信息战（四）——战场演练（线段树，树状数组）-优快云博客

本文介绍了一种利用线段树和树状数组解决特定查询更新问题的方法。通过这两种数据结构，可以高效地处理数组中元素的更新操作及求第k小元素等问题。文章提供了完整的代码实现。

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两种做法：线段树和树状数组

TLE了几次= = 主要是cout

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 30000+10;
const int maxm = 65535*3+1;
int n,num[maxn],C[maxm];
inline int lowbit(int x){
	return x&(-x);
}
int sum(int x){
	int ret = 0;
	while(x > 0){
		ret += C[x];
		x -= lowbit(x);
	}
	return ret;
}
void add(int x,int d){
	while(x <= maxm) {
		C[x] += d;
		x += lowbit(x);
	}
}
int find(int x){
	int L = 1,R = maxm;
	while(L<=R){
		int mid = (L+R)>>1;
		if(sum(mid)<=x){
			L = mid+1;
		}else{
			R = mid-1;
		}
	} 
	return L;
}
int main(){	
	scanf("%d",&n);
	memset(C,0,sizeof(C));
	for(int i = 1; i <= n; i++){
		scanf("%d",&num[i]);
		add(num[i],1);
	}
	int m;
	scanf("%d",&m);
	while(m--){
		char task[40];
		int a,b;
		scanf("%s",task);
		if(task[0]=='Q'){
			scanf("%d",&a);
			if(a>n)	printf("-1\n");
			else printf("%d\n",find(n-a));
		}
		else if(task[0]=='A'){
			scanf("%d%d",&a,&b);
			if(num[a]>0){
				add(num[a],-1);
				num[a]-=b;
				if(num[a]>0) add(num[a],1);
				else  n--;	
			}
		}
		else{
			scanf("%d%d",&a,&b);
			if(num[a]>0){
				add(num[a],-1);
				num[a]+=b;
				add(num[a],1);	
			}
		} 
	}
	printf("%d\n",n);
	return 0;
}

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 30000+10;
const int maxm = 65535*3+1;
struct node{
	int lson,rson,sum;
	int mid(){
		return lson+(rson-lson)/2;
	}
}tree[maxm*4];
int n,num[maxn];
void pushup(int pos){
	tree[pos].sum = tree[pos*2+1].sum+tree[pos*2].sum;
}
void build(int l,int r,int pos){
	tree[pos].lson = l;
	tree[pos].rson = r;
	tree[pos].sum = 0;
	if(l == r)	return;
	int mid = tree[pos].mid();
	build(l,mid,pos*2);
	build(mid+1,r,pos*2+1);
}
void update(int x,int pos,int k){
	if(tree[pos].lson==tree[pos].rson && tree[pos].lson==x){
		tree[pos].sum += k;
	}else{
		int mid = tree[pos].mid();
		if(x<=mid) update(x,pos*2,k);
		else update(x,pos*2+1,k);
		pushup(pos);
	}
}
int query(int L,int R,int pos){
	if(tree[pos].lson >= L && tree[pos].rson<=R) return tree[pos].sum;
	int mid = tree[pos].mid();
	if(R <= mid)  return query(L,R,pos*2);
	else if(L > mid) return query(L,R,pos*2+1);
	else return query(L,mid,pos*2)+query(mid+1,R,pos*2+1);
}
int find(int x){
	int L = 1,R = maxm;
	while(L<=R){
		int mid = (L+R)/2;
		if(query(1,mid,1)<=x){
			L = mid+1;
		}else{
			R = mid-1;
		}
	} 
	return L;
}
int main(){	
	scanf("%d",&n);
	int maxk = 0;
	build(1,maxm,1);
	for(int i = 1; i <= n; i++){
		scanf("%d",&num[i]);
		update(num[i],1,1);
	}
	int m;
	scanf("%d",&m);
	while(m--){
		char task[40];
		int a,b;
		scanf("%s",task);
		if(task[0]=='Q'){
			scanf("%d",&a);
			if(a>n)	printf("-1\n");
			else printf("%d\n",find(n-a));
		}
		else if(task[0]=='A'){
			scanf("%d%d",&a,&b);
			if(num[a]>0){
				update(num[a],1,-1);
				num[a]-=b;
				if(num[a]>0) update(num[a],1,1);
				else  n--;	
			}
		}
		else{
			scanf("%d%d",&a,&b);
			if(num[a]>0){
				update(num[a],1,-1);
				num[a]+=b;
				update(num[a],1,1);	
			}
		} 
	}
	printf("%d\n",n);
	return 0;
}

转载于:https://www.cnblogs.com/pangblog/p/3367630.html