BZOJ 1070: [SCOI2007]修车(费用流)

又是一道水题= =，突然发现第一页下面好多水题，等我A了它们= =

可以发现ans=sigma(t[i]*k) k表示是第几个修，然后将m拆成m*n个点，表示第几次修，从s向n*m个点连流量为1的边，从n*m个点分别向车连权值为t[i]*k的边，然后再从车向T连边就行了（和noi的某道题一样还有GDKOI2014T2一样）

CODE：

#include<cstdio> #include<iostream> #include<cstring> #include<algorithm> #include<queue> #define maxn 650 #define maxm 180000 #define inf 0x7fffffff using namespace std; struct edges{     int to,cap,dist,next; }edge[maxm]; int s,t,next[maxn],l; int addedge(int from,int to,int cap,int dist){     l++;     edge[l*2]=(edges){to,cap,dist,next[from]};     edge[l*2+1]=(edges){from,0,-dist,next[to]};     next[from]=l*2;next[to]=l*2+1;     return 0; } bool b[maxn]; int dist[maxn],way[maxn]; queue<int> q; bool spfa(){     for (int i=1;i<=t;i++) dist[i]=inf;     memset(b,0,sizeof(b));     dist[s]=0;     q.push(s);     while (!q.empty()){         int u=q.front();q.pop();         b[u]=0;         for (int i=next[u];i;i=edge[i].next)             if (edge[i].cap&&dist[u]+edge[i].dist<dist[edge[i].to]) {                 dist[edge[i].to]=dist[u]+edge[i].dist;                 way[edge[i].to]=i;                 if (!b[edge[i].to]){                     b[edge[i].to]=1;q.push(edge[i].to);                 }             }     }     if (dist[t]==inf) return 0;     return 1; } int mcmf(){     int cost=0;     while (spfa()){         cost+=dist[t];         int x=t;         while (x!=s){             edge[way[x]].cap-=1;             edge[way[x]^1].cap+=1;             x=edge[way[x]^1].to;         }     }     return cost; } int n,m; int main(){     scanf("%d%d",&m,&n);     s=n*m+n+1;t=n*m+n+2;     for (int i=1;i<=n;i++)         for (int j=1;j<=m;j++){             int x;             scanf("%d",&x);             for (int k=1;k<=n;k++) addedge((k-1)*m+j,n*m+i,1,x*k);         }     for (int i=1;i<=n*m;i++) addedge(s,i,1,0);     for (int i=1;i<=n;i++) addedge(n*m+i,t,1,0);     printf("%.2lf\n",mcmf()*1.0/n);     return 0; }

 

转载于:https://www.cnblogs.com/New-Godess/p/4348965.html