194. Reactor Cooling

time limit per test: 0.5 sec.
memory limit per test: 65536 KB

input: standard
output: standard

The terrorist group leaded by a well known international terrorist Ben Bladen is buliding a nuclear reactor to produce plutonium for the nuclear bomb they are planning to create. Being the wicked computer genius of this group,you are responsible for developing the cooling system for the reactor.

The cooling system of the reactor consists of the number of pipes that special cooling liquid flows by. Pipes are connected at special points,called nodes,each pipe has the starting node and the end point. The liquid must flow by the pipe from its start point to its end point and not in the opposite direction.

Let the nodes be numbered from 1 to N. The cooling system must be designed so that the liquid is circulating by the pipes and the amount of the liquid coming to each node (in the unit of time) is equal to the amount of liquid leaving the node. That is,if we designate the amount of liquid going by the pipe from i-th node to j-th as f _ij,(put f _ij= 0 if there is no pipe from node i to node j),for each i the following condition must hold:


sum(j=1..N,f _ij) = sum(j=1..N,f _ji)


Each pipe has some finite capacity,therefore for each i and j connected by the pipe must be f _ij≤ c _ijwhere c _ijis the capacity of the pipe. To provide sufficient cooling,the amount of the liquid flowing by the pipe going from i-th to j-th nodes must be at least l _ij,thus it must be f _ij≥ l _ij.

Given c _ijand l _ijfor all pipes,find the amount f _ij,satisfying the conditions specified above.

Input

The first line of the input file contains the number N (1 ≤ N ≤ 200) - the number of nodes and and M — the number of pipes. The following M lines contain four integer number each - i,j,l _ijand c _ijeach. There is at most one pipe connecting any two nodes and 0 ≤ l _ij≤ c _ij≤ 10 ⁵for all pipes. No pipe connects a node to itself. If there is a pipe from i-th node to j-th,there is no pipe from j-th node to i-th.

Output

On the first line of the output file print YES if there is the way to carry out reactor cooling and NO if there is none. In the first case M integers must follow,k-th number being the amount of liquid flowing by the k-th pipe. Pipes are numbered as they are given in the input file.

Sample test(s)

Input

Test #14 61 2 1 22 3 1 23 4 1 24 1 1 21 3 1 24 2 1 2Test #24 61 2 1 32 3 1 33 4 1 34 1 1 31 3 1 34 2 1 3

Output

Test #1

NO

Test #2

YES
1
2
3
2
1

1

开始抄书吧。。。

基本的有上下界的网络流模型。

在原图的基础上构造超级源点S和超级汇点T，并添加边和修改容量。blabla...

#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 = 2e2 + 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,0});
    edges.push_back((Edge){to,from,0});
    m = edges.size();
    G[from].push_back(m-2);
    G[to].push_back(m-1);
  }

  bool BFS() {
    memset(vis,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,sizeof(cur));
      flow += DFS(s,INF);
    }
    return flow;
  }

  bool check(int n){
    for(int i = 0;i < G[0].size();i++){
        Edge& e = edges[G[0][i]];
        if(e.flow != e.cap)
            return false;
    }
    for(int i = 1;i <= n;i++){
        Edge& e = edges[G[i][0]];
        if(e.to == n+1 && e.flow != e.cap){
            return false;
        }
    }
    return true;
  }
};

Dinic g;

int xiajie[maxn*10000],s[maxn];
vector<int> task;

int main(){
    int n,m;
    while(scanf("%d%d",&n,&m) != EOF){
        g.ClearAll(n+5);
        int source = 0,sink = n+1;
        memset(s,sizeof s);
        task.clear();
        for(int i = 1;i <= m;i++){
            int u,v,l,c;
            scanf("%d%d%d%d",&u,&v,&l,&c);
            xiajie[i] = l;
            s[v] += l;
            s[u] -= l;
            g.AddEdge(u,c-l);
            task.push_back(g.edges.size()-2);
        }
        for(int i = 1;i <= n;i++){
            if(s[i] > 0){
                g.AddEdge(source,i,s[i]);
            }
            else{
                g.AddEdge(i,sink,-s[i]);
            }
        }
        g.Maxflow(source,sink);
        if(!g.check(n))
            cout << "NO" << endl;
        else{
            cout << "YES" << endl;
            for(int i = 0;i < task.size();i++){
                cout << xiajie[i+1]+g.edges[task[i]].flow << endl;
            }
        }
    }
    return 0;
}