題目鏈接

思路：線段樹維護區間和，當k=1時，詢問二分詢問[x-(x~n-1)]找到最小位置，復雜度n*logn*logn卡過

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<math.h>
#include<set>
#include<stack>
#include<vector>
#include<map>
#include<queue>
#define myself i,l,r
#define lson i<<1
#define rson i<<1|1
#define Lson i<<1,l,mid
#define Rson i<<1|1,mid+1,r
#define half (l+r)/2
#define inff 0x3f3f3f3f
#define lowbit(x) x&(-x)
#define PI 3.14159265358979323846
#define min4(a,b,c,d) min(min(a,b),min(c,d))
#define min3(x,y,z) min(min(x,y),min(y,z))
#define pii make_pair
#define pr pair<int,int>
const int dir[4][2]= {0,-1,-1,0,0,1,1,0};
typedef long long ll;
const ll inFF=9223372036854775807;
typedef unsigned long long ull;
using namespace std;
const int maxn=5e4+5;
int n,m;
int tree[maxn<<2],lazy[maxn<<2];
void pushup(int i)
{tree[i]=tree[lson]+tree[rson];
}
void pushdown(int i,int l,int r)
{int mid=half;if(lazy[i]!=-1){tree[lson]=(mid-l+1)*lazy[i];tree[rson]=(r-mid)*lazy[i];lazy[lson]=lazy[rson]=lazy[i];lazy[i]=-1;}
}
void build(int i,int l,int r)
{lazy[i]=-1;tree[i]=1;if(l==r) return ;int mid=half;build(Lson);build(Rson);pushup(i);
}
int query(int i,int l,int r,int ql,int qr)
{if(ql<=l&&qr>=r) return tree[i];int mid=half;pushdown(i,l,r);if(qr<=mid) return query(Lson,ql,qr);else if(ql>=mid+1) return query(Rson,ql,qr);else return query(Lson,ql,mid)+query(Rson,mid+1,qr);
}
void update(int i,int l,int r,int ql,int qr,int val)
{if(ql<=l&&qr>=r){tree[i]=val*(r-l+1);lazy[i]=val;return;}pushdown(i,l,r);int mid=half;if(qr<=mid) update(Lson,ql,qr,val);else if(ql>=mid+1) update(Rson,ql,qr,val);else update(Lson,ql,mid,val),update(Rson,mid+1,qr,val);pushup(i);
}
int erfen(int x,int r,int val)
{int l=x,mid,ans;while(l<=r){mid=half;if(query(1,0,n-1,x,mid)>=val){ans=mid;r=mid-1;}elsel=mid+1;}return ans;
}
int main()
{int t,op,x,y,val;cin>>t;while(t--){scanf("%d %d",&n,&m);build(1,0,n-1);while(m--){scanf("%d %d %d",&op,&x,&y);if(op==2){val=query(1,0,n-1,x,y);printf("%d\n",y-x+1-val);update(1,0,n-1,x,y,1);}else{val=query(1,0,n-1,x,n-1);if(val==0) printf("Can not put any one.\n");else{int l=x,r=n-1,s,e;s=erfen(x,n-1,1);if(val<=y)e=erfen(x,n-1,val);elsee=erfen(x,n-1,y);printf("%d %d\n",s,e);update(1,0,n-1,x,e,0);}}}printf("\n");}
}

?

總結

以上是生活随笔為你收集整理的HDU - 4614 Vases and Flowers 线段树+二分的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。