BZOJ 2424: [HAOI2010]訂貨 費用流

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2424: [HAOI2010]訂貨

Time Limit: 1 Sec  

Memory Limit: 256 MB

題目串連http://www.lydsy.com/JudgeOnline/problem.php?id=2424
Description

某公司估計市場在第i個月對某產品的需求量為Ui，已知在第i月該產品的訂貨單價為di，上個月月底未銷完的單位產品要付存貯費用m，假定第一月月初的庫存量為零，第n月月底的庫存量也為零，問如何安排這n個月訂購計劃，才能使成本最低？每月月初訂購，訂購後產品立即到貨，進庫並供應市場，於當月被售掉則不必付存貯費。假設倉庫容量為S。

Input第1行：n, m, S (0<=n<=50, 0<=m<=10, 0<=S<=10000)第2行：U1 , U2 , ... , Ui , ... , Un (0<=Ui<=10000)第3行：d1 , d2 , ..., di , ... , dn (0<=di<=100)Output只有1行，一個整數，代表最低成本Sample Input

3 1 1000
2 4 8
1 2 4

Sample Output34HINT 

題意

 

題解：

費用流，S--月--T

                 |

　　　　  下個月

代碼：

#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <cmath>using namespace std;const int MAXN = 10000;const int MAXM = 100000;const int INF = 0x3f3f3f3f;struct Edge{    int to, next, cap, flow, cost;    int x, y;} edge[MAXM],HH[MAXN],MM[MAXN];int head[MAXN],tol;int pre[MAXN],dis[MAXN];bool vis[MAXN];int N, M;char map[MAXN][MAXN];void init(){    N = MAXN;    tol = 0;    memset(head, -1, sizeof(head));}void addedge(int u, int v, int cap, int cost)//左端點，右端點，容量，花費{    edge[tol]. to = v;    edge[tol]. cap = cap;    edge[tol]. cost = cost;    edge[tol]. flow = 0;    edge[tol]. next = head[u];    head[u] = tol++;    edge[tol]. to = u;    edge[tol]. cap = 0;    edge[tol]. cost = -cost;    edge[tol]. flow = 0;    edge[tol]. next = head[v];    head[v] = tol++;}bool spfa(int s, int t){    queue<int>q;    for(int i = 0; i < N; i++)    {        dis[i] = INF;        vis[i] = false;        pre[i] = -1;    }    dis[s] = 0;    vis[s] = true;    q.push(s);    while(!q.empty())    {        int u = q.front();        q.pop();        vis[u] = false;        for(int i = head[u]; i != -1; i = edge[i]. next)        {            int v = edge[i]. to;            if(edge[i]. cap > edge[i]. flow &&                    dis[v] > dis[u] + edge[i]. cost )            {                dis[v] = dis[u] + edge[i]. cost;                pre[v] = i;                if(!vis[v])                {                    vis[v] = true;                    q.push(v);                }            }        }    }    if(pre[t] == -1) return false;    else return true;}//返回的是最大流， cost存的是最小費用int minCostMaxflow(int s, int t, int &cost){    int flow = 0;    cost = 0;    while(spfa(s,t))    {        int Min = INF;        for(int i = pre[t]; i != -1; i = pre[edge[i^1]. to])        {            if(Min > edge[i]. cap - edge[i]. flow)                Min = edge[i]. cap - edge[i]. flow;        }        for(int i = pre[t]; i != -1; i = pre[edge[i^1]. to])        {            edge[i]. flow += Min;            edge[i^1]. flow -= Min;            cost += edge[i]. cost * Min;        }        flow += Min;    }    return flow;}int main(){    int n, m, ss;    scanf("%d%d%d",&n,&m,&ss);    int ch = 0, cm = 0;    init();//注意    int beg = 0;//超級起點    int end = 3000;//超級匯點    for(int i=1;i<=n;i++)    {        int x;scanf("%d",&x);        addedge(i,end,x,0);    }    for(int i=1;i<=n;i++)    {        int x;scanf("%d",&x);        addedge(0,i,9999999,x);    }    for(int i=1;i<n;i++)addedge(i,i+1,ss,m);    int ans = 0;    minCostMaxflow(beg,end,ans);    printf("%d\n",ans);}

 

BZOJ 2424: [HAOI2010]訂貨 費用流