POJ 3468 继续线段树

这次写的是接着上次的线段树写的，上次只写了线段树的单点修改，区间求最大值。

然而之后又继续学习了线段树的区间修改以及区间求和的问题

然后呢，这次肯定就是关于；

线段树的区间修改，区间求最大值：

学完之后感觉是和之前的那些差不多，都是一个思想，只不过对于一个区间的修改，也就是区间内每一个值都需要加上相同的值，时间上也就是区间修改的总的sum和，就是区间内元素的个数乘上add数组的值， add数组保存的是在某个区间要对没个区间元素的增加值。（大体上的思路都是修改之后通过递归和二分的思想去维护每一个区间）。

题目链接：点击打开链接

这也是一个纯模版题，只要会模版就可以AC，哈哈

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

#include <iostream>
#include <cstdio>
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 10;

#define lson ri<<1,l,m       //这个是我在网上学的一个小技巧
#define rson ri<<1|1,m+1,r

ll sum[maxn<<2], add[maxn<<2];

struct Node {
    int left ,right;
    int mid (){
        return (left + right)>>1;
    }
}tree[maxn<<2];

void PushUp (int ri){
    sum[ri] = sum[ri<<1] + sum[ri<<1|1];  // sum[ri<<1|1] 相当于sum[(ri<<1)+1]
}

void PushDown (int ri, int m){
    if (add[ri]){
        add[ri<<1] += add[ri];
        add[ri<<1|1] += add[ri];
        sum[ri<<1] += add[ri] * (m-(m>>1));
        sum[ri<<1|1] += add[ri] *(m>>1);
        add[ri] = 0;
    }
}

void build (int ri, int l, int r){
    //cout << "text\n";
    tree[ri].left = l;
    tree[ri].right = r;
    add[ri] = 0;
    if (l == r){
        //cin >> sum[ri];
        scanf("%I64d",&sum[ri]);
        return ;
    }
    int m = tree[ri].mid();
    build (lson);
    build (rson);
    PushUp (ri);
}

void Update (int key,int l, int r, int ri){
    if (tree[ri].left == l && tree[ri].right == r){
        add[ri] += key;
        sum[ri] +=(ll) key * (r - l + 1);
        return ;
    }
    if (tree[ri].left == tree[ri].right)  return ;
    PushDown(ri,tree[ri].right - tree[ri].left + 1);
    int m = tree[ri].mid();
    if (r <= m){
        Update (key,l,r,ri<<1);
    }
    else if (l > m){
        Update (key, l, r, ri <<1|1);
    }
    else{
        Update (key,l, m , ri<<1);
        Update (key, m+1, r, ri<<1|1);
    }
    PushUp (ri);
}

ll search (int l, int r, int ri){
    if (tree[ri].left == l && tree[ri].right == r){
        return sum[ri];
    }
    PushDown (ri, tree[ri].right - tree[ri].left + 1);
    int m = tree[ri].mid();
    ll res = 0;
    if (r <= m){
        res += search (l, r, ri<<1);
    }
    else if (l > m){
        res += search (l, r, ri<<1|1);
    }
    else{
        res += search (l,m,ri<<1);
        res += search (m+1, r, ri<<1|1);
    }
    return res;
}

int main (){
    //ios::sync_with_stdio(false);
    int n, m;
    while (~scanf("%d%d",&n,&m)){
        build (1, 1, n);
        //cout << "text\n";
        while (m--){
            //cout << "text\n";
            char ch[2];
            int a,b,c;
            //cin >> ch;
            scanf("%s",ch);
            if (ch[0] == 'Q'){
//                cin >>a >> b;
//                cout << search (a,b,1) <<endl;
                scanf("%d%d",&a,&b);
                printf("%I64d\n",search(a,b,1));
            }
            else{
                //cin >> a >> b >> c;
                scanf("%d%d%d",&a,&b,&c);
                Update (c, a, b, 1);
            }
        }
    }
    return 0;
}

POJ 3468 继续线段树

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