【网络流24题】【LOJ6013】负载平衡（环形纸牌均分，最小费最大流）

problem

• 一个有n个数的环
• 每次只能向相邻的数移动，移动一个数代价为1
• 求让所有数相等的最小代价

solution

• 从s向每个点连容量为库存量，费用为0的边
• 从每个点向t连容量为平均库存量，费用为0的边
• 在相邻两个点之间连容量为inf，费用为1的边
• 然后跑最小费用最大流即可

codes

#include<iostream>
#include<algorithm>
#include<queue>
#include<cstring>
using namespace std;

const int N = 110*2+10, M = 1000+10, inf = 1<<30;

//Grape
struct Edge{
    int from, to, cap, flow, cost;
}e[M];
int tot=1, head[N], Next[M];
void AddEdge(int u, int v, int w, int c){
    //正向边，初始容量w，单位费用c
    e[++tot].from = u, e[tot].to = v, e[tot].cap = w, e[tot].flow = 0, e[tot].cost = c;
    Next[tot] = head[u], head[u] = tot;
    //反向边，初始容量0，单位费用-c，与正向边成对存储
    e[++tot].from = v, e[tot].to = u, e[tot].cap = 0, e[tot].flow = 0, e[tot].cost = -c;
    Next[tot] = head[v], head[v] = tot;
}

//Cost flow
int s, t, incf[N], pre[N];
int dist[N], vis[N];
bool spfa(){
    queue<int>q;
    memset(dist,0x3f,sizeof(dist));//inf
    memset(vis,0,sizeof(vis));
    q.push(s); dist[s]=0; vis[s]=1;
    incf[s] = inf; //到s为止的增广路上各边的最小的剩余容量
    while(q.size()){
        int x = q.front(); q.pop(); vis[x] = 0;
        for(int i = head[x]; i; i = Next[i]){
            if(e[i].flow==e[i].cap)continue; //剩余容量为0，不再残量网络中，不遍历
            int y = e[i].to;
            if(dist[y]>dist[x]+e[i].cost){
                dist[y] = dist[x]+e[i].cost;
                incf[y] = min(incf[x], e[i].cap-e[i].flow);
                pre[y] = i;//记录前驱，用于找方案
                if(!vis[y])vis[y]=1, q.push(y);
            }
        }
    }
    if(dist[t] == 0x3f3f3f3f)return false;//汇点不可达，已求出最大流
    return true;
}
int maxflow(){
    int flow = 0, cost = 0;
    while(spfa()){
        int x = t;
        while(x != s){
            int i = pre[x];
            e[i].flow += incf[t];
            e[i^1].flow -= incf[t];
            x = e[i].from;
        }
        flow += incf[t];
        cost += dist[t]*incf[t];
    }
    return cost;
}

//Timu
int n, a[110], sum, ans;

int main(){
    ios::sync_with_stdio(false);
    cin>>n; for(int i = 1; i <= n; i++)cin>>a[i],sum+=a[i];  sum/=n;
    s = 0, t = n+1;
    for(int i = 1; i <= n; i++){
        AddEdge(s,i,a[i],0);
        AddEdge(i,t,sum,0);
        AddEdge(i,i+1>n?1:i+1,inf,1);
        AddEdge(i,i-1<1?n:i-1,inf,1);
    }
    cout<<maxflow()<<'\n';
    return 0;
}
  

 

  
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文章来源: gwj1314.blog.csdn.net，作者：小哈里，版权归原作者所有，如需转载，请联系作者。

原文链接：gwj1314.blog.csdn.net/article/details/80658216