UVa 12501 Bulky process of bulk reduction(线段树 + lazy思想 + 相对位置)

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3945

解题报告人： GHQ（SpingWater）

题意：给一个序列，有两种操作：

1、query i j : a[i]*1+a[i+1]*2+a[i+2]*3+....+a[j]*(j+1)的值。

2、change i j u：把区间[i,j]的值都加上u。

思路：用线段树维护两个值sum， ans：

sum为当前区间[l,r]的a[l]*1+a[l+1]*2+a[l+2]*3+....+a[r]*(r+1)的值。

ans为当前区间[l,r]的a[l]+a[l+1]+a[l+2]+....a[r]的和。

这样如果包含有某个区间。那就这样区间的值就是ans+sum* x，x为这个区间【l，r】的相对位置。

代码如下：

#include<stdio.h>
#include<stdlib.h>
#define MAXN 100005
#define NOT -1
struct Info{
	long long l, r;
	long long store;
	long long ans;
	long long sum;
}info[4 * MAXN];
long long build(long long root, long long l, long long r)
{
	long long mid = (l + r) >> 1;
	if(l == r){	info[root].l = info[root].r = l; info[root].store = info[root].ans = info[root].sum = 0; return 0;}

	info[root].l =l;
	info[root].r = r;
	info[root].store = info[root].ans = info[root].sum  = 0;
	build(root<<1, l, mid);
	build(root<<1|1, mid + 1, r);
	return 0;
}
long long update(long long root, long long l, long long r, long long num)
{
	long long tl = root << 1;
	long long tr = tl | 1;
	long long mid = (info[root].l + info[root].r) >> 1;
	long long temp = r - l + 1;
	info[root].sum += num * temp;
	info[root].ans += num * temp*(temp + 2 * (l - info[root].l) + 1) / 2;
	if(info[root].l == l && info[root].r == r)info[root].store += num;
	else
	{
		if(info[root].store)
		{
			info[tl].store += info[root].store;
			temp = info[tl].r - info[tl].l + 1;
			info[tl].sum += info[root].store * temp;
			info[tl].ans += info[root].store * temp * (temp + 1) / 2;

			info[tr].store += info[root].store;
			temp = info[tr].r - info[tr].l + 1;
			info[tr].sum += info[root].store * temp;
			info[tr].ans += info[root].store * temp * (temp + 1) / 2;
			info[root].store = 0;
		}
		if(l > mid)update(tr, l, r, num);
		else if(r <= mid)update(tl, l, r, num);
		else
		{
			update(tl, l,mid, num);
			update(tr, mid + 1, r, num);
		}
	}
	return 0;
}
long long query(long long root, long long l, long long r, long long x)
{
	long long tl = root << 1;
	long long tr = tl | 1;
	long long temp;
	long long mid = (info[root].l + info[root].r) >> 1;
	if(info[root].l == l && info[root].r == r) return info[root].ans + x * info[root].sum;
	else
	{
		if(info[root].store)
		{
			info[tl].store += info[root].store;
			temp = info[tl].r - info[tl].l + 1;
			info[tl].sum += info[root].store * temp;
			info[tl].ans += info[root].store * temp * (temp + 1) / 2;

			info[tr].store += info[root].store;
			temp = info[tr].r - info[tr].l + 1;
			info[tr].sum += info[root].store * temp;
			info[tr].ans += info[root].store * temp * (temp + 1) / 2;
			info[root].store = 0;
		}
		if(l > mid)return query(tr, l,r, x);
		else if(r <= mid)return query(tl, l, r, x);
		else return  query(tl, l, mid, x) + query(tr, mid + 1, r, x + mid + 1 -l);
	}
}
int main()
{
	long long T, N, M, l, r;
	long long cas = 0;
	char op[20];
	long long num;
	for(scanf("%lld", &T); T--;)
	{
		printf("Case %lld:\n", ++cas);
		scanf("%lld%lld", &N, &M);
		build(1, 1, N);
		while(M--)
		{
			scanf("%s%lld%lld",op, &l, &r);
			if(op[0] == 'c')
			{
				scanf("%lld", &num);
				update(1, l, r, num);
			}
			else
				printf("%lld\n", 50 * (r - l + 1)*(r - l + 2) + query(1, l, r, 0));
		}
	}
	return 0;
}