UVa 11082 Matrix Decompressing (网络流)

题意：给定一个矩阵的每行每列的前缀和，矩阵的元素是1-20，求这个矩阵。

析：一个网络流题，首先先把每个点的数减1，那么元素就成了0-19，这样就是一个普通的网络流了，建立一个源点和汇点，源点向每行连一条边，

汇点向每列连一条边，每个行向每个列连一条容量为19的边，其他的边都是相应的容量。最后跑一次最大流就行了。

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>
#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 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 = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-5;
const int maxn = 500 + 10;
const int mod = 1e6;
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];
  bool vis[maxn];
  int d[maxn];
  int cur[maxn];

  void init(){
    edges.clear();
    for(int i = 0; i < maxn; ++i)  G[i].clear();
  }

  bool bfs(){
    memset(vis, 0, sizeof vis);
    queue<int> q;
    q.push(s);
    d[s] = 0;  vis[s] = true;
    while(!q.empty()){
      int x = q.front();  q.pop();
      for(int i = 0; i < G[x].size(); ++i){
        Edge &e = edges[G[x][i]];
        if(!vis[e.to] && e.cap > e.flow){
          vis[e.to] = 1;
          d[e.to] = d[x] + 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].size(); ++i){
      Edge& e = edges[G[x][i]];
      if(d[x] + 1 == d[e.to] && (f = dfs(e.to, min(a, e.cap-e.flow))) > 0){
        e.flow += f;
        edges[G[x][i]^1].flow -= f;
        flow += f;
        a -= f;
        if(!a)  break;
      }
    }
    return flow;
  }

  int maxflow(int s, int t){
    this->s = s;  this->t = t;
    int flow = 0;
    while(bfs()){
      memset(cur, 0, sizeof cur);
      flow += dfs(s, INF);
    }
    return flow;
  }

  void addEdge(int from, int to, int cap){
    edges.push_back(Edge{from, to, cap, 0});
    edges.push_back(Edge{to, from, 0, 0});
    m = edges.size();
    G[from].push_back(m-2);
    G[to].push_back(m-1);
  }
};
int a[maxn], b[maxn];
Dinic dinic;

void solve(){
  dinic.init();
  int s = 455;
  int t = 456;
  a[0] = b[0] = 0;
  for(int i = 1; i <= n; ++i)  dinic.addEdge(s, i, a[i]-a[i-1]-m);
  for(int i = n+1; i <= m+n; ++i)  dinic.addEdge(i, t, b[i-n]-b[i-1-n]-n);
  for(int i = 1; i <= n; ++i)
    for(int j = n+1; j <= m+n; ++j)
      dinic.addEdge(i, j, 19);

  dinic.maxflow(s, t);
  for(int i = 1; i <= n; ++i)
    for(int j = n+1; j <= m+n; ++j){
      for(int k = 0; k < dinic.G[i].size(); ++k){
        Edge&e = dinic.edges[dinic.G[i][k]];
        if(e.to == j){
            printf("%d%c", e.flow+1, j == m+n ? '
' : ' ');
            break;
        }
      }
    }
}

int main(){
  int T;  cin >> T;
  for(int kase = 1; kase <= T; ++kase){
    scanf("%d %d", &n, &m);
    for(int i = 1; i <= n; ++i)  scanf("%d", a+i);
    for(int i = 1; i <= m; ++i)  scanf("%d", b+i);
    printf("Matrix %d
", kase);
    solve();
    if(kase < T)  puts("");
  }
  return 0;
}