uva 11082（網絡流）

      題目鏈接：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=&problem=2023&mosmsg=Submission+received+with+ID+12094940

題意是已知一個矩陣的行數和列數和前i行的和還有前j列的和， 矩陣每個元素的值都介於1~20要求構造出這個矩陣，將每一行看成一個點， 每一列看成一個點， 源點和行相連容量爲該行的和， 每個列的點和匯點相連容量爲該列的和， 每個行和每個列也相連容量爲20， 行和列形成一個完全二分圖的結構， 對應的行連到對應的列的流量即爲該行該列的數值， 由於每個數要大於1所以實現時每條弧的容量還要減去某一個值， 詳見代碼。

#include <iostream>
#include <string>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <ctime>
#include <queue>
#include <stack>
#include <list>
#include <algorithm>

using namespace std;

const int N = 505;
const int M = 12005;
const int INF = 1000000000;

int res[N][N];
typedef int LL;

struct Dinic {

	struct Edge {
		int v;
		LL cap, flow;
		Edge* next, * pair;

		void init(int a, LL b, Edge* e1, Edge* e2) {
			v = a, cap = b, flow = 0, next = e1, pair = e2;
		}    
	};

	Edge* head[N], * used[N];
	Edge* it;
	int lev[N], que[N];
	Edge E[M];
	int n, s, t;
	LL maxFlow;

	void init(int n, int s, int t) {
		it = E;
		this->n = n;
		this->s = s, this->t = t;
		for (int i = 0; i < n; i++)
			head[i] = 0;
	}

	void add(int u, int v, LL c) {
		it->init(v, c, head[u], it + 1);
		head[u] = it++;
		it->init(u, 0, head[v], it - 1);
		head[v] = it++;
	}

	bool bfs() {
		for (int i = 0; i < n; lev[i++] = -1);
		lev[s] = 0;
		int st = 0, ed = 0;
		que[ed++] = s;
		while (st < ed) {
			int u = que[st++];
			for (Edge* e = head[u]; e;  e = e->next) {
				int v = e->v;
				if (lev[v] == -1 && e->cap > e->flow) {
					lev[v] = lev[u] + 1;
					que[ed++] = v;
				}
			}
		}
		return lev[t] != -1;    
	}

	LL dfs(int u, LL f) {
		if (u == t) return f;
		for (Edge* & e = used[u]; e; e = e->next) {
			int v = e->v;
			if (e->cap > e->flow && lev[v] == lev[u] + 1) {
				LL tmp = dfs(v, min(e->cap - e->flow, f));
				if (tmp > 0) {
					e->flow += tmp;
					e->pair->flow -= tmp;
					return tmp;
				}
			}
		}
		return 0;
	}

	void run() {
		maxFlow = 0;
		while (bfs()) {
			for (int i = 0; i < n; i++)
				used[i] = head[i];
			LL f = 1;
			while (f) {
				f = dfs(s, INF);
				maxFlow += f;
			}
		}
	}
	
	void solve(int r) {
		for (int u = 1; u <= r; u++) {
			for (Edge* e = head[u]; e; e = e->next) {
				if (e->v > r) {
					res[u][e->v - r] = e->flow + 1;
				}
			}	
		}		
	}
}G;

int main() {
	int i,j, r, c;
	int T;
	scanf("%d",&T);
	int test=0;

	while(T--)
	{
		scanf("%d %d",&r,&c);
		int s,t;
		s=0,t=r+c+1;
		G.init(t + 1, s, t);
		int pre=0;
		for(i=1; i<=r; i++)
		{
			int temp;
			scanf("%d",&temp);
			G.add(s,i,temp-pre-c);
			for(j=1; j<=c; j++)
			{
				G.add(i,r+j,19);
			}
			pre=temp;
		}

		pre=0;
		for(i=1; i<=c; i++)
		{
			int temp;
			scanf("%d",&temp);
			G.add(r+i,t,temp-pre-r);
			pre=temp;
		}

		printf("Matrix %d\n",++test);
		G.run();
		G.solve(r);


		for (i = 1; i <= r; i++) {
			for (j = 1; j <= c - 1; j++)
				printf("%d ", res[i][j]);
			printf("%d\n", res[i][c]);
		}
		puts("");
	}
	return 0;
}