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poj3468 线段树

时间:2016-04-15 22:00:59      阅读:204      评论:0      收藏:0      [点我收藏+]

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  http://poj.org/problem?id=3468 题目链接, 很经典的线段树的应用, 这里复习一下, 再写一遍, 代码如下:

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 100000 + 10;
int N, Q;
int a[maxn];
struct Segment{
    int l, r;
    long long sum;
    int lazy;
}seg[3*maxn];

void build(int rt, int l, int r){
    seg[rt].l = l; seg[rt].r = r;
    seg[rt].lazy = 0;
    if(l==r){
        seg[rt].sum = a[l];
        return ;
    }
    int chl = 2*rt, chr = 2*rt+1;
    int mid = (l+r)/2;
    build(chl, l, mid);
    build(chr, mid+1, r);
    seg[rt].sum = seg[chl].sum + seg[chr].sum;
}

void push_down(int rt){
    int chl = 2*rt, chr = 2*rt+1;
    if(seg[rt].lazy){
        seg[chl].sum += (long long)seg[rt].lazy*(long long)(seg[chl].r-seg[chl].l+1);
        seg[chl].lazy += seg[rt].lazy;
        seg[chr].sum += (long long)seg[rt].lazy*(long long)(seg[chr].r-seg[chr].l+1);
        seg[chr].lazy += seg[rt].lazy;
        seg[rt].lazy = 0;
    }
}

void update(int rt, int l, int r, int c){
    if(seg[rt].l==l && seg[rt].r==r){
        seg[rt].sum += (long long)c*(r-l+1);
        seg[rt].lazy += c;
        return ;
    }
    push_down(rt);
    int mid = (seg[rt].l+seg[rt].r)/2;
    if(r <= mid)
        update(2*rt, l, r, c);
    else if(l>mid)
        update(2*rt+1, l, r, c);
    else{
        update(2*rt, l, mid, c);
        update(2*rt+1, mid+1, r, c);
    }
    seg[rt].sum = seg[2*rt].sum + seg[2*rt+1].sum;
}

long long query(int rt, int l, int r){
    if(seg[rt].l==l && seg[rt].r==r){
        return seg[rt].sum;
    }
    push_down(rt);
    int mid = (seg[rt].l+seg[rt].r)/2;
    if(r <= mid)
        return query(2*rt, l, r);
    else if(l > mid)
        return query(2*rt+1, l, r);
    else {
        long long v1 = query(2*rt, l, mid);
        long long v2 = query(2*rt+1, mid+1, r);
        return v1 + v2;
    }
    seg[rt].sum = seg[2*rt].sum + seg[2*rt+1].sum;
}
int main() {
    scanf("%d%d", &N, &Q);
    for(int i=1; i<=N; i++)
        scanf("%d", &a[i]);
    build(1, 1, N);
    for(int i=0; i<Q; i++){
        char s[6];
        scanf("%s", s);
        if(s[0] == Q){
            int l, r;
            scanf("%d%d", &l, &r);
            printf("%lld\n", query(1, l, r));
        }else{
            int l, r, c;
            scanf("%d%d%d", &l, &r, &c);
            update(1, l, r, c);
        }
    }
    return 0;
}

 

poj3468 线段树

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原文地址:http://www.cnblogs.com/xingxing1024/p/5396966.html

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