【loj】#6008. 「网络流 24 题」餐巾计划(最小费用流)

记录一个菜逼的成长。。

题目链接

#include <bits/stdc++.h>using namespace std;#define rep(i,l,r) for( int i = l; i <= r; i++ )#define rep0(i,l,r) for( int i = l; i < r; i++ )#define ALL(v) (v).begin(),(v).end()#define cl(a,b) memset(a,b,sizeof(a))#define clr clear()#define pb push_back#define mp make_pair#define fi first#define se secondconst int INF = 0x3f3f3f3f;const int MAX_V = 2000 + 10;struct edge{    int to,cap,cost,rev;    edge(){}    edge(int _to,int _cap,int _cost,int _rev):to(_to),cap(_cap),cost(_cost),rev(_rev){}};int V;                  //顶点数vector<edge>G[MAX_V];   //图的邻接表示int prevv[MAX_V],preve[MAX_V];//最短路中的前驱节点和对应的边int dist[MAX_V];        //最短距离void add(int from,int to,int cap,int cost){    G[from].push_back(edge(to,cap,cost,G[to].size()));    G[to].push_back(edge(from,0,-cost,G[from].size() - 1));}//求解从s到t流量为f的最小费用流//如果不能再增广则返回-1int min_cost_flow(int s,int t,int f){    int res = 0;    while(f > 0){        //利用Bellman-Ford算法求s到t的最短路        fill(dist,dist+V,INF);        dist[s] = 0;        bool update = true;        while(update){            update = false;            for( int v = 0; v < V; v++ ){                if(dist[v] == INF)continue;                for( int i = 0; i < G[v].size(); i++ ){                    edge &e = G[v][i];                    if(e.cap > 0 && dist[e.to] > dist[v] + e.cost){                        dist[e.to] = dist[v] + e.cost;                        prevv[e.to] = v;                        preve[e.to] = i;                        update = true;                    }                }            }        }        if(dist[t] == INF){            //不能再增广            return -1;        }        //沿s到t的最短路尽量增广        int d = f;        for( int v = t; v != s; v = prevv[v] ){            d = min(d,G[prevv[v]][preve[v]].cap);        }        f -= d;        res += d * dist[t];        for( int v = t; v != s; v = prevv[v]){            edge &e = G[prevv[v]][preve[v]];            e.cap -= d;            G[v][e.rev].cap += d;        }    }    return res;}int a[MAX_V];int main(){  int n,P,M,F,N,S;  scanf("%d%d%d%d%d%d",&n,&P,&M,&F,&N,&S);  rep(i,1,n)scanf("%d",a+i);  int s = 0,t = n + n + 1;V = t + 1;  int sum = 0;  rep(i,1,n){    sum += a[i];    if(i + M <= n)add(i,i+M+n,INF,F);    if(i + N <= n)add(i,i+N+n,INF,S);    if(i + 1 <= n)add(i,i+1,INF,0);//第i天多余的新餐巾,可以留给下一天，建一条边    add(s,i,a[i],0);    add(s,i+n,INF,P);    add(i+n,t,a[i],0);  }  printf("%d\n",min_cost_flow(s,t,sum));  return 0;}