POJ-3468-线段树-A Simple Problem with Integers

来源：互联网发布：虚拟设计软件编辑：程序博客网时间：2024/06/03 20:24

题意
求区间和
C abc表示给区间[a,b]里的数都加上c
思路
线段树+区间更新

区间更新就是在update时不一下全部更新 而是在区间上留下标记当查询区间时，再将标记向下传递 更新

#include <iostream>#include <cstdio>#include <cstdlib>#include <algorithm>#include <vector>#include <string.h>using namespace std;const int maxn = 100100;long long sum[maxn << 2];long long  addmark[maxn << 2];void pushup(int root){    sum[root] = sum[root << 1] + sum[root << 1 | 1];}void pushdown(int root,int m){    if (addmark[root])    {        addmark[root << 1] += addmark[root];        addmark[root << 1 | 1] += addmark[root];        sum[root << 1] += addmark[root] * (m - (m >> 1));        sum[root << 1 | 1] += addmark[root] * (m >> 1);        addmark[root] = 0;    }}void build(int l, int r, int root){    addmark[root] = 0;    if (r == l)    {        scanf("%lld", &sum[root]);        return;    }    int m = (l + r) >> 1;    build(l, m, root << 1);    build(m + 1, r, root << 1 | 1);    pushup(root);}void update(int L, int R, int l, int r, int root,int c){    if (L <= l&&r <= R)    {        addmark[root] += c;        sum[root] += (long long)c*(r - l + 1);        return;    }    pushdown(root, r - l + 1);    int m = (l + r) >> 1;    if (L <= m) update(L, R, l, m, root << 1,c);    if (R > m) update(L, R, m + 1, r, root << 1 | 1, c);    pushup(root);}long long  query(int L, int R, int l, int r, int root){    if (L <= l&&r <= R)    {        return sum[root];    }    pushdown(root, r - l + 1);    int m = (l + r) >> 1;    long long ans = 0;    if (L <= m) ans += query(L, R, l, m, root << 1);    if (m < R) ans += query(L, R, m + 1, r, root << 1 | 1);    return ans;}int main(void){    int n,q;    cin >> n >> q;    build(1, n, 1);    while (q--)    {        char s[5];        int a, b, c;        scanf("%s", s);        if (s[0] == 'Q')        {            scanf("%d%d", &a, &b);            cout << query(a, b, 1, n, 1) << endl;        }        else        {            scanf("%d%d%d", &a, &b, &c);            update(a, b, 1, n, 1, c);        }    }    return 0;}

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