USACO 2018JAN Silver

A.lifeguards（线段树+枚举）
枚举删掉哪条线段，离散化+线段树维护线段覆盖长度qaq
B.rental（贪心+二分+枚举）
枚举留几头牛，贪心算收益qaq
C.mootube（并查集+暴力）
类似最大生成树的并查集+暴力处理出dis，然后暴力回答询问qaq

A

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
#define N 100010
inline int read(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();return x*f;
}
int n,a[N<<1],L[N],R[N],num=0,m,ans=0;
struct node{int len,covlen,cov;
}tr[N<<4];
inline void pushup(int p){if(tr[p].cov) return;tr[p].covlen=tr[p<<1].covlen+tr[p<<1|1].covlen;
}
inline void build(int p,int l,int r){if(l==r){tr[p].len=a[l+1]-a[l];return;}int mid=l+r>>1;build(p<<1,l,mid);build(p<<1|1,mid+1,r);tr[p].len=tr[p<<1].len+tr[p<<1|1].len;
}
inline void cover(int p,int l,int r,int x,int y){if(x<=l&&r<=y){tr[p].cov++;if(tr[p].cov==1) tr[p].covlen=tr[p].len;return;}int mid=l+r>>1;if(x<=mid) cover(p<<1,l,mid,x,y);if(y>mid) cover(p<<1|1,mid+1,r,x,y);pushup(p);
}
inline void del(int p,int l,int r,int x,int y){if(x<=l&&r<=y){tr[p].cov--;if(tr[p].cov==0){if(l==r) tr[p].covlen=0;else pushup(p);}return;}int mid=l+r>>1;if(x<=mid) del(p<<1,l,mid,x,y);if(y>mid) del(p<<1|1,mid+1,r,x,y);pushup(p);
}
int main(){freopen("lifeguards.in","r",stdin);freopen("lifeguards.out","w",stdout);n=read();for(int i=1;i<=n;++i){a[++num]=L[i]=read();a[++num]=R[i]=read();}sort(a+1,a+num+1);m=unique(a+1,a+num+1)-a-1;build(1,1,m-1);for(int i=1;i<=n;++i){L[i]=lower_bound(a+1,a+m+1,L[i])-a;R[i]=lower_bound(a+1,a+m+1,R[i])-a;cover(1,1,m-1,L[i],R[i]-1);}for(int i=1;i<=n;++i){del(1,1,m-1,L[i],R[i]-1);ans=max(ans,tr[1].covlen);cover(1,1,m-1,L[i],R[i]-1);}printf("%d\n",ans);return 0;
}

B

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
#define N 100010
inline int read(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();return x*f;
}
int n,m,r,a[N],b[N];
ll suma[N],sumb[N],sumx[N],sumc[N],ans=0;
struct node{int x,c;
}s[N];
inline bool cmp(int x,int y){return x>y;}
inline bool cmp1(node x,node y){return x.c>y.c;}
int main(){freopen("rental.in","r",stdin);freopen("rental.out","w",stdout);n=read();m=read();r=read();for(int i=1;i<=n;++i) a[i]=read();sort(a+1,a+n+1,cmp);for(int i=1;i<=n;++i) suma[i]=suma[i-1]+a[i];for(int i=1;i<=m;++i) s[i].x=read(),s[i].c=read();sort(s+1,s+m+1,cmp1);for(int i=1;i<=m;++i) sumx[i]=sumx[i-1]+s[i].x,sumc[i]=sumc[i-1]+(ll)s[i].c*s[i].x;for(int i=1;i<=r;++i) b[i]=read();sort(b+1,b+r+1,cmp);for(int i=1;i<=r;++i) sumb[i]=sumb[i-1]+b[i];for(int i=0;i<=n;++i){//留i头牛的最优收益if(n-i>r) continue;ll res=sumb[n-i];int pos=upper_bound(sumx+1,sumx+m+1,suma[i])-sumx-1;res+=sumc[pos];if(pos!=m) res+=(ll)s[pos+1].c*(suma[i]-sumx[pos]);ans=max(ans,res);}printf("%lld\n",ans);return 0;
}

C

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
#define N 5010
inline int read(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();return x*f;
}
int n,m,dis[N][N],fa[N],sz[N];
vector<int>a[N];
struct edge{int x,y,val;
}e[N];
inline bool cmp(edge x,edge y){return x.val>y.val;}
inline int find(int x){return x==fa[x]?x:fa[x]=find(fa[x]);}
int main(){freopen("mootube.in","r",stdin);freopen("mootube.out","w",stdout);n=read();m=read();for(int i=1;i<=n;++i) fa[i]=i,a[i].push_back(i);for(int i=1;i<n;++i) e[i].x=read(),e[i].y=read(),e[i].val=read();sort(e+1,e+n,cmp);for(int ii=1;ii<=n-1;++ii){int xx=find(e[ii].x),yy=find(e[ii].y);if(a[xx].size()>a[yy].size()) swap(xx,yy);fa[xx]=yy;for(int i=0;i<a[xx].size();++i)for(int j=0;j<a[yy].size();++j){int x=a[xx][i],y=a[yy][j];dis[x][y]=dis[y][x]=e[ii].val;}for(int i=0;i<a[xx].size();++i) a[yy].push_back(a[xx][i]);}while(m--){int k=read(),x=read(),ans=0;for(int i=1;i<=n;++i) if(dis[x][i]>=k) ans++;printf("%d\n",ans);}return 0;
}

USACO 2018JAN Silver相关推荐

一中OJ #3509 七的倍数 [USACO Jan16,洛谷P3131] | 同余前缀和 | 解题报告
一中OJ | #3509 七的倍数 [USACO Jan16 Silver , Subsequences Summing to Sevens] 时限 1000MS/Case 内存 128MB/Case ...
2021 USACO Dec 题解
前言:最近开始刷usaco的silver了,希望12月份时可以过silver (害怕) P1 Closest Cow Wins (滑动窗口,数学): FN和FJ是死对头,在一条数轴有 k 块草,第 i ...
USACO翻译：USACO 2014 DEC Silver三题
USACO 2014 DEC SILVER 一.题目概览中文题目名称回程马拉松奶牛慢跑英文题目名称 piggyback marathon cowjog 可执行文件名 piggyback ma ...
USACO翻译：USACO 2014 FEB SILVER 三题
USACO 2014 FEB SILVER 一.题目概览中文题目名称自动打字路障神秘代码英文题目名称 auto rblock scode 可执行文件名 auto rblock scode 输 ...
USACO翻译：USACO 2012 FEB Silver三题
USACO 2012 FEB SILVER 一.题目概览中文题目名称矩形草地奶牛IDs 搬家英文题目名称 planting cowids relocate 可执行文件名 planting co ...
【USACO 2020 January Silver】Loan Repayment 题解
[USACO 2020 January Silver]Loan Repayment 题目: 题目描述 Farmer John 欠了 Bessie N 加仑牛奶(1≤N≤10^12).他必须在 K 天内 ...
USACO 2018 FEBURARY CONTEST :SILVER T1
USACO 2018 February Contest, Silver Problem 1. Rest Stops Farmer John and his personal trainer Bessi ...
USACO 1月 2021-2022 January Contest Silver银组题解
你好啊我又又又来了要准备usaco的铁铁们可以参考这个文章哦! 想刷好USACO--看这篇文章就够了_GeekAlice的博客-CSDN博客我最近是发现了一个很好用的网站https://blog.c ...
USACO 2012 January Contest, Silver Division Solution
T1是一道构图然后跑最短路的题. T2是一道裸的dp 然而我看了以后觉得爆搜是一定可以过掉的== 于是我先来一发二进制枚举,然后two points 维护答案.. T3是一道神题首先要分类讨论,然后 ...

USACO 2018JAN Silver

A

B

C

USACO 2018JAN Silver相关推荐

最新文章

热门文章