SGU 194 Reactor Cooling (有容量和下界的可行流)

题意：给定上一个有容量和下界的网络，让你求出一组可行解。

析：先建立一个超级源点 s 和汇点 t ，然后在输入时记录到每个结点的下界的和，建边的时候就建立c - b的最后再建立 s 和 t ， 在建立时，如果 i 结点的输入的大于输出的，那么就是从 s 建立一条边，否则 i 与 t 建立，然后跑一次最大流，就OK了，注意求出的流量是没有下界，再加上下界的就好了。

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>
#include <list>
#include <assert.h>
#include <bitset>
#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 fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(x,n)  for(int i = (x); i < (n); ++i)
#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 LL LNF = 1e16;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 200 + 50;
const int mod = 1000;
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(int n){
    this-> n = n;
    edges.clear();
    for(int i = 0; i < n; ++i)  G[i].clear();
  }

  void addEdge(int from, int to, LL 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);
  }

  bool bfs(){
    memset(vis, 0, sizeof vis);
    queue<int> q;
    q.push(s);
    d[s] = 0;
    vis[s] = 1;
    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 dinic;
int in[maxn*maxn], out[maxn*maxn];
int B[maxn*maxn];

int main(){
  scanf("%d %d", &n, &m);
  int s = 0, t = n + 1;
  for(int i = 0; i < m; ++i){
    int u, v, b, c;
    scanf("%d %d %d %d", &u, &v, &b, &c);
    dinic.addEdge(u, v, c - b);
    B[i] = b;
    in[v] += b;  out[u] += b;
  }
  int ans = 0;
  for(int i = 1; i <= n; ++i){
    int c = in[i] - out[i];
    if(c > 0)  dinic.addEdge(s, i, c), ans += c;
    else dinic.addEdge(i, t, -c);
  }
  if(dinic.maxFlow(s, t) != ans){ puts("NO");  return 0; }
  puts("YES");
  for(int i = 0; i < m; ++i)
    printf("%d\n", dinic.edges[i<<1].flow + B[i]);
  return 0;
}