CodeForces 296C Greg and Array （線段樹）

Greg and Array

time limit per test

1.5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Greg has an array a = a1, a2, ..., an and m operations. Each operation looks as: li, ri, di, (1 ≤ li ≤ ri ≤ n). To apply operation i to the array means to increase all array elements with numbers li, li + 1, ..., ri by value di.

Greg wrote down k queries on a piece of paper. Each query has the following form: xi, yi, (1 ≤ xi ≤ yi ≤ m). That means that one should apply operations with numbers xi, xi + 1, ..., yi to the array.

Now Greg is wondering, what the array a will be after all the queries are executed. Help Greg.

Input

The first line contains integers n, m, k (1 ≤ n, m, k ≤ 105). The second line contains n integers: a1, a2, ..., an (0 ≤ ai ≤ 105) — the initial array.

Next m lines contain operations, the operation number i is written as three integers: li, ri, di, (1 ≤ li ≤ ri ≤ n), (0 ≤ di ≤ 105).

Next k lines contain the queries, the query number i is written as two integers: xi, yi, (1 ≤ xi ≤ yi ≤ m).

The numbers in the lines are separated by single spaces.

Output

On a single line print n integers a1, a2, ..., an — the array after executing all the queries. Separate the printed numbers by spaces.

Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams of the %I64dspecifier.

Sample test(s)

input

3 3 3
1 2 3
1 2 1
1 3 2
2 3 4
1 2
1 3
2 3

output

9 18 17

input

1 1 1
1
1 1 1
1 1

output

2

input

4 3 6
1 2 3 4
1 2 1
2 3 2
3 4 4
1 2
1 3
2 3
1 2
1 3
2 3

output

5 18 31 20

題意：長度爲n的序列；m種操作：a b c表示第a～b個數的值加上c；k次執行：a b表示執行第a～b種操作。最後輸出執行完所有操作後的數列。

先用線段樹計算出每種操作執行的次數，將操作中要加的數乘以次數更新。再用線段樹求出每個位置的數需要加的最終數值。感覺兩個都是簡單線段樹，就習慣性套模板，而且兩個線段樹又是先後分開執行，所以寫了一個線段樹就夠了。

#include <cstdio>
#include <iostream>
#include <algorithm>
#define ls node << 1
#define rs node << 1 | 1
#define lson l,mid,ls
#define rson mid + 1,r,rs
#define maxn 100005

using namespace std;

long long a[maxn];
int n,m,k,x,y;
long long cnt[maxn];

struct Operation {
    int l;
    int r;
    long long d;
} o[maxn];

struct Query {
    int l;
    int r;
    long long c;
    long long num;
} q[maxn << 2];

void build(int l, int r, int node)
{
    q[node].l = l;
    q[node].r = r;
    q[node].num = 0;
    q[node].c = 0;
    if(l == r)
        return;
    int mid = (l + r) >> 1;
    build(lson);
    build(rson);
}

void pushdown(int node,int len)
{
    if(q[node].c) {
        q[ls].c += q[node].c;
        q[rs].c += q[node].c;
        q[ls].num += q[node].c * (len - (len >> 1));
        q[rs].num += q[node].c * (len >> 1);
        q[node].c = 0;
    }
}

void update(int x,int y,long long c,int l,int r,int node)
{
    if(x <= l && y >= r) {
        q[node].c += c;
        q[node].num += c * (r - l + 1);
        return;
    }
    pushdown(node, r - l + 1);
    int mid = (l + r) >> 1;
    if(x <= mid)
        update(x,y,c,lson);
    if(y > mid)
        update(x,y,c,rson);
}

long long findnum(int x, int l, int r, int node)
{
    if(x == l && x == r)
        return q[node].num;
    pushdown(node,r - l + 1);
    int mid = (l + r) >> 1;
    if(x > mid)
        return findnum(x,rson);
    if(x <= mid)
        return findnum(x,lson);
}

int main()
{
    //freopen("in.txt","r",stdin);
    while(scanf("%d%d%d",&n,&m,&k) != EOF) {
        for(long long i = 1; i <= n; i++)
            scanf("%lld",&a[i]);
        for(long long i = 1; i <= m; i++)
            scanf("%d%d%lld",&o[i].l,&o[i].r,&o[i].d);
        build(1,m,1);
        for(long long i = 1; i <= k; i++) {
            scanf("%d%d",&x,&y);
            update(x,y,1,1,m,1);
        }
        for(int i = 1; i <= m; i++) {
            cnt[i] = findnum(i,1,m,1);
            //cout << cnt[i] << endl;
            o[i].d *= cnt[i];
        }
        build(1,n,1);
        for(int i = 1; i <= m; i++)
            update(o[i].l,o[i].r,o[i].d,1,n,1);
        for(int i = 1; i < n; i++)
            printf("%lld ",a[i] + findnum(i,1,n,1));
        printf("%lld\n",a[n] + findnum(n,1,n,1));
    }
}