题意：给定一个矩阵的每行的和和每列的和，以及每个格子的限制，让你求出原矩阵。
析：把行看成X，列看成Y，其实就是二分图，然后每个X到每个Y边一条边，然后加一个超级源点和汇点分别向X和Y连边，这样就形成了一个有源汇有上下界的网络，如果有最大流，那么这个矩阵就存在。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n) for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 250 + 10;
const int maxm = 1e5 + 10;
const int mod = 50007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r >= 0 && r < n && c >= 0 && c < m;
}
struct Edge{
int from, to, cap, flow;
};
struct Dinic{
int n, m, s, t;
vector<Edge> edges;
vector<int> G[maxn];
int d[maxn];
bool vis[maxn];
int cur[maxn];
void init(int n){
this-> n = n;
for(int i = 0; i < n; ++i) G[i].cl;
edges.cl;
}
void addEdge(int from, int to, int cap){
edges.pb((Edge){from, to, cap, 0});
edges.pb((Edge){to, from, 0, 0});
m = edges.sz;
G[from].pb(m-2); G[to].pb(m-1);
}
bool bfs(){
ms(vis, 0); vis[s] = 1;
d[s] = 1;
queue<int> q;
q.push(s);
while(!q.empty()){
int x = q.front(); q.pop();
for(int i = 0; i < G[x].sz; ++i){
Edge &e = edges[G[x][i]];
if(!vis[e.to] && e.cap > e.flow){
d[e.to] = d[x] + 1;
vis[e.to] = 1;
q.push(e.to);
}
}
}
return vis[t];
}
int dfs(int x, int a){
if(x == t || a == 0) return a;
int flow = 0, f;
for(int &i = cur[x]; i < G[x].sz; ++i){
Edge &e = edges[G[x][i]];
if(d[e.to] == d[x] + 1 && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0){
e.flow += f;
edges[G[x][i]^1].flow -= f;
a -= f;
flow += f;
if(a == 0) break;
}
}
return flow;
}
int maxflow(int s, int t){
this->s = s; this-> t = t;
int flow = 0;
while(bfs()){ ms(cur, 0); flow += dfs(s, INF); }
return flow;
}
};
Dinic dinic;
int in[maxn];
int down[maxn][maxn], up[maxn][maxn];
bool ok;
void judge(int i, int j, int v){
if(v < down[i][j] || v > up[i][j]) ok = false;
}
int main(){
int T; cin >> T;
while(T--){
scanf("%d %d", &n, &m);
int s = 0, t = n + m + 1;
int S = t + 1, T = S + 1;
FOR(i, 1, S) FOR(j, 1, S){
down[i][j] = -1000;
up[i][j] = 1000;
}
ms(in, 0);
dinic.init(T + 2);
for(int i = 1; i <= n; ++i){
int x; scanf("%d", &x);
in[s] -= x; in[i] += x;
dinic.addEdge(s, i, 0);
}
for(int i = 1; i <= m; ++i){
int x; scanf("%d", &x);
in[i+n] -= x; in[t] += x;
dinic.addEdge(i+n, t, 0);
}
int num; scanf("%d", &num);
char op[5]; int u, v, c;
ok = true;
while(num--){
scanf("%d %d %s %d", &u, &v, op, &c);
if(!ok) continue;
if(u && v){
if(op[0] == '='){ judge(u, v+n, c); up[u][v+n] = down[u][v+n] = c; }
else if(op[0] == '<') up[u][v+n] = min(up[u][v+n], c - 1);
else down[u][v+n] = max(down[u][v+n], c + 1);
}
else if(u){
for(int j = 1; j <= m; ++j){
if(op[0] == '='){ judge(u, j+n, c); up[u][j+n] = down[u][j+n] = c; }
else if(op[0] == '<') up[u][j+n] = min(up[u][j+n], c - 1);
else down[u][j+n] = max(down[u][j+n], c + 1);
}
}
else if(v){
for(int i = 1; i <= n; ++i){
if(op[0] == '='){ judge(i, v+n, c); up[i][v+n] = down[i][v+n] = c; }
else if(op[0] == '<') up[i][v+n] = min(up[i][v+n], c - 1);
else down[i][v+n] = max(down[i][v+n], c + 1);
}
}
else{
for(int i = 1; i <= n; ++i)
for(int j = 1; j <= m; ++j){
if(op[0] == '='){ judge(i, j+n, c); up[i][j+n] = down[i][j+n] = c; }
else if(op[0] == '<') up[i][j+n] = min(up[i][j+n], c - 1);
else down[i][j+n] = max(down[i][j+n], c + 1);
}
}
}
int ans = 0;
if(!ok){ puts("IMPOSSIBLE"); goto A; }
for(int i = 1; i <= n; ++i)
for(int j = 1; j <= m; ++j){
if(up[i][j+n] < down[i][j+n]) ok = false;
dinic.addEdge(i, j+n, up[i][j+n] - down[i][j+n]);
in[i] -= down[i][j+n];
in[j+n] += down[i][j+n];
}
dinic.addEdge(t, s, INF);
for(int i = 0; i <= t; ++i){
if(in[i] > 0) dinic.addEdge(S, i, in[i]), ans += in[i];
if(in[i] < 0) dinic.addEdge(i, T, -in[i]);
}
if(!ok || ans != dinic.maxflow(S, T)){ puts("IMPOSSIBLE"); goto A; }
for(int i = 1; i <= n; ++i){
int cur = 0;
for(int j = 0; dinic.edges[dinic.G[i][j]].to != n+1; ++j, cur = j);
for(int j = cur, cnt = 0; cnt < m; ++cnt, ++j)
printf("%d%c", dinic.edges[dinic.G[i][j]].flow + down[i][cnt+n+1], "
"[cnt+1==m]);
}
A:;
if(T) puts("");
}
return 0;
}