题目链接：https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=0&problem=2023&mosmsg=Submission+received+with+ID+26557955

建立一个二分图，每行对应一个 \(X\) 结点，每列对应一个 \(Y\) 结点，从 \(s\) 向 \(X_I\) 连容量为 \(r[i] - C\)，从 \(Y_i\) 向 \(t\) 连容量为 \(c[i] - R\),
每个结点 (X_i,Y_j), 从 \(X_i\) 向 \(Y_j\) 连容量为 \(19\)，最大流每条边对应的流量就是矩阵的元素大小

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const int maxn = 405;
const int INF = 1000000007;

int T, n, m, ans;
int R[maxn], C[maxn], r[maxn], c[maxn];
int id[maxn][maxn];

int h[maxn], cnt = 1;
struct E{
	int to, next, cap;
}e[maxn << 1];
void add(int u, int v, int c){
	e[++cnt].to = v;
	e[cnt].cap = c;
	e[cnt].next = h[u];
	h[u] = cnt;
}

int d[maxn];
queue<int> q;
int s, t;

int bfs(){
	while(!q.empty()) q.pop();
	memset(d,0,sizeof(d));
	d[s] = 1;
	q.push(s);
	while(!q.empty()){
		int u = q.front(); q.pop();
		for(int i = h[u] ; i != -1 ; i = e[i].next){
			int v = e[i].to;
			if(!d[v] && e[i].cap){
				d[v] = d[u] + 1;
				q.push(v);
			}
		}
	}
	return d[t];
} 

int dfs(int u,int lim){
	if(u==t) return lim;
	int f=0,tmp;
	for(int i=h[u];i!=-1;i=e[i].next){
		int v=e[i].to;
		if(d[v]==d[u]+1&&e[i].cap&&(tmp=dfs(v,min(lim,e[i].cap)))){
			e[i].cap-=tmp; e[i^1].cap+=tmp;
			f+=tmp; lim-=tmp;
			if(!lim) return f;
		}
	}
	if(lim) d[u]=0;
	return f;
}

void dinic(){
	ans=0;
	while(bfs()){
		ans+=dfs(s, INF);
	}
}
 
ll read(){ ll s = 0, f = 1; char ch = getchar(); while(ch < ‘0‘ || ch > ‘9‘){ if(ch == ‘-‘) f = -1; ch = getchar(); } while(ch >= ‘0‘ && ch <= ‘9‘){ s = s * 10 + ch - ‘0‘; ch = getchar(); } return s * f; }

int main(){
	int kase = 0;
	T = read();
	while(T--){
		memset(h, -1, sizeof(h)); cnt = 1;
		
		n = read(), m = read();
		for(int i = 1 ; i <= n ; ++i){
			R[i] = read();
		}
		for(int i = 1 ; i <= n ; ++i){
			r[i] = R[i] - R[i - 1];
		}
		
		for(int i = 1 ; i <= m ; ++i){
			C[i] = read();
		}
		for(int i = 1 ; i <= m ; ++i){
			c[i] = C[i] - C[i - 1];
		}
		
		s = n + m + 1, t = n + m + 2;
		for(int i = 1 ; i <= n ; ++i){ // 源点 -> 行 
			add(s, i, r[i] - m);
			add(i, s, 0);
		}
		
		for(int i = 1 ; i <= m ; ++i){ // 列 -> 汇点 
			add(n + i, t, c[i] - n);
			add(t, n + 1, 0);
		}
		
		for(int i = 1 ; i <= n ; ++i){ // (i,j) 
			for(int j = 1 ; j <= m ; ++j){
				add(i, n + j, 19);
				id[i][j] = cnt;
				add(n + j, i, 0);
			}
		}
		
//		for(int i = 1 ; i <= n ; ++i){ // (i,j) 
//			for(int j = 1 ; j <= m ; ++j){
//				printf("%d ", id[i][j]);
//			} printf("\n");
//		}
		
		dinic();
		
		printf("Matrix %d\n", ++kase);
		for(int i = 1 ; i <= n ; ++i){
			for(int j = 1 ; j <= m ; ++j){
				printf("%d ", 19 - e[id[i][j]].cap + 1);
			} printf("\n");
		}
	}
	
	return 0;
}

UVa 11082 - Matrix Decompressing (网络流)