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，求判断是否有解并求一组解

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建图显然

\[s \rightarrow _{[行和,行和]} x \rightarrow _{格子(x,y)的限制[l,r]} y \rightarrow_{[列和,列和]} t\]

有源汇上下界可行流

注意是非负整数

#include 
    
     #include 
     
      #include 
      
       #include 
       
        using namespace std;#define fir first#define sec secondtypedef long long ll;const int N=2005, M=4e5+5, INF=1e9;inline ll read(){    char c=getchar();ll x=0,f=1;    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}    return x*f;}int n, m, x, extra[N], s, t, tot, a[205][205]; pair
        
          g[205][205];char op[3];inline bool addCons(int x, int y, char c, int val) {    pair
         
           &now = g[x][y]; if(c == '=') { if(now.fir <= val && val <= now.sec) now = make_pair(val, val); else return false; } else if(c == '<') { val--; if(now.fir <= val) now.sec = min(now.sec, val); else return false; } else { val++; if(now.sec >= val) now.fir = max(now.fir, val); else return false; } return true;}struct edge{int v, c, f, ne, lower;}e[M];int cnt=1, h[N];inline int ins(int u, int v, int c, int b=0) { e[++cnt]=(edge){v, c, 0, h[u], b}; h[u]=cnt; e[++cnt]=(edge){u, 0, 0, h[v], b}; h[v]=cnt; return cnt-1;}int q[N], head, tail, vis[N], d[N], cur[N];bool bfs(int s, int t) { memset(vis, 0, sizeof(vis)); head=tail=1; q[tail++]=s; d[s]=0; vis[s]=1; while(head!=tail) { int u=q[head++]; for(int i=h[u];i;i=e[i].ne) if(!vis[e[i].v] && e[i].c>e[i].f) { vis[e[i].v]=1; d[e[i].v]=d[u]+1; q[tail++]=e[i].v; if(e[i].v == t) return true; } } return false;}int dfs(int u, int a, int t) { if(u==t || a==0) return a; int flow=0, f; for(int &i=cur[u];i;i=e[i].ne) if(d[e[i].v]==d[u]+1 && (f=dfs(e[i].v, min(a, e[i].c-e[i].f), t))>0) { flow+=f; e[i].f+=f; e[i^1].f-=f; a-=f; if(a==0) break; } if(a) d[u]=-1; return flow;}int dinic(int s, int t) { int flow=0; while(bfs(s, t)) { for(int i=0; i<=tot; i++) cur[i]=h[i]; flow+=dfs(s, INF, t); } return flow;}int main() { freopen("in","r",stdin); int T=read(); while(T--) { n=read(); m=read(); s=0; t=n+m+1; cnt=1; memset(h,0,sizeof(h)); memset(extra, 0, sizeof(extra)); for(int i=1; i<=n; i++) x=read(), ins(s, i, 0, x), extra[s]-=x, extra[i]+=x; for(int i=1; i<=m; i++) x=read(), ins(n+i, t, 0, x), extra[n+i]-=x, extra[t]+=x; for(int i=1; i<=n; i++) for(int j=1; j<=m; j++) g[i][j] = make_pair(0, INF); int cons=read(), flag=1; while(cons--) { int x=read(), y=read(); scanf("%s",op); int val=read(); if(x==0 && y==0) for(int i=1; i<=n; i++) for(int j=1; j<=m; j++) flag &= addCons(i, j, op[0], val); else if(x==0) for(int i=1; i<=n; i++) flag &= addCons(i, y, op[0], val); else if(y==0) for(int j=1; j<=m; j++) flag &= addCons(x, j, op[0], val); else flag &= addCons(x, y, op[0], val); } if(!flag) puts("IMPOSSIBLE"); else { for(int i=1; i<=n; i++) for(int j=1; j<=m; j++) { int u=i, v=j+n, b=g[i][j].fir, c=g[i][j].sec; a[i][j]=ins(u, v, c-b, b); extra[u]-=b; extra[v]+=b; } ins(t, s, INF); int ss=t+1, tt=t+2, sum=0; tot=t+2; for(int i=s; i<=t; i++) { if(extra[i]>0) ins(ss, i, extra[i]), sum+=extra[i]; if(extra[i]<0) ins(i, tt, -extra[i]); } int flow=dinic(ss, tt); if(flow != sum) puts("IMPOSSIBLE"); else { for(int i=1; i<=n; i++) for(int j=1; j<=m; j++) printf("%d%c",e[a[i][j]].f + e[a[i][j]].lower, j==m?'\n':' '); } } puts(""); }}