1070: [SCOI2007]修车

1070: [SCOI2007]修车

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

分析：

　　每个第几次修车等的时间都不一样，当前第i个人修理的车的队列是a1,a2...ak，那么对于ak等待的时间就是前面所有车的修理时间之和。如果这样建图显然空间开不下。考虑换种建图方式，a1对于后面所有车的等待时间都会加上它的修理时间，每个修理的汽车，都会对后面所有修的车加上它的修理时间。

　　所以每个人拆成n个点，第i个点表示倒数第i个修的车，那么所有的花费都应乘以i。

代码：

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<iostream>
#include<cctype>
#include<set>
#include<vector>
#include<queue>
#include<map>
#define fi(s) freopen(s,"r",stdin);
#define fo(s) freopen(s,"w",stdout);
using namespace std;
typedef long long LL;

inline int read() {
    int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
    for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
}

const int INF = 1e9;
const int N = 1005;

struct Edge{
    int u, to, nxt, f, c;
    Edge () {} 
    Edge (int _1,int _2,int _3,int _4,int _5) { u = _1, to = _2, f = _3, c = _4, nxt = _5; }
}e[100005];
int head[N], En = 1, S, T;

namespace MaxflowMincost{
    int pre[N], dis[N], q[100005], Maxflow, Mincost; // q[N]！！！ 
    bool vis[N];
    bool spfa() {
        for (int i=1; i<=T; ++i) dis[i] = INF, vis[i] = false, pre[i] = 0;
        int L = 1, R = 0;
        q[++R] = S, dis[S] = 0, vis[S] = true, pre[S] = 0;
        while (L <= R) {
            int u = q[L ++]; vis[u] = false;
            for (int i=head[u]; i; i=e[i].nxt) {
                int v = e[i].to;
                if (dis[v] > dis[u] + e[i].c && e[i].f > 0) {
                    dis[v] = dis[u] + e[i].c;
                    pre[v] = i;
                    if (!vis[v]) q[++R] = v, vis[v] = true;
                }
            }
        }
        return dis[T] != INF;
    }
    void mcf() {
        int zf = INF;
        for (int i=T; i!=S; i=e[pre[i]].u) 
            zf = min(zf, e[pre[i]].f);
        for (int i=T; i!=S; i=e[pre[i]].u) 
            e[pre[i]].f -= zf, e[pre[i] ^ 1].f += zf;
        Maxflow += zf, Mincost += dis[T] * zf;
    }
    void solve() {
        Mincost = Maxflow = 0;
        while (spfa()) 
            mcf();
    }
}

void add_edge(int u,int v,int f,int c) {
    ++En; e[En] = Edge(u, v, f, c, head[u]); head[u] = En;
    ++En; e[En] = Edge(v, u, 0, -c, head[v]); head[v] = En;
}

int main() {
    int m = read(), n = read();
    S = n + m * n + 1, T = S + 1;
    
    for (int i=1; i<=n; ++i) add_edge(S, i, 1, 0);
    for (int i=1; i<=n; ++i) { // 第i辆车 
        for (int j=1; j<=m; ++j) { // 第j个人 
            int w = read();
            for (int k=1; k<=n; ++k) // 第k次修 
                add_edge(i, j * n + k, 1, k * w); // i+n, j*n+k !!!
        }
    }
    for (int i=n+1,lim=(n+m*n); i<=lim; ++i) add_edge(i, T, 1, 0);
    
    MaxflowMincost::solve();
    int ans = MaxflowMincost::Mincost;
    printf("%.2lf", (1.0 * ans) / (1.0 * n));
    return 0;
}