制
来，无限制的自己添加）。由于s->网络->t，若有解，总流量f必为
矩阵所有数之和，故添加边t-S,容量上下线为矩阵和sum，这样保证了每个点出入流相等，满足流量平衡，
转化为无源无汇有上下线可行流问题，在建立超级源点汇点ss,tt，按无源无汇法做一遍ss->tt最大流，
判断可行即可。可行输出方案。PS，若题目改为求有源有汇有限制最大流问题，可以在残量网络再走一次
s->t（原图源汇）最大流（solo可行的后悔边啥~）
此题，有源有汇有限制可行流问题，主要是建图，就是一个字烦！其一：0的时候要所有，行列为0要
讨论，其二，>,<,=符号的判定，讨论下，其三，重边，这个不是链前星锁解决那个重边，而是重边
是“且”的关系，如，先给上限，再来一边给下限！坑爹啊这是要！无法，hash判重吧，重了，再修改原来的
边，要搞清楚反向边（建图时跑第一次（第一次判断有无解）前）流量始终为0（提供后悔的，和之后没有
半毛钱关系）！至于算法，不再是问题，dinic跑的。
#include<iostream> //47ms，1A
#include<cstdio>
#include<queue>
#include<vector>
using namespace std;
int n,m;const int inf=0x3f3f3f3f;
int suminf[300];int sumouf[300];
int e[20000][5]; int head[300];int level[300];int vis[300];
int hash[250][300];int ans[250][300];
int nume=0;
bool bfs() //dinic“d,bfs”，不用说了
{
for(int i=0;i<n+m+4;i++)
level[i]=vis[i]=0;
vis[n+m+2]=1;
queue<int>q;q.push(n+m+2);
while(!q.empty())
{
int cur=q.front();q.pop();
for(int i=head[cur];i!=-1;i=e[i][1])
{
int v=e[i][0];
if(!vis[v]&&e[i][2]>0)
{
level[v]=level[cur]+1;
vis[v]=1;
if(v==m+n+3)return 1;
q.push(v);
}
}
}
return vis[n+m+3];
}
int dfs(int u,int minf)
{
if(u==n+m+3||minf==0){return minf;}
int sumf=0,f;
for(int i=head[u];i!=-1&&minf;i=e[i][1])
{
int v=e[i][0];
if(level[v]==level[u]+1&&e[i][2]>0)
{
f=dfs(v,minf<e[i][2]?minf:e[i][2]);
if(f==0)continue;
e[i][2]-=f;e[i^1][2]+=f;
minf-=f;sumf+=f;
}
}
return sumf;
}
void dinic()
{
int sumflow=0;
while(bfs())
{
sumflow+=dfs(n+m+2,inf);
}
}
bool check() //无源无汇的，出发点ss（超级源点）的边有一条不满流就无解，原因简单必需保证流量平衡（为之提供下限）
{
for(int i=head[n+m+2];i!=-1;i=e[i][1])
{
if(e[i][2]!=0)return 0;
}
return 1;
}
int main()
{
int N;
scanf("%d",&N);
while(N--) //N组输入
{
scanf("\n%d%d",&n,&m); //忽略空行
nume=0;
int sum=0,temp;
bool mark=0;
for(int i=0;i<=n+m+5;i++)
{
head[i]=-1;
suminf[i]=sumouf[i]=0;
}
for(int i=0;i<=n;i++)
for(int j=n+1;j<=n+m+1;j++)
ans[i][j]=hash[i][j]=0;
for(int i=1;i<=n;i++) //有源有汇的源点s
{
scanf("%d",&temp);sum+=temp;
e[nume][0]=i;e[nume][1]=head[0];head[0]=nume;
e[nume][2]=0;e[nume][3]=temp;e[nume++][4]=temp;
suminf[i]+=temp;sumouf[0]+=temp;
e[nume][0]=0;e[nume][1]=head[i];head[i]=nume;
e[nume++][2]=0;
}
for(int i=n+1;i<=m+n;i++) //汇点t
{
scanf("%d",&temp);
e[nume][0]=n+1+m;e[nume][1]=head[i];head[i]=nume;
e[nume][2]=0;e[nume][3]=temp;e[nume++][4]=temp;
suminf[n+m+1]+=temp;sumouf[i]+=temp;
e[nume][0]=i;e[nume][1]=head[n+m+1];head[n+m+1]=nume;
e[nume++][2]=0;
}
e[nume][0]=0;e[nume][1]=head[n+m+1];head[n+m+1]=nume; //添加t->s
e[nume][2]=0;e[nume][3]=sum;e[nume++][4]=sum;
suminf[0]+=sum;sumouf[n+m+1]+=sum;
e[nume][0]=n+m+1;e[nume][1]=head[0];head[0]=nume;
e[nume++][2]=0;
int ca;scanf("%d",&ca); int row,col,lim;char cc;
for(int i=0;i<ca;i++) //按条件建图
{
scanf("%d%d %c %d",&row,&col,&cc,&lim);
if(col!=0&&row==0)
{
if(cc=='=')
for(int i=1;i<=n;i++)
{
if(hash[i][n+col]) //判重，下面一致
{
int pp;
for( pp=head[i];e[pp][0]!=n+col;pp=e[pp][1])
;
suminf[n+col]-=e[pp][3]; sumouf[i]-=e[pp][3];
if(lim>e[pp][3])e[pp][3]=lim;
if(lim<e[pp][4])e[pp][4]=lim;
if(e[pp][3]>e[pp][4])mark=1;
e[pp][2]=e[pp][4]-e[pp][3];
suminf[n+col]+=e[pp][3]; sumouf[i]+=e[pp][3];
}
else
{
e[nume][0]=n+col;e[nume][1]=head[i];head[i]=nume;
e[nume][2]=0;e[nume][3]=lim;e[nume++][4]=lim; suminf[n+col]+=lim; sumouf[i]+=lim;hash[i][n+col]=1;
e[nume][0]=i;e[nume][1]=head[n+col];head[n+col]=nume;
e[nume++][2]=0;
}
}
else if(cc=='>')
{
for(int i=1;i<=n;i++)
{
if(hash[i][n+col])
{ int pp;
for( pp=head[i];e[pp][0]!=n+col;pp=e[pp][1])
;
suminf[n+col]-=e[pp][3]; sumouf[i]-=e[pp][3];
if(lim+1>e[pp][3])e[pp][3]=lim+1;
if(inf<e[pp][4])e[pp][4]=inf;
if(e[pp][3]>e[pp][4])mark=1;
e[pp][2]=e[pp][4]-e[pp][3];
suminf[n+col]+=e[pp][3]; sumouf[i]+=e[pp][3];
}
else
{ e[nume][0]=n+col;e[nume][1]=head[i];head[i]=nume;
e[nume][2]=inf-lim-1;e[nume][3]=lim+1;e[nume++][4]=inf;
suminf[n+col]+=lim+1; sumouf[i]+=lim+1;hash[i][n+col]=1;
e[nume][0]=i;e[nume][1]=head[n+col];head[n+col]=nume;
e[nume++][2]=0;
}
}
}
else if(cc=='<')
{
for(int i=1;i<=n;i++)
if(hash[i][n+col])
{ int pp;
for(pp=head[i];e[pp][0]!=n+col;pp=e[pp][1])
;
suminf[n+col]-=e[pp][3]; sumouf[i]-=e[pp][3];
if(0>e[pp][3])e[pp][3]=0;
if(lim-1<e[pp][4])e[pp][4]=lim-1;
if(e[pp][3]>e[pp][4])mark=1;
e[pp][2]=e[pp][4]-e[pp][3];
suminf[n+col]+=e[pp][3]; sumouf[i]+=e[pp][3];
}
else
{
e[nume][0]=n+col;e[nume][1]=head[i];head[i]=nume;
e[nume][2]=lim-1;e[nume][3]=0;e[nume++][4]=lim-1;hash[i][n+col]=1;
//inf[n+rol]+=lim; sumouf[i]+=lim;
e[nume][0]=i;e[nume][1]=head[n+col];head[n+col]=nume;
e[nume++][2]=0;
}
} }
else if(row!=0&&col==0)
{
if(cc=='>')
{
for(int i=n+1;i<=n+m;i++)
{ if(hash[row][i])
{ int pp;
for( pp=head[row];e[pp][0]!=i;pp=e[pp][1])
;
suminf[i]-=e[pp][3]; sumouf[row]-=e[pp][3];
if(lim+1>e[pp][3])e[pp][3]=lim+1;
if(inf<e[pp][4])e[pp][4]=inf;
if(e[pp][3]>e[pp][4])mark=1;
e[pp][2]=e[pp][4]-e[pp][3];
suminf[i]+=e[pp][3]; sumouf[row]+=e[pp][3];
}
else
{
e[nume][0]=i;e[nume][1]=head[row];head[row]=nume;
e[nume][2]=inf-lim-1;e[nume][3]=lim+1;e[nume++][4]=inf;
suminf[i]+=lim+1; sumouf[row]+=lim+1;hash[row][i]=1;
e[nume][0]=row;e[nume][1]=head[i];head[i]=nume;
e[nume++][2]=0;
}
}
}
else if(cc=='=')
for(int i=n+1;i<=n+m;i++)
{
if(hash[row][i])
{ int pp;
for(pp=head[row];e[pp][0]!=i;pp=e[pp][1])
;
suminf[i]-=e[pp][3]; sumouf[row]-=e[pp][3];
if(lim>e[pp][3])e[pp][3]=lim;
if(lim<e[pp][4])e[pp][4]=lim;
if(e[pp][3]>e[pp][4])mark=1;
e[pp][2]=e[pp][4]-e[pp][3];
suminf[i]+=e[pp][3]; sumouf[row]+=e[pp][3];
}
else
{ e[nume][0]=i;e[nume][1]=head[row];head[row]=nume;
e[nume][2]=0;e[nume][3]=lim;e[nume++][4]=lim;
suminf[i]+=lim; sumouf[row]+=lim;hash[row][i]=1;
e[nume][0]=row;e[nume][1]=head[i];head[i]=nume;
e[nume++][2]=0;
}
}
else if(cc=='<')
for(int i=n+1;i<=n+m;i++)
{
if(hash[row][i])
{ int pp;
for( pp=head[row];e[pp][0]!=i;pp=e[pp][1])
;
suminf[i]-=e[pp][3]; sumouf[row]-=e[pp][3];
if(0>e[pp][3])e[pp][3]=0;
if(lim-1<e[pp][4])e[pp][4]=lim-1;
if(e[pp][3]>e[pp][4])mark=1;
e[pp][2]=e[pp][4]-e[pp][3];
suminf[i]+=e[pp][3]; sumouf[row]+=e[pp][3];
}
else
{
e[nume][0]=i;e[nume][1]=head[row];head[row]=nume;
e[nume][2]=lim-1;e[nume][3]=0;e[nume++][4]=lim-1;hash[row][i]=1;
//inf[i]+=lim; sumouf[row]+=lim;
e[nume][0]=row;e[nume][1]=head[i];head[i]=nume;
e[nume++][2]=0;
}
}
}
else if(row==0&&col==0)
{
if(cc=='=')
{
for(int i=1;i<=n;i++)
for(int j=n+1;j<=n+m;j++)
{
if(hash[i][j])
{ int pp;
for( pp=head[i];e[pp][0]!=j;pp=e[pp][1])
;
suminf[j]-=e[pp][3]; sumouf[i]-=e[pp][3];
if(lim>e[pp][3])e[pp][3]=lim;
if(lim<e[pp][4])e[pp][4]=lim;
if(e[pp][3]>e[pp][4])mark=1;
e[pp][2]=e[pp][4]-e[pp][3];
suminf[j]+=e[pp][3]; sumouf[i]+=e[pp][3];
}
else
{
e[nume][0]=j;e[nume][1]=head[i];head[i]=nume;
e[nume][2]=0;e[nume][3]=lim;e[nume++][4]=lim;
suminf[j]+=lim; sumouf[i]+=lim;hash[i][j]=1;
e[nume][0]=i;e[nume][1]=head[j];head[j]=nume;
e[nume++][2]=0;
}
}
}
if(cc=='>')
{
for(int i=1;i<=n;i++)
for(int j=n+1;j<=n+m;j++)
{
if(hash[i][j])
{ int pp;
for( pp=head[i];e[pp][0]!=j;pp=e[pp][1])
;
suminf[j]-=e[pp][3]; sumouf[i]-=e[pp][3];
if(lim+1>e[pp][3])e[pp][3]=lim+1;
if(inf<e[pp][4])e[pp][4]=inf;
if(e[pp][3]>e[pp][4])mark=1;
e[pp][2]=e[pp][4]-e[pp][3];
suminf[j]+=e[pp][3]; sumouf[i]+=e[pp][3];
}
else
{
e[nume][0]=j;e[nume][1]=head[i];head[i]=nume;
e[nume][2]=inf-lim-1;e[nume][3]=lim+1;e[nume++][4]=inf;
suminf[j]+=lim+1; sumouf[i]+=lim+1;hash[i][j]=1;
e[nume][0]=i;e[nume][1]=head[j];head[j]=nume;
e[nume++][2]=0;
}
}
}
if(cc=='<')
for(int i=1;i<=n;i++)
for(int j=n+1;j<=n+m;j++)
{
if(hash[i][j])
{ int pp;
for( pp=head[i];e[pp][0]!=j;pp=e[pp][1])
;
suminf[j]-=e[pp][3]; sumouf[i]-=e[pp][3];
if(0>e[pp][3])e[pp][3]=0;
if(lim-1<e[pp][4])e[pp][4]=lim-1;
if(e[pp][3]>e[pp][4])mark=1;
e[pp][2]=e[pp][4]-e[pp][3];
suminf[j]+=e[pp][3]; sumouf[i]+=e[pp][3];
}
else
{
e[nume][0]=j;e[nume][1]=head[i];head[i]=nume;
e[nume][2]=lim-1;e[nume][3]=0;e[nume++][4]=lim-1;hash[i][j]=1;
//inf[j]+=lim; sumouf[i]+=lim;
e[nume][0]=i;e[nume][1]=head[j];head[j]=nume;
e[nume++][2]=0;
}
}
}
else
{
if(cc=='=')
{
if(hash[row][n+col])
{ int pp;
for(pp=head[row];e[pp][0]!=n+col;pp=e[pp][1])
;
suminf[n+col]-=e[pp][3]; sumouf[row]-=e[pp][3];
if(lim>e[pp][3])e[pp][3]=lim;
if(lim<e[pp][4])e[pp][4]=lim;
if(e[pp][3]>e[pp][4])mark=1;
e[pp][2]=e[pp][4]-e[pp][3];
suminf[n+col]+=e[pp][3]; sumouf[row]+=e[pp][3];
}
else
{
e[nume][0]=col+n;e[nume][1]=head[row];head[row]=nume;
e[nume][2]=0;e[nume][3]=lim;e[nume++][4]=lim;
suminf[col+n]+=lim; sumouf[row]+=lim;hash[row][col+n]=1;
e[nume][0]=row;e[nume][1]=head[col+n];head[col+n]=nume;
e[nume++][2]=0;
}
}
else if(cc=='>')
{
if(hash[row][n+col])
{ int pp;
for(pp=head[row];e[pp][0]!=n+col;pp=e[pp][1])
;
suminf[n+col]-=e[pp][3]; sumouf[row]-=e[pp][3];
if(lim+1>e[pp][3])e[pp][3]=lim+1;
if(inf<e[pp][4])e[pp][4]=inf;
if(e[pp][3]>e[pp][4])mark=1;
e[pp][2]=e[pp][4]-e[pp][3];
suminf[n+col]+=e[pp][3]; sumouf[row]+=e[pp][3];
}
else
{
e[nume][0]=col+n;e[nume][1]=head[row];head[row]=nume;
e[nume][2]=inf-lim-1;e[nume][3]=lim+1;e[nume++][4]=inf;
suminf[col+n]+=lim+1; sumouf[row]+=lim+1;hash[row][col+n]=1;
e[nume][0]=row;e[nume][1]=head[col+n];head[col+n]=nume;
e[nume++][2]=0;
}
}
else if(cc=='<')
{
if(hash[row][n+col])
{ int pp;
for( pp=head[row];e[pp][0]!=n+col;pp=e[pp][1])
;
suminf[n+col]-=e[pp][3]; sumouf[row]-=e[pp][3];
if(0>e[pp][3])e[pp][3]=0;
if(lim-1<e[pp][4])e[pp][4]=lim-1;
if(e[pp][3]>e[pp][4])mark=1;
e[pp][2]=e[pp][4]-e[pp][3];
suminf[n+col]+=e[pp][3]; sumouf[row]+=e[pp][3];
}
else
{ e[nume][0]=col+n;e[nume][1]=head[row];head[row]=nume;
e[nume][2]=lim-1;e[nume][3]=0;e[nume++][4]=lim-1;hash[row][col+n]=1;
// inf[col]+=lim; sumouf[row]+=lim;
e[nume][0]=row;e[nume][1]=head[col+n];head[col+n]=nume;
e[nume++][2]=0;
}
}
}
}
for(int i=1;i<=n;i++) //有些边没有限制，不要忘记加边！！
for(int j=n+1;j<=n+m;j++)
if(!hash[i][j])
{
e[nume][0]=j;e[nume][1]=head[i];head[i]=nume;
e[nume][4]=inf;e[nume++][2]=inf;
e[nume][0]=i;e[nume][1]=head[j];head[j]=nume;
e[nume++][2]=0;
}
//无源无汇添加ss,tt
for(int i=0;i<=n+m+1;i++)
{
e[nume][0]=i;e[nume][1]=head[n+m+2];head[n+m+2]=nume;
e[nume++][2]=suminf[i];
e[nume][0]=n+m+2;e[nume][1]=head[i];head[i]=nume;
e[nume++][2]=0; e[nume][0]=n+m+3;e[nume][1]=head[i];head[i]=nume;
e[nume++][2]=sumouf[i];
e[nume][0]=i;e[nume][1]=head[n+m+3];head[n+m+3]=nume;
e[nume++][2]=0;
}
dinic(); //从ss->tt，判断有无解
if(mark||!check())printf("IMPOSSIBLE\n");
else //输出答案
{
for(int i=1;i<=n;i++)
for(int j=head[i];j!=-1;j=e[j][1])
ans[i][e[j][0]]=e[j][4]-e[j][2]; //原来上限-残留量，即为所用（上限-下限-残留+下限）
for(int i=1;i<=n;i++)
for(int j=n+1;j<=n+m;j++)
{
if(j!=n+m)printf("%d ",ans[i][j]);
else printf("%d\n",ans[i][j]);
}
}
if(N!=0)printf("\n");
}
return 0;
}