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预处理出右边Aj，然后扫一遍即可NYOJ要用 long long...#include#include#includeusing namespace std;const int N = 100000 + 10;typedef __int64 ll;struct node{ ll val, pos;}p[N];ll a[N];int main(){ i

题意略。首先树剖拆出链后建线段树，然后在线段树上维护，线段树节点记录的信息为最左和最右端的颜色（0～1e9,要从-1开始初始化），和该段区间里面的所记录的颜色段数量。因为考虑到，左子树的最右端和右子树的最左端颜色可能相同，那么相同时当前节点记录的颜色段数量便要-1，即( query( lson ) + query( rson ) - ( tr[rt<<1].rc == tr[rt<<1|1].

#include using namespace std;const int MAXN=10005;struct Edge{ int to, next;}edge[MAXN<<1];int head[MAXN], tot;int top[MAXN]; //子重链起点int fa[MAXN]; //父节点int deep[MAXN]; //深度int num[MAXN]

#include #include #include #include #include #include #include #include #include #include #include using namespace std;#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1 | 1#

第一次碰到这种题目，模板啊orz#include #include #include #include #include #include #include #include #include #include #include using namespace std;#define lson l, mid, rt << 1#define rson mid + 1

题目描述 Alex is a great magic-user, he can use his magic to attack monster, Today he has learn N number of magic, The i magic need Vi magic power and it can attack Wi damage.Alex's magic power is l