bzoj 1070 SCOI2007 修车

    好久没写网络流了……

    一开始以为是DP，没想出来，看题解发现是网络流。

    构图蛮有意思的。

    把维修人员拆成n个点，每个分点都与那n个点连边，费用为 c[i][j] * (1..n) 这是表示修了这个车后以后的人会增加这么些费用。

    上代码：

#include <cstdio>

#include <cstring>

#include <cstdlib>

#include <iostream>

#include <algorithm>

#include <queue>

#define N 65

#define M 15

#define inf 0x7f7f7f7f

using namespace std;



int n, m, S, T;

int t[N][M], fa[N*M];

int p[N*M], next[N*M*200], v[N*M*200], f[N*M*200], c[N*M*200], bnum = -1;

int dis[N*M], vis[N*M];

queue<int> q;



void addbian(int x, int y, int fl, int co)

{

    bnum++; next[bnum] = p[x]; p[x] = bnum;

    v[bnum] = y; f[bnum] = fl; c[bnum] = co;

    bnum++; next[bnum] = p[y]; p[y] = bnum;

    v[bnum] = x; f[bnum] = 0; c[bnum] = -co;

}



bool bfs()

{

    for (int i = 1; i <= T; ++i) {dis[i] = inf; vis[i] = 0;}

    vis[S] = 1; q.push(S); dis[S] = 0; fa[S] = -1;

    while (!q.empty())

    {

        int j = q.front(); q.pop();

        int k = p[j];

        while (k != -1)

        {

            if (f[k] && dis[v[k]] > dis[j] + c[k])

            {

                fa[v[k]] = k;

                dis[v[k]] = dis[j] + c[k];

                if (!vis[v[k]])

                {

                    vis[v[k]] = 1;

                    q.push(v[k]);

                }

            }

            k = next[k];

        }

        vis[j] = 0;

    }

    if (dis[T] == inf) return false;

    else return true;

}



void dinic()

{

    int ans = 0;

    while (bfs())

    {

        int k = fa[T];

        while (k != -1)

        {

            if (!f[k])

                printf("now\n");

            f[k] --;

            ans += c[k];

            f[k^1] ++;

            k = fa[v[k^1]];

        }

    }

    printf("%.2lf\n", (double)ans/(double)n);

}



int main()

{

    scanf("%d%d", &m, &n);

    for (int i = 1; i <= n; ++i)

        for (int j = 1; j <= m; ++j)

            scanf("%d", &t[i][j]);

    S = n*m+n+1; T = S+1;

    for (int i = 1; i <= T; ++i) p[i] = -1;

    for (int i = 1; i <= n*m; ++i) addbian(i, T, 1, 0);

    for (int i = 1; i <= n; ++i) addbian(S, n*m+i, 1, 0);

    for (int i = 1; i <= n; ++i)

        for (int j = 1; j <= n*m; ++j)

            addbian(n*m+i, j, 1, t[i][(j-1)/n+1]*(j%n+1));

    dinic();

    return 0;

}