HDU 6118 度度熊的交易计划 (最小费用流)

Posted 2020-10-09 dwtfukgv

题意：。

析：最小费用流，建立一个超级源点 s 和汇点 t，然后从s 向每个地区边一条容量是 b，费用是a，从每个地区从 t 连一条容量为 d，费用为 -c的边，注意是 -c，然后每个地区有路的就直接连上就好，然后在增广的时候，增广的时候到正数的时候就停止。

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(i,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 double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1000 + 10;
const int maxm = 1e5 + 10;
const int mod = 30007;
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, cost;
};

struct Mcmf{
  int n, m, s, t;
  vector<Edge> edges;
  vector<int> G[maxn];
  bool inq[maxn];
  int d[maxn];
  int p[maxn];
  int a[maxn];

  void init(int n){
    this-> n = n;
    for(int i = 0; i < n; ++i)  G[i].cl;
    edges.cl;
  }

  void addEdge(int from, int to, int cap, int cost){
    edges.pb((Edge){from, to, cap, 0, cost});
    edges.pb((Edge){to, from, 0, 0, -cost});
    m = edges.sz;
    G[from].pb(m-2);
    G[to].pb(m-1);
  }

  bool bellman(int &flow, int &cost){
    ms(inq, 0);  for(int i = 0; i <= n; ++i)  d[i] = -INF;
    d[s] = 0;  inq[s] = 1;  p[s] = 0;  a[s] = INF;
    queue<int> q;
    q.push(s);
    while(!q.empty()){
      int u = q.front();  q.pop();
      inq[u] = 0;
      for(int i = 0; i < G[u].sz; ++i){
        Edge &e = edges[G[u][i]];
        if(e.cap > e.flow && d[e.to] < d[u] + e.cost){
          d[e.to] = d[u] + e.cost;
          p[e.to] = G[u][i];
          a[e.to] = min(a[u], e.cap - e.flow);
          if(!inq[e.to]){ q.push(e.to);  inq[e.to] = 1; }
        }
      }
    }
//    printf("%d %d %d\n", cost, d[t], a[t]);
    if(d[t] < 0)  return false;
    int u = t;
    while(u != s){
      edges[p[u]].flow += a[t];
      edges[p[u]^1].flow -= a[t];
      u = edges[p[u]].from;
    }
    cost += d[t] * a[t];
    flow += a[t];
    return true;
  }

  int mincost(int s, int t){
    int flow = 0, cost = 0;
    this->s = s;  this-> t = t;
    while(bellman(flow, cost));
    return cost;
  }
};

Mcmf mcmf;

int main(){
  while(scanf("%d %d", &n, &m) == 2){
    int s = 0, t = n + 1;
    mcmf.init(t + 2);
    for(int i = 1; i <= n; ++i){
      int a, b, c, d;
      scanf("%d %d %d %d", &a, &b, &c, &d);
      mcmf.addEdge(s, i, b, -a);
      mcmf.addEdge(i, t, d, c);
    }
    while(m--){
      int u, v, c;
      scanf("%d %d %d", &u, &v, &c);
      if(u == v)  continue;
      mcmf.addEdge(u, v, INF, -c);
      mcmf.addEdge(v, u, INF, -c);
    }
    printf("%d\n", mcmf.mincost(s, t));
  }
  return 0;
}