problem:
給n個節點,有很多命令,兩種命令,一種是更新節點的值,一種是區間求和:
solution:
利用線段樹堆區間求和
note:
要關閉cin的同步,不然會超時。
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using
namespace
std;
#define ms(s) memset(s,0,sizeof(s))
typedef
unsigned
long
long ull;
typedef
long
long ll;
const
double pi = 3.141592653589;
const
int inf = 0x3fffffff;
#define maxn 50000
struct node
};node tree[maxn*4];
int value[maxn+10];
int flag;
//初始化樹,根節點是1
void init_tree(int root, int l, int r)
}//查詢和
int query_tree(int root, int l, int r)
}void update_tree(int root, int idx, int v)
}int main()
else
if(s == "sub")
else
if(s == "query")
else
break;}}
return
0;}
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