sgu 326. Perspective （最大流）

326. Perspective

Time limit per test: 0.25 second(s)
Memory limit: 65536 kilobytes

input: standard
output: standard

Breaking news! A Russian billionaire has bought a yet undisclosed NBA team. He's planning to invest huge effort and money into making that team the best. And in fact he's been very specific about the expected result: the first place.

Being his advisor, you need to determine whether it's possible for your team to finish first in its division or not.

More formally, the NBA regular season is organized as follows: all teams play some games, in each game one team wins and one team loses. Teams are grouped into divisions, some games are between the teams in the same division, and some are between the teams in different divisions.

Given the current score and the total number of remaining games for each team of your division, and the number of remaining games between each pair of teams in your division, determine if it's possible for your team to score at least as much wins as any other team in your division.

Input

The first line of input contains  N  (2 ≤  N  ≤ 20) — the number of teams in your division. They are numbered from 1 to  N , your team has number 1.

The second line of input contains  N  integers  w ₁ ,  w ₂ ,...,  w _N , where  w _i  is the total number of games that  i ^th  team has won to the moment.

The third line of input contains  N  integers  r ₁ ,  r ₂ ,...,  r _N , where  r _i  is the total number of remaining games for the  i ^th  team (including the games inside the division).

The next  N  lines contain  N  integers each. The  j ^th  integer in the  i ^th  line of those contains  a _ij  — the number of games remaining between teams  i  and  j . It is always true that  a _ij =a _ji  and  a _ii =0, for all  i   a _i1  +  a _i2  +... +  a _iN  ≤  r _i .

All the numbers in input are non-negative and don't exceed 10\,000.

Output

On the only line of output, print "

YES

" (without quotes) if it's possible for the team 1 to score at least as much wins as any other team of its division, and "

NO

" (without quotes) otherwise.

Example(s)

sample input	sample output
3 1 2 2 1 1 1 0 0 0 0 0 0 0 0 0	YES

sample input	sample output
3 1 2 2 1 1 1 0 0 0 0 0 1 0 1 0	NO

类似之前做过的一道用网络流解构造性问题方案的题目。竟然还是wa了好久，重新换一种编号方式就过了。。。以后做网络流的题目一定要把编号想的特别清楚。

#include<cstdio>
#include<map>
#include<queue>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<list>
#include<set>
#include<cmath>
using namespace std;
const int maxn = 1e3 + 5;
const int INF = 1e9;
const double eps = 1e-6;
typedef unsigned long long ULL;
typedef long long LL;
typedef pair<int, int> P;
#define fi first
#define se second

struct Edge {
  int from, to, cap, flow;
};

struct Dinic {
  int n, m, s, t;
  vector<Edge> edges;    // 边数的两倍
  vector<int> G[maxn];   // 邻接表，G[i][j]表示结点i的第j条边在e数组中的序号
  bool vis[maxn];        // BFS使用
  int d[maxn];           // 从起点到i的距离
  int cur[maxn];         // 当前弧指针

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

  void ClearFlow() {
    for(int i = 0; i < edges.size(); i++) edges[i].flow = 0;
  }

  void AddEdge(int from, int to, int cap) {
    //cout << from << ' ' << to << ' ' << cap << endl;
    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);
  }

  bool BFS() {
    memset(vis, 0, sizeof(vis));
    queue<int> Q;
    Q.push(s);
    vis[s] = 1;
    d[s] = 0;
    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 == 0) 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;
  }
};

Dinic g;
int won[maxn];
int match[maxn][maxn];

int main(){
    int n;
    while(scanf("%d", &n) != EOF){
        bool ans = true;
        for(int i = 1;i <= n;i++)
            cin >> won[i];
        for(int i = 1;i <= n;i++){
            int x;
            cin >> x;
            if(i == 1)
                won[i] += x;
            else{
                if(won[i] > won[1])
                    ans = false;
            }
        }
        for(int i = 1;i <= n;i++){
            for(int j = 1;j <= n;j++){
                cin >> match[i][j];
            }
        }
        int source = 0, sink = n*n+n+1;
        g.ClearAll(sink + 5);
        int cnt = 1;
        int total = 0;
        for(int i = 2;i <= n;i++){
            for(int j = i+1;j <= n;j++){
                g.AddEdge(source, cnt, match[i][j]);
                g.AddEdge(cnt, n*n+i, INF);
                g.AddEdge(cnt, n*n+j, INF);
                cnt++;
                total += match[i][j];
            }
        }
        for(int i = 2;i <= n;i++){
            g.AddEdge(n*n+i, sink, won[1]-won[i]);
        }
        if(g.Maxflow(source, sink) != total)
            ans = false;
        if(ans)
            puts("YES");
        else
            puts("NO");
    }
    return 0;
}