HDU-1166.敌兵布阵

问题描述

线段树裸题

单点更新即可。

代码

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#include <bits/stdc++.h>
#define ll long long
#define IOS {ios::sync_with_stdio(false);cin.tie(0);}
using namespace std;

const int maxn = 50000 + 5;
ll T, N, a[maxn], d[maxn << 2];
int _a, _b;
void push_up(int p)
{
d[p] = d[p << 1] + d[p << 1 | 1];
}
void build(int l, int r, int p)
{
if(l == r)
d[p] = a[l];
else
{
int m = (l + r) >> 1;
build(l, m, p << 1);
build(m + 1, r, p << 1 | 1);
push_up(p);
}
}
void update(int x, int v, int l, int r, int p)
{
if(l == r)
d[p] += v;
else
{
int m = (l + r) >> 1;
if(x <= m)
update(x, v, l, m, p << 1);
else
update(x, v, m + 1, r, p << 1 | 1);
push_up(p);
}
}
int query(int L, int R, int l, int r, int p)
{
if(L <= l && R >= r)
return d[p];
int ans = 0;
int m = (l + r) >> 1;
if(L <= m)
{
ans += query(L, R, l, m, p << 1);
}
if(R > m)
{
ans += query(L, R, m + 1, r, p << 1 | 1);
}
return ans;
}
int main(void)
{
IOS
char str[10];
cin >> T;
for(int Case = 1; Case <= T; Case++)
{
memset(a, 0, sizeof(a));
memset(d, 0, sizeof(d));
cout << "Case " << Case << ":" << endl;
cin >> N;
for(int i = 1; i <= N; i++)
cin >> a[i];
build(1, N, 1);
while(cin >> str && str[0] != 'E')
{
cin >> _a >> _b;
if(str[0] == 'Q')
cout << query(_a, _b, 1, N, 1) << endl;
else if(str[0] == 'A')
update(_a, _b, 1, N, 1);
else if(str[0] == 'S')
update(_a, -_b, 1, N, 1);
}
}
return 0;
}

注意点

  1. $build$中的赋值,$l == r$的具体含义
  2. $update$中对于$l = = r$的操作
  3. $query$函数中$flag$或$ans$的赋值
  4. 多组数据的话 注意要$memset$相关数组。

参考资料

  1. https://www.cnblogs.com/Yokel062/p/11352683.html
  2. https://www.cnblogs.com/jason2003/p/9676729.html
  3. https://blog.csdn.net/huangzihaoal/article/details/81813454
  4. https://www.cnblogs.com/xenny/p/9801703.html
  5. https://blog.csdn.net/zearot/article/details/52280189

一个非常好的线段树模板可以参考,据说是某位神牛所写

https://www.cnblogs.com/Destiny-Gem/articles/3875243.html