【贪心+线段树】Codeforces 557C Arthur and Table

题面在这里

其实很简单……

暴枚最长桌脚的长度l，然后长度比l长的桌脚全部都要砍掉

长度比l短的桌脚选择代价前k小的砍掉

用线段树维护……

示例程序 ：

#include<cstdio>#include<vector>#include<algorithm>using namespace std;typedef long long ll;inline char nc(){    static char buf[100000],*p1=buf,*p2=buf;    return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;}inline int red(){    int res=0,f=1;char ch=nc();    while (ch<'0'||'9'<ch) {if (ch=='-') f=-f;ch=nc();}    while ('0'<=ch&&ch<='9') res=res*10+ch-48,ch=nc();    return res*f;}const int maxn=100005;int n,m,l[maxn];ll ans=1ll<<62,sum=0;vector<int> a[maxn];struct node{    node *l,*r;    int L,R,s;ll w;    node () {}    node (int _L,int _R):L(_L),R(_R),s(0),w(0) {}    void pushup() {s=l->s+r->s; w=l->w+r->w;}}nil,bace[maxn<<1];typedef node* P_node;P_node null,len,Rot;void init(){    nil=node(0,0);null=&nil;    null->l=null->r=null;len=bace;}P_node newnode(int L,int R){    *len=node(L,R);    len->l=len->r=null;    return len++;}P_node build(int L,int R){    P_node x=newnode(L,R);    if (L==R) return x;    int mid=L+R>>1;    x->l=build(L,mid); x->r=build(mid+1,R);    return x;}void insert(P_node x,int k,int wi){    if (x->L==x->R) {x->s+=wi;x->w=(ll)x->s*x->L;return;}    int mid=x->L+x->R>>1;    if (k<=mid) insert(x->l,k,wi);else insert(x->r,k,wi);    x->pushup();}ll query(P_node x,int k){    if (k> x->s) return 1ll<<62;    if (x->L==x->R) return (ll)x->L*k;    if (k<= x->l->s) return query(x->l,k);    return query(x->r,k- x->l->s)+ x->l->w;}int main(){    n=red();m=100000;    for (int i=1;i<=n;i++) l[i]=red();    for (int i=1,x;i<=n;i++) x=red(),sum+=x,a[l[i]].push_back(x);    for (int i=1;i<=m;i++){        int Max=0,cd=0;        for (int j : a[i])         if (j>=Max) cd=Max,Max=j;else         if (j>cd) cd=j;        if (Max) ans=min(ans,sum-Max);        if (cd) ans=min(ans,sum-Max-cd);    }    init();Rot=build(1,m);    for (int i=1,all=0;i<=m;i++){        for (int j: a[i]) sum-=j,all++;        if (all==a[i].size()&&a[i].size()>2) ans=min(ans,sum);        if (a[i].size()>1&&all>=a[i].size()*2-1) ans=min(ans,sum+query(Rot,all-a[i].size()*2+1)) ;        for (int j: a[i]) insert(Rot,j,1);    }    printf("%I64d",ans);    return 0; }