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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\n", EdmondsKarp(st, des)); } return 0; }
餐巾计划问题(费用流)
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