POJ 3468 A Simple Problem with Integers(线段树区间修改)

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A Simple Problem with Integers

Time Limit: 5000MS

Memory Limit: 131072KB

64bit IO Format: %lld & %llu

Submit Status

Description

You have N integers, A₁, A₂, ... , A_N. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.

Input

The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A₁, A₂, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a₊₁, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a₊₁, ... , A_b.

Output

You need to answer all Q commands in order. One answer in a line.

Sample Input

10 5
1 2 3 4 5 6 7 8 9 10
Q 4 4
Q 1 10
Q 2 4
C 3 6 3
Q 2 4

Sample Output

Hint

The sums may exceed the range of 32-bit integers.

大意：对于每个C询问，将区间[a,b]上的元素都加c；对于每个Q询问，输出区间[a,b]的和。

思路：简单的线段树区间更新。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<cmath>
#include<vector>
#include<algorithm>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lc rt<<1
#define rc rt<<1|1
using namespace std;

typedef long long ll;
const int maxn = 200000 + 10;
const int INF = 1e9 + 10;
ll sum[maxn],add[maxn],ans;

void PushUp(int rt)
{
    sum[rt] = sum[lc] + sum[rc];
}

void PushDown(int rt, int m)
{
    if (add[rt]) {
        add[lc] += add[rt];
        add[rc] += add[rt];
        sum[lc] += add[rt] * (m - (m >> 1));
        sum[rc] += add[rt] * (m >> 1);
        add[rt] = 0;
    }
}

void build(int l, int r, int rt)
{
    if (l == r) scanf("%lld",&sum[rt]);
    else {
        int m = (l + r) >> 1;
        build(lson);
        build(rson);
        PushUp(rt);
    }
}

void update(int ql, int qr, int c, int l, int r, int rt)
{
    if (ql <= l && r <= qr) { //如果完全覆盖，那么暂时不更新子结点，因为暂时不会用到，以节省时间。
        add[rt] += c;
        sum[rt] += (ll)c*(r-l+1);
        return;
    }
    //如果没有完全覆盖，则要考虑左右子结点，这时要先把左右子结点更新一下再操作
    PushDown(rt, r-l+1);
    int m = (l + r) >> 1;
    if (ql <= m) update(ql, qr, c, lson);
    if (m < qr) update(ql, qr, c, rson);
    //因为子结点已经被修改，最后更新父节点
    PushUp(rt);
}

ll query(int ql, int qr, int l, int r, int rt)
{
    if (ql <= l && r <= qr) {
        return sum[rt];
    }

    PushDown(rt, r-l+1);
    int m = (l + r) >> 1;
    ll ans = 0;
    if (ql <= m) ans += query(ql, qr, lson);
    if (m < qr) ans += query(ql, qr, rson);
    //这里不用再更新sum[rt],因为这时的sum[rt]就是真实的区间和
    return ans;
}

int main()
{
    int n,q;
    scanf("%d%d",&n,&q);
    build(1,n,1);
    while(q--) {
        char s[2];
        scanf("%s",s);
        int a,b,c;
        if (s[0] == 'Q') {
            scanf("%d%d",&a,&b);
            ll ans = query(a,b,1,n,1);
            printf("%lld\n",ans);
        }
        else {
            scanf("%d%d%d",&a,&b,&c);
            update(a,b,c,1,n,1);
        }
    }
}