HDU 1166 (segment tree)

最新推荐文章于 2020-10-11 14:32:41 发布

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最新推荐文章于 2020-10-11 14:32:41 发布

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分类专栏： acm 文章标签： segment_tree hdu

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本文介绍了一个简单的段树实现问题，旨在帮助读者理解并实践段树数据结构的应用。文章提供了完整的C++代码示例，展示了如何构建段树、更新节点值以及查询区间和等操作。

An easy problem aiming at practicing the implementation of the segment tree. Apparently, we should use segment tree to solve it. And within the tree structure is the sum.

Note that the number of the nodes is much larger than the size of the integer array because of the "tree" structure. Otherwise, it may cause Runtime Error (ACCESS_VIOLATION).

AC Code:

#include <iostream>
#include <cstdio>
#include <cstring>
#define INF 0x3f3f3f3f
#define N 50010
#define MST(a,b) memset(a,b,sizeof(a))
#define FOR(i,a,b) for(int i=a;i<=b;i++)
using namespace std;

struct TREE {
    int left,right;
    int suml;
} tree[900000];//*** Large Enough? ***

void Built(int root,int l,int r){
    tree[root].left = l;
    tree[root].right = r;
    tree[root].suml=0;
    if(l == r)return;
    int mid = (l+r) >> 1;
    Built ( root<<1   , l , mid );
    Built ( root<<1|1 , 1+mid , r);
}

void Fix(int root,int left,int right,int del) {
    if (left == tree[root].left && tree[root].right == right) {
        tree[root].suml += del;
        return;
    }    if (left < tree[root].left || right > tree[root].right)
        return;
    Fix(root << 1, left, right, del);
    Fix(root << 1 | 1, left, right, del);
    tree[root].suml += del;
}

int Query_sum(int root,int left,int right) {
    if (left == tree[root].left && tree[root].right == right)
        return tree[root].suml;
    if (tree[root << 1 | 1].left > right)//left only
        return Query_sum(root << 1, left, right);
    if (left > tree[root << 1].right)//right only
        return Query_sum(root << 1 | 1, left, right);
    int s1 = 0, s2 = 0;
    s1 = Query_sum(root << 1, left, tree[root << 1].right);
    s2 = Query_sum(root << 1 | 1, tree[root << 1 | 1].left, right);
    return s1 + s2; 
    if (left < tree[root].left || right > tree[root].right)
        return -INF;
}


int main(){
    int t,n,a,b;
    cin>>t;
    FOR(ncase,1,t) {
        printf("Case %d:\n", ncase);
        cin >> n;
        Built(1, 1, n);
        FOR(i, 1, n) {
            cin >> a;
            Fix(1, i, i, a);
        }
        char s[10];
        cin >> s;
        while (s[0] != 'E') {
            cin >> a >> b;
            if (s[0] == 'A') {
                Fix(1, a, a, b);
            }
            else if (s[0] == 'S') {
                Fix(1, a, a, -b);
            }
            else
                printf("%d\n", Query_sum(1, a, b));
            cin>>s;
        }
    }
    return 0;
}