BZOJ 1070: [SCOI2007]修车（费用流）

http://www.lydsy.com/JudgeOnline/problem.php?id=1070

题意：

思路：

神奇的构图。

因为排在后面的人需要等待前面的车修好，这里将每个技术人员拆成n个点，第k个点表示该技术人员倒数第k的顺序来修理该车，此时它的时间对于答案的贡献就是kw。

最后跑一遍最小费用流。

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<vector>
#include<stack>
#include<queue>
#include<cmath>
#include<map>
#include<set>
using namespace std;
typedef long long ll;
typedef pair<int,int> pll;
const int INF = 0x3f3f3f3f;
const int maxn = 1000 + 5;

int n,m;

struct Edge
{
    int from, to, cap, flow, cost;
    Edge(int u, int v, int c, int f, int w) :from(u), to(v), cap(c), flow(f), cost(w) {}
};

struct MCMF
{
    int n, m;
    vector<Edge> edges;
    vector<int> G[maxn];
    int inq[maxn];
    int d[maxn];
    int p[maxn];
    int a[maxn];

    void init(int n)
    {
        this->n = n;
        for (int i = 0; i<n; i++) G[i].clear();
        edges.clear();
    }

    void AddEdge(int from, int to, int cap, int cost)
    {
        edges.push_back(Edge(from, to, cap, 0, cost));
        edges.push_back(Edge(to, from, 0, 0, -cost));
        m = edges.size();
        G[from].push_back(m - 2);
        G[to].push_back(m - 1);
    }

    bool BellmanFord(int s, int t, int &flow, int & cost)
    {
        for (int i = 0; i<n; i++) d[i] = INF;
        memset(inq, 0, sizeof(inq));
        d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = INF;

        queue<int> Q;
        Q.push(s);
        while (!Q.empty()){
            int u = Q.front(); Q.pop();
            inq[u] = 0;
            for (int i = 0; i<G[u].size(); i++){
                Edge& e = edges[G[u][i]];
                if (e.cap>e.flow && d[e.to]>d[u] + e.cost){
                    d[e.to] = d[u] + e.cost;
                    p[e.to] = G[u][i];
                    a[e.to] = min(a[u], e.cap - e.flow);
                    if (!inq[e.to]) { Q.push(e.to); inq[e.to] = 1; }
                }
            }
        }

        if (d[t] == INF) return false;
        flow += a[t];
        cost += d[t] * a[t];
        for (int u = t; u != s; u = edges[p[u]].from)
        {
            edges[p[u]].flow += a[t];
            edges[p[u] ^ 1].flow -= a[t];
        }
        return true;
    }

    int MincostMaxdflow(int s, int t){
        int flow = 0, cost = 0;
        while (BellmanFord(s, t, flow, cost));
        return cost;
    }
}t;


int main()
{
    //freopen("in.txt","r",stdin);
    while(~scanf("%d%d",&m,&n))
    {
        int src=0,dst=n*m+n+1;
        t.init(dst+1);
        for(int i=1;i<=n;i++)
        {
            t.AddEdge(src,i,1,0);
            for(int j=1;j<=m;j++)
            {
                int x; scanf("%d",&x);
                for(int k=1;k<=n;k++)
                    t.AddEdge(i,j*n+k,1,k*x);
            }
        }
        for(int j=n+1;j<=m*n+n;j++)
            t.AddEdge(j,dst,1,0);
        int ans = t.MincostMaxdflow(src,dst);
        printf("%.2f
",(double)ans/n);
    }
    return 0;
}