BZOJ 1070 [SCOI2007]修车 (费用流)

计算出每一辆车的离开的时间有些难。
可以考虑每一辆车对时间造成的贡献。
首先把每一个工作人员拆成n个点。
重点来了：第i个点代表这个人修的车中倒数第i辆
由于我们无法确定是该人修多少辆车。
我们反过来设就没有问题。
然后将每辆车向这些点连边，流量为1费用为**F(i,j)*i**
源点连向车，工作人员连向汇点。
跑最小费用最大流。

/*header*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#include <map>
#include <queue>
#define gc getchar()
#define pc putchar
#define ll long long
#define mk make_pair
#define fi first
#define se second
using namespace std;
inline int gi() {
  int x = 0,f = 1;char c = gc;
  while(c < '0' || c > '9') {if(c == '-')f = -1;c = gc;}
  while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = gc;}return x * f;
}

const int maxN = 5000 + 7;
const int maxM = 100000 + 7 ;
const int inf = 0x3f3f3f3f;
int n,m,s,t;
namespace FF {
    int ans,maxflow,head[maxN];
    struct Node {
        int u,v,flow,spend,nex;
    } Map[maxM];

    int dis[maxN],vis[maxN],num,path[maxN];

    void init() {
        s = n * m + n + 1;
        t = s + 1;
        num = -1;
        memset(head,-1,sizeof(head));
        return;
    }

    void add_Node(int u,int v,int w,int spend) {
        Map[++ num] = (Node) {u , v, w, spend, head[u]};head[u] = num;
        Map[++ num] = (Node) {v , u, 0, -spend, head[v]};head[v] = num;
        return ;
    }

    bool spfa() {
        queue<int>q;
        q.push(s);
        memset(dis,0x3f,sizeof(dis));
        memset(path,0,sizeof(path));
        dis[s] = 0;
        vis[s] = true;
        while(!q.empty()) {
            int p = q.front();
            q.pop();
            vis[p] = false;
            for(int i = head[p]; i != -1; i = Map[i].nex) {
                int v = Map[i].v;
                if(dis[v] > dis[p] + Map[i].spend && Map[i].flow) {
                    dis[v] = dis[p] + Map[i].spend;
                    path[v] = i;
                    if(!vis[v]) {
                        q.push(v);
                        vis[v] = true;
                    }
                }
            }
        }
        if(dis[t] == 0x3f3f3f3f) return false;
        return true;
    }
    int min(int a,int b) {
        return a > b ? b : a ;
    }
    void f() {
        int mn = 0x7fffffff;
        for(int i = t; i != s; i = Map[path[i]].u)
            mn = min(mn,Map[path[i]].flow);
        ans += mn;

        for(int i = t; i != s; i = Map[path[i]].u) {
            Map[path[i]].flow -= mn;
            Map[path[i] ^ 1].flow += mn;
            maxflow += mn * Map[path[i]].spend;
        }
    }
    
    void EK() {
        while(spfa())
            f();
        double anss = 1.0 * maxflow;
        printf("%.2lf",anss / n);
        return ;
    }
}

int a[100][700];

int main() {
    m = gi();n = gi();
    FF::init();
    for(int i = 1;i <= n;++ i) for(int j = 1;j <= m;++ j) a[j][i] = gi();
    for(int i = 1;i <= m;++ i) {
        for(int j = 1;j <= n;++ j) {
            for(int k = 1;k <= n;++ k) {
                FF::add_Node(n * m + k,(i - 1) * n + j, 1,a[i][k] * j);
            }
        }
    }
    for(int i = 1;i <= n;++ i) FF::add_Node(s , n * m + i, 1, 0);
        for(int i = 1;i <= m;++ i) {
        for(int j = 1;j <= n;++ j) {
                FF::add_Node((i - 1) * n + j, t, 1,0);
        }
    }
    FF::EK();
    return 0;
}

标签：mn,Map,include,int,dis,1070,path,SCOI2007,BZOJ
来源： https://www.cnblogs.com/gzygzy/p/10358942.html