[SCOI 2007] 修车

[题目链接]

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

[算法]

首先 ， 我们发现 ， 在倒数第i个修车会对答案产生i * k的贡献

将每辆车建一个点 ， 每名技术人员建n个点 ，将车与技术人员连边 ， 第i个技术人员的第j个点与第k辆车连边表示k是i修的倒数第j辆车

然后在这张图上求最小费用最大流 ， 即可

时间复杂度 ： O(Costflow(NM , N ^ 2M))

[代码]

#include<bits/stdc++.h>
using namespace std;
#define MAXN 550
const int INF = 1e9;

struct edge
{
        int to , w , cost , nxt;
} e[MAXN * MAXN * ];

int n , m , tot , ans , S , T;
int a[MAXN][MAXN];
int head[MAXN * MAXN] , dist[MAXN * MAXN] , pre[MAXN * MAXN] , incf[MAXN * MAXN];
bool inq[MAXN * MAXN];

template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void read(T &x)
{
    T f = ; x = ;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << ) + (x << ) + c - '';
    x *= f;
}
inline void addedge(int u , int v , int w , int cost)
{
        ++tot;
        e[tot] = (edge){v , w , cost , head[u]};
        head[u] = tot;
        ++tot;
        e[tot] = (edge){u ,  , -cost , head[v]};
        head[v] = tot;
}
inline bool spfa()
{
        int l , r;
        static int q[MAXN * MAXN];
        for (int i = ; i <= T; i++)
        {
                pre[i] = ;
                dist[i] = INF;
                incf[i] = INF;
                inq[i] = false;
        }
        q[l = r = ] = S;
        dist[S] = ;
        inq[S] = true;
        while (l <= r)
        {
                int cur = q[l++];
                inq[cur] = false;
                for (int i = head[cur]; i; i = e[i].nxt)
                {
                        int v = e[i].to , w = e[i].w , cost = e[i].cost;
                        if (w >  && dist[cur] + cost < dist[v])
                        {
                                dist[v] = dist[cur] + cost;
                                incf[v] = min(incf[cur] , w);
                                pre[v] = i;
                                if (!inq[v])
                                {
                                        q[++r] = v;
                                        inq[v] = true;
                                }
                        }
                }
        }
        if (dist[T] != INF) return true;
        else return false;
}
inline void update()
{
        int now = T , pos;
        while (now != S)
        {
                pos = pre[now];
                e[pos].w -= incf[T];
                e[pos ^ ].w += incf[T];
                now = e[pos ^ ].to;
        }
        ans += dist[T] * incf[T];
}

int main()
{

        read(m); read(n);
        for (int i = ; i <= n; i++)
        {
                for (int j = ; j <= m; j++)
                {
                        read(a[i][j]);
                }
        }
        tot = ;
        S = n * m + n +  , T = S + ;
        for (int i = ; i <= n; i++) addedge(S , i ,  , );
        for (int i = ; i <= n; i++)
        {
                for (int j = ; j <= m; j++)
                {
                        for (int k = ; k <= n; k++)
                        {
                                addedge(i , n + (j - ) * n + k ,  , a[i][j] * k);
                        }
                }
        }
        for (int i = ; i <= n * m; i++) addedge(i + n , T ,  , );
        while (spfa()) update();
        printf("%.2lf\n" , 1.0 * ans / n);

        return ;

}