题意:n个人排队买票,要把他们拆成k条队到k个窗口买,可以有队伍为空,每条队的顺序保持拆之前的顺序。如果某人和他前一个人买的票相同,就可以打八折,求最小花费。
题解:拆成k条队意味着只有[n-k,n-1]组前后关系,那么可以转成二分图最大权匹配,流的时候限制流量在[n-k,n-1]间就可以了。
#include<map>
#include<queue>
#include<string>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define MAXN 5300
#define MAXM 35000
using namespace std;struct na{
int y,z,f,ne;
};
int n,m,k,l[MAXN],r[MAXN],num=,p,ch,S,T,dis[MAXN],mi[MAXN],ro[MAXN],qi[MAXN],no,v[MAXN],R[MAXN],Ro[][],li[];
na b[MAXM];
bool bo[MAXN];
const int INF=1e9;
int mmh=,an=;
queue<int> q;
inline int min(int x,int y){return x>y?y:x;}
inline void spfa(){
register int i;
q.push(S);
bo[S]=;
for (i=;i<=no;i++) dis[i]=INF;
mi[S]=INF;dis[S]=;dis[T]=INF;
while(!q.empty()){
int k=q.front();q.pop();bo[k]=;
if (k==T) continue;
for (i=l[k];i;i=b[i].ne){
if (b[i].z>&&dis[b[i].y]>b[i].f+dis[k]){
dis[b[i].y]=b[i].f+dis[k];
mi[b[i].y]=min(mi[k],b[i].z);
ro[b[i].y]=i;
qi[b[i].y]=k;
if (!bo[b[i].y]){
bo[b[i].y]=;
q.push(b[i].y);
}
}
}
}
}
inline int read(){
p=;ch=getchar();
while (ch<''||ch>'') ch=getchar();
while (ch>=''&&ch<='') p=p*+ch-, ch=getchar();
return p;
}
inline void add(int x,int y,int z,int f){
num++;
if (l[x]==) l[x]=num;else b[r[x]].ne=num;
b[num].y=y;b[num].z=z;b[num].f=f;r[x]=num;
}
inline void in(int x,int y,int z,int f){
add(x,y,z,f);add(y,x,,-f);
}
map<string,int> ma;
char s[];
int main(){
register int i,j;
n=read();m=read();k=read();S=;no=T=n+n+;
for (i=;i<=k;i++){
scanf("%s",s);
ma[string(s)]=i;
v[i]=read();
}
for (i=;i<=n;i++) scanf("%s",s),R[i]=ma[string(s)],mmh+=v[R[i]],in(S,i,,),in(i+n,T,,);
for (i=;i<=n;i++)
for (j=i+;j<=n;j++) in(i,j+n,,R[i]==R[j]?-v[R[j]]:),Ro[i][j]=num-;
for (int f=;;f++){
spfa();
if (dis[T]==INF) break;
if (dis[T]>=&&f>=n-m) break;
an+=mi[T]*dis[T];
for (i=T;i;i=qi[i]) b[ro[i]].z-=mi[T],b[((ro[i]-)^)+].z+=mi[T];
}
printf("%lf\n",.*mmh+0.2*an);
int s=;
for (i=;i<=n;i++){
for (j=;j<i;j++)
if (b[Ro[j][i]].z==) break;
if (j==i) li[i]=++s;else li[i]=li[j];
printf("%d\n",li[i]);
}
}