标签：inpu sort font mes 各路 mario 离散化 lse number
InputThe first line follows an integer T, the number of test data.
For each test data:
The first line contains two integers n, m (1 <= n <=10^5, 1 <= m <= 10^5), n is the length of the road, m is the number of queries.
Next line contains n integers, the height of each brick, the range is [0, 1000000000].
Next m lines, each line contains three integers L, R,H.( 0 <= L <= R < n 0 <= H <= 1000000000.)OutputFor each case, output "Case X: " (X is the case number starting from 1) followed by m lines, each line contains an integer. The ith integer is the number of bricks Mario can hit for the ith query.
Sample Input
1 10 10 0 5 2 7 5 4 3 8 7 7 2 8 6 3 5 0 1 3 1 1 9 4 0 1 0 3 5 5 5 5 1 4 6 3 1 5 7 5 7 3
Sample Output
Case 1: 4 0 0 3 1 2 0 1 5 1
题意：求区间[l-r]的<=k的个数
裸的主席树，离散化一下就行了。
#include<bits/stdc++.h> using namespace std; #define Debug(x) cout<<#x<<":"<<(x)<<endl typedef long long ll; const int maxn = 100000+5; struct node { int l,r,num; }tree[maxn*20]; int a[maxn],b[maxn],rt[maxn],cnt; void pushup(int k) { tree[k].num=tree[tree[k].l].num+tree[tree[k].r].num; } void build(int l,int r,int &x) { x=++cnt; tree[x].num=0; if(l==r) return ; int m=(l+r)>>1; build(l,m,tree[x].l); build(m+1,r,tree[x].r); pushup(x); } int query(int l,int r,int pre,int now,int x,int y) { if(x<=l&&r<=y) return tree[now].num-tree[pre].num; int m=(l+r)>>1; int ans=0; if(x<=m) ans+=query(l,m,tree[pre].l,tree[now].l,x,y); if(y>m) ans+=query(m+1,r,tree[pre].r,tree[now].r,x,y); return ans; } void updata(int l,int r,int pre,int &now,int k) { now=++cnt; tree[now]=tree[pre]; if(l==r) { tree[now].num++; return ; } int m=(l+r)>>1; if(k<=m) updata(l,m,tree[pre].l,tree[now].l,k); else updata(m+1,r,tree[pre].r,tree[now].r,k); pushup(now); } int main() { int n,m,t,p=1; scanf("%d",&t); while(t--) { printf("Case %d:\n",p++); scanf("%d %d",&n,&m); { cnt=0; for(int i=1; i<=n; i++) scanf("%d",&a[i]),b[i]=a[i]; sort(b+1,b+1+n); int len=unique(b+1,b+1+n)-(b+1); for(int i=1; i<=n; i++) a[i]=lower_bound(b+1,b+1+len,a[i])-b; build(1,len,rt[0]); for(int i=1; i<=n; i++) updata(1,len,rt[i-1],rt[i],a[i]); while(m--) { int l,r,k1,k; scanf("%d %d %d",&l,&r,&k); l++; r++; k1=lower_bound(b+1,b+1+len,k)-b; if(b[k1]!=k) k1--; if(k1==0)//不判断会出现MLE { printf("0\n"); continue; } printf("%d\n",query(1,len,rt[l-1],rt[r],1,k1)); } } } }
PS：摸鱼怪的博客分享，欢迎感谢各路大牛的指点~
标签：inpu sort font mes 各路 mario 离散化 lse number
原文地址：https://www.cnblogs.com/MengX/p/9104963.html
