POJ 3468 A Simple Problem with Integers 成段更新-优快云博客

本文链接：https://blog.youkuaiyun.com/waitfor_/article/details/7239596

本文介绍了一种使用线段树进行区间更新与查询的优化方法，通过懒惰传播避免重复计算，减少时间复杂度。适用于解决大规模数据集上的区间操作问题。

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中间乘法溢出了，好纠结(囧)..

#include<cstdio>
#include<cstdlib>
#include<cstring>
typedef long long ll;
int in[100000];
struct Tree{
	int s;
	int t;
	ll sum;
	int add;
}tree[270010];

void build(int s,int t,int id){
	tree[id].s=s;
	tree[id].t=t;
	tree[id].add=0;
	if(s==t)
		tree[id].sum=in[s];
	else{
		int mid=(tree[id].s+tree[id].t)>>1;
		build(s,mid,id*2);
		build(mid+1,t,id*2+1);
		tree[id].sum=tree[id*2].sum+tree[id*2+1].sum;
	}
}

void insert(int s,int t,int id,int colour){
	if(tree[id].s==s && tree[id].t==t){
		tree[id].add+=colour;
		tree[id].sum+=(ll)(tree[id].t-tree[id].s+1)*colour;
		return ;
	}
	tree[id].sum+=(t-s+1)*colour;
	if(tree[id].add!=0){
		tree[id*2].add+=tree[id].add;
		tree[id*2].sum+=(ll)(tree[id*2].t-tree[id*2].s+1)*tree[id].add;        //
		tree[id*2+1].add+=tree[id].add;
		tree[id*2+1].sum+=(ll)(tree[id*2+1].t-tree[id*2+1].s+1)*tree[id].add;  //
		tree[id].add=0;
	}
	int mid=(tree[id].s+tree[id].t)>>1;
	if(mid<s)
		insert(s,t,id*2+1,colour);
	else if(mid>=t)
		insert(s,t,id*2,colour);
	else{
		insert(s,mid,id*2,colour);
		insert(mid+1,t,id*2+1,colour);
	}
}


ll query(int s,int t,int id){
	if(tree[id].s==s && tree[id].t==t)
		return tree[id].sum;
	if(tree[id].add!=0){
		tree[id*2].add+=tree[id].add;
		tree[id*2].sum+=(ll)(tree[id*2].t-tree[id*2].s+1)*tree[id].add;        //
		tree[id*2+1].add+=tree[id].add;
		tree[id*2+1].sum+=(ll)(tree[id*2+1].t-tree[id*2+1].s+1)*tree[id].add;  //
		tree[id].add=0;
	}
	int mid=(tree[id].s+tree[id].t)>>1;
	if(mid<s)
		return query(s,t,id*2+1);
	else if(mid>=t)
		return query(s,t,id*2);
	else
		return query(s,mid,id*2)+query(mid+1,t,id*2+1);
}

int main(){
	int n,i,a,b,c,m,cc;
	scanf("%d %d",&n,&m);
	for(i=1;i<=n;i++)
		scanf("%d",&in[i]);
	build(1,n,1);
	for(i=1;i<=m;i++){
		char ss[5];
		scanf("%s",ss);
		if(ss[0]=='C'){
			scanf("%d %d %d",&a,&b,&c);	
			insert(a,b,1,c);
		}
		else{
			scanf("%d %d",&a,&b);
			printf("%lld\n",query(a,b,1));
		}
	}
}