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  • 餐巾计划问题(费用流)

    拆点,建二分图,Xi表示第i天用完的餐巾,Yi表示第i天需要的餐巾,求费用流。

    //http://www.cnblogs.com/IMGavin/
    #include <iostream>
    #include <stdio.h>
    #include <cstdlib>
    #include <cstring>
    #include <queue>
    #include <vector>
    #include <map>
    #include <stack>
    #include <set>
    #include <bitset>
    #include <algorithm>
    using namespace std;
    
    typedef long long LL;
    #define gets(A) fgets(A, 1e8, stdin)
    const int INF = 0x3F3F3F3F, N = 2008, M = 2000000;
    
    const double EPS = 1e-6;
    struct Node{
    	int u, v, cap, cost;
    	int next;
    }edge[M];//有向图,u到v的容量,费用
    int tot;
    int head[N], pre[N], path[N], dis[N];
    bool inq[N];
    
    void init(){
    	tot = 0;
    	memset(head, -1, sizeof(head));
    }
    
    void add(int u, int v, int cap, int cost){
    	edge[tot].u = u;
    	edge[tot].v = v;
    	edge[tot].cap = cap;
    	edge[tot].cost = cost;
    	edge[tot].next = head[u];
    	head[u] = tot++;
    	edge[tot].u = v;
    	edge[tot].v = u;
    	edge[tot].cap = 0;
    	edge[tot].cost = -cost;
    	edge[tot].next = head[v];
    	head[v] = tot++;
    }
    
    bool SPFA(int st, int des){//计算最小费用
    	memset(inq, 0, sizeof(inq));
    	memset(dis, 0x3f, sizeof(dis));
    	queue <int> q;
    	q.push(st);
    	dis[st] = 0;
    	inq[st] = true;
    	while(!q.empty()){
    		int u = q.front();
    		q.pop();
    		inq[u] = false;
    		for(int i = head[u]; ~i; i = edge[i].next){
    			int v = edge[i].v;
    			if(edge[i].cap > 0 && dis[v] > dis[u] + edge[i].cost){
    				dis[v] = dis[u] + edge[i].cost;
    				pre[v] = u;
    				path[v] = i;
    				if(!inq[v]){
    					inq[v] = true;
    					q.push(v);
    				}
    			}
    		}
    	}
    	return dis[des] < INF;
    }
    
    int EdmondsKarp(int st, int des){//最小费用最大流
    	int mincost = 0, flow = 0;//最小费用与流量
    	while(SPFA(st, des)){
    		int f = INF;
    		for(int i = des; i != st; i = pre[i]){
                if(f > edge[path[i]].cap){
                    f = edge[path[i]].cap;
                }
    		}
    		for(int i = des; i != st; i = pre[i]){
    			edge[path[i]].cap -= f;
    			edge[path[i]^1].cap += f;
    		}
    		mincost += f * dis[des];
    		flow += f;
    	}
    	return mincost;
    }
    int main(){
    	int day, p, m, f, n, s;
    	while(cin >> day >> p >> m >> f >> n >> s){
    		init();
    		int st = 0, des = 2 * day + 1;
    		for(int i = 1; i <= day; i++){
    			int r;
    			scanf("%d", &r);
    			add(st, i, r, 0);
    			add(i + day, des, r, 0);
    			add(st, i + day, INF, p);
    			if(i + m <= day){
    				add(i, i + m + day, INF, f);
    			}
    			if(i + n <= day){
    				add(i, i + n + day, INF, s);
    			}
    			if(i < day){
    				add(i, i + 1, INF, 0);
    			}
    		}
    		printf("%d
    ", EdmondsKarp(st, des));
    	}
    	return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/IMGavin/p/6390248.html
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