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;
}