Bzoj 1070: [SCOI2007] Repair car

Keyword: How to build a map? How to transform the network stream;

If you think about network flow, you know it's a cost stream.

Then how to build the map;

Here are a few notes:

1. Join the source point, Meeting Point

2. Add (0,x,1,0) to each source point and customer connection

3: The question is how to connect with customers and skilled workers:

First we know that each customer can practice a day cost (I,J) line with each worker, but there is a waiting time

The enumeration wait time is the edge of the connection K*cost (I,J),

But we're not customers. Even a worker is connected to a meeting point, because we know that a worker can only do 1 k*cost (I,J), for example, x this skilled worker can only be the first to even Y, the second even z;

So a clever idea is: to form n+1-n*m+n A new point, let 1-n a customer with these points. (In fact, 1:30 will not say)

In short, we want to be the first customer to choose the younger brother I technical worker's words and only when the contact once.

#include <iostream> #include <algorithm> #include <cstdio> #include <cstring> #include < string> #include <cstdlib> #include <vector> #include <queue> #include <map> #define N 100000# Define INF 0x3f3f3fusing namespace std;struct edge{int v,next,flow,cost,cap;}    E[n<<2];int head[n],tot;int pre[n],dis[n];int vis[n];int n,m;int cnt;void init () {tot=0; memset (head,-1,sizeof (Head));}    void Add (int u,int v,int cap,int cost) {e[tot].v=v;    E[tot].cap=cap;    E[tot].cost=cost;    e[tot].flow=0;    E[tot].next=head[u];;    head[u]=tot++;    E[tot].v=u;    E[tot].cap=0;    E[tot].cost=-cost;    e[tot].flow=0;    E[TOT].NEXT=HEAD[V]; head[v]=tot++;}    int SPFA (int s,int t) {queue<int>q;        for (int i=0;i<=cnt;i++) {dis[i]=inf;        vis[i]=0;    Pre[i]=-1;    } dis[s]=0;    Vis[s]=1;    Q.push (s);        while (!q.empty ()) {int U=q.front ();        Q.pop ();        vis[u]=0; for (int i=head[u];i!=-1;i=e[I].next) {int v=e[i].v;                if (e[i].cap>e[i].flow&&dis[v]>dis[u]+e[i].cost) {dis[v]=dis[u]+e[i].cost;                Pre[v]=i;                    if (!vis[v]) {vis[v]=1;                Q.push (v); }}}} return pre[t]!=-1;}        int mincost (int s,int t,int &cost) {int flow=0;    cost=0;        while (SPFA (s,t)) {int min=inf;        for (int i=pre[t];i!=-1;i=pre[e[i^1].v]) min=min (Min,e[i].cap-e[i].flow);            for (int i=pre[t];i!=-1;i=pre[e[i^1].v]) {e[i].flow+=min;            E[i^1].flow-=min;        Cost+=e[i].cost*min;    } flow+=min; } return flow;}   int Mp[123][123];int Main () {init ();   scanf ("%d%d", &m,&n);   int s=0;   for (int i=1;i<=n;i++) for (int j=1;j<=m;j++) scanf ("%d", &mp[i][j]);   for (int i=1;i<=n;i++) Add (s,i,1,0);   Cnt=n; for (int i=1;i<=m;i++)//mainly here for (int k=1;k<=n;k++) {++cnt;   for (int j=1;j<=n;j++) Add (J,cnt,1,k*mp[j][i]);   } for (int i=n+1;i<=cnt;i++) Add (i,cnt+1,1,0);   cnt++;   int t=cnt;   int ans;   int Flow=mincost (S,t,ans);   printf ("%d\n", ans);   printf ("%.2lf\n", (double) ans/n); return 0;}

　　

Bzoj 1070: [SCOI2007] Repair car