hdoj1556 Color the ball【線段樹+區間更新】

Problem Description

N個氣球排成一排，從左到右依次編號爲1,2,3....N.每次給定2個整數a b(a <= b),lele便爲騎上他的“小飛鴿"牌電動車從氣球a開始到氣球b依次給每個氣球塗一次顏色。但是N次以後lele已經忘記了第I個氣球已經塗過幾次顏色了，你能幫他算出每個氣球被塗過幾次顏色嗎？

Input

每個測試實例第一行爲一個整數N,(N <= 100000).接下來的N行，每行包括2個整數a b(1 <= a <= b <= N)。
當N = 0，輸入結束。

Output

每個測試實例輸出一行，包括N個整數，第I個數代表第I個氣球總共被塗色的次數。

Sample Input

3

1 1

2 2

3 3

3

1 1

1 2

1 3

0

Sample Output

1 1 1

3 2 1

比較裸的模板題: 線段樹+區間更新

#include<bits/stdc++.h>
using namespace std;

#define lid (id << 1)
#define rid (id << 1 | 1)
const int N = 100005;

int a[N];

struct Segtree{
    int l, r;
    int sum, mx;
    int lazy;
}tr[N * 4];

void push_up(int id){
    tr[id].sum = tr[rid].sum + tr[lid].sum;
    //tr[id].mx = max(tr[rid].mx, tr[lid].mx);
}
void push_down(int id){
    if(tr[id].lazy){
        tr[lid].lazy += tr[id].lazy;
        tr[rid].lazy += tr[id].lazy;
        tr[lid].sum += tr[id].lazy * (tr[lid].r - tr[lid].l + 1);
        tr[rid].sum += tr[id].lazy * (tr[rid].r - tr[rid].l + 1);
        tr[id].lazy = 0;
    }
}
void build(int id, int l, int r){
    tr[id].l = l;
    tr[id].r = r;
    tr[id].lazy = 0;
    tr[id].sum = 0;
    if(l == r){
        tr[id].sum  = a[l];
        return;
    }
    int mid = (l + r) >> 1;
    build(lid, l, mid);
    build(rid, mid + 1, r);
    push_up(id);
}
void update(int id, int l, int r, int z){
    if(tr[id].l == l && tr[id].r == r){
        tr[id].sum += z * (tr[id].r - tr[id].l + 1);
        tr[id].lazy += z;
        return ;
    }
    push_down(id);
    int mid = (tr[id].l + tr[id].r) >> 1;
    if(r <= mid)
        update(lid, l, r, z);
    else if(l > mid)
        update(rid, l, r, z);
    else{
        update(lid, l, mid, z);
        update(rid, mid + 1, r, z);
    }
    push_up(id);
}
int query(int id, int l, int r){
    if(tr[id].l == l && tr[id].r == r)
        return tr[id].sum;
    push_down(id);//區間查詢的時候需要更新子節點
    int mid = (tr[id].l + tr[id].r) >> 1;
    if(r <= mid)
        return query(lid, l, r);
    if(l > mid)
        return query(rid, l, r);
    return query(lid, l, mid) + query(rid, mid + 1, r);
}
int main(){
    int n;
    while(scanf("%d", &n) != EOF){
        if(n == 0)
            break;
        memset(a, 0, sizeof(a));
        build(1, 1, n);
        for(int i = 0; i < n; i++){
            int x, y;
            cin >> x >> y;
            update(1, x, y, 1);
        }
        printf("%d", query(1, 1, 1));
        for(int i = 2; i <= n; i++)
            printf(" %d", query(1, i, i));
        cout << endl;
    }
    return 0;
}