[bzoj]1520: [POI2006]Szk-Schools

对于每座学校，都将它与可更换的值连上一条容量为1，权值为c*|a-m|的边。

然后跑一遍费用流就行了。

#include <cstdio>#include <queue>#include <cstring>#include <algorithm>using namespace std;const int M=1000005,N=2005,Inf=1<<29;int n,m,a,b,c;struct edge{int nxt,to,ro,c,flow;}e[M];int s,t;int tot,head[N];inline void add(int u,int v,int ro,int c){e[tot].to=v,e[tot].ro=ro,e[tot].c=c,e[tot].nxt=head[u],head[u]=tot,tot++;e[tot].to=u,e[tot].ro=0,e[tot].c=-c,e[tot].nxt=head[v],head[v]=tot,tot++;}bool vis[N];int dis[N],prp[N],prd[N],qaq[N];inline bool spfa(int s,int t){register int i,j;for (i=0;i<=n+n+1;i++) dis[i]=Inf;vis[s]=1;dis[s]=0;prp[s]=prd[s]=1,qaq[s]=Inf;queue<int> q;q.push(s);while (!q.empty()){int x=q.front();q.pop(); vis[x]=false;for (i=head[x];i!=-1;i=e[i].nxt){if (e[i].ro>e[i].flow&&dis[x]+e[i].c<dis[e[i].to]){dis[e[i].to]=dis[x]+e[i].c;prp[e[i].to]=i;qaq[e[i].to]=min(qaq[x],e[i].ro-e[i].flow); prd[e[i].to]=x;if (!vis[e[i].to]) q.push(e[i].to),vis[e[i].to]=1;;}}}return dis[t]==Inf?0:1;}inline void MCMF(int s,int t){int flow=0,cost=0;while (spfa(s,t)){flow+=qaq[t];cost+=dis[t]*qaq[t];for (int i=t;i!=s;i=prd[i]){e[prp[i]].flow+=qaq[t];e[prp[i]^1].flow-=qaq[t];//printf("%d ",i);}//printf("\n");}if (flow<n) printf("NIE");else printf("%d",cost);}int main(){register int i,j;scanf("%d",&n);memset(head,-1,sizeof(head));s=0,t=n+n+1;for (i=1;i<=n;i++){add(s,i,1,0);add(i+n,t,1,0);scanf("%d %d %d %d",&m,&a,&b,&c);for (j=a;j<=b;j++)add(i,n+j,1,c*abs(m-j));}MCMF(s,t);}