POJ 3468 A Simple Problem with Integers 解题报告 线段树 区间修改
解题思路:用lazy标记延迟更新,以此来提高效率,不然会tle。
具体讲就是如果要改的区间比当前区间范围小,那么标记这个区间,直接把值加上去。如果之后用到了其子区间,再往下更新。
#include<iostream>
#include<math.h>
#include<iomanip>
#include<algorithm>
#include<queue>
#include<cstring>
#include<string>
#include<string.h>
#include<map>
#include<stack>
#include<stdio.h>
#include<cstdio>
#include<stdlib.h>
#include<fstream>
#include<sstream>
#include<set>
#pragma warning(disable:4996)
#define INF 0x3f3f3f3f
#define ll long long
#define PI acos(-1.0)
const int N = 100005;
const int maxn = 1e9;
using namespace std;
int n, t, m;
ll a[100005];
ll tree[400020];//tree里放区间和
ll lazy[400020];
void pushdown(int id, int l, int r)
{
if (lazy[id])
{
lazy[id << 1] += lazy[id];
lazy[id << 1 | 1] += lazy[id];
tree[id << 1] += lazy[id] * l;
tree[id << 1 | 1] += lazy[id] * r;
lazy[id] = 0;
}
}
void build(int id, int l, int r)
{
lazy[id] = 0;
if (l == r)
{
tree[id] = a[l];
return;
}
int mid = (l + r) >> 1;
build(id << 1, l, mid);
build(id << 1 | 1, mid + 1, r);
tree[id] = tree[id << 1] + tree[id << 1 | 1];
}
ll int query(int id, int l, int r, int x, int y)
{
if (l > r || x > r || y < l)
return 0;
if (x <= l && y >= r)
{
return tree[id];
}
int mid = (l + r) >> 1;
pushdown(id, mid - l + 1, r - mid);
ll sum = 0;
if (l <= mid)
sum += query(id << 1, l, mid, x, y);
if (r > mid)
sum += query(id << 1 | 1, mid + 1, r, x, y);
return sum;
}
void Add(int id, int l, int r, int x, int y, int z)
{
if (l > r || x > r || y < l)
return;
if (l >= x && r <= y)
{
tree[id] += z * (r - l + 1);
lazy[id] += z;
return;
}
int mid = (l + r) >> 1;
pushdown(id, mid - l + 1, r - mid);
if (l <= mid)
Add(id << 1, l, mid, x, y, z);
if (r > mid)
Add(id << 1 | 1, mid + 1, r, x, y, z);
tree[id] = tree[id << 1] + tree[id << 1 | 1];
}
int main()
{
scanf("%d%d", &n, &m);
memset(a, 0, sizeof(a));
for (int i = 1; i <= n; i++)
{
scanf("%lld", &a[i]);
}
build(1, 1, n);
char ch[10] = "";
while (m--)
{
scanf("%s", ch);
if (strcmp(ch, "Q") == 0)
{
int x, y;
scanf("%d%d", &x, &y);
printf("%lld\n", query(1, 1, n, x, y));
}
else if (strcmp(ch, "C") == 0)
{
int x, y, z;
scanf("%d%d%d", &x, &y, &z);
Add(1, 1, n, x, y, z);
}
memset(ch, 0, sizeof(0));
}
}