点击打开链接hdu1556
思路:线段树成段更新
分析:
1 简单的线段树的成段更新,我们把它看成区间的更改和区间的求和即可,那这样我们只要建立好线段树然后每一次进行更新,最后对每一个[i , i]区间进行求和即可
代码:
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; const int MAXN = 100010; struct Node{ int left; int right; int mark;//延时标记 int sum; }; Node node[4*MAXN]; int N; //向上更新 void push_up(int pos){ node[pos].sum = node[pos<<1].sum + node[(pos<<1)+1].sum; } //向下更新i void push_down(int pos){ if(node[pos].mark){ node[pos<<1].mark += node[pos].mark; node[(pos<<1)+1].mark += node[pos].mark; node[pos<<1].sum += node[pos].mark*(node[pos<<1].right-node[pos<<1].left+1); node[(pos<<1)+1].sum += node[pos].mark*(node[(pos<<1)+1].right-node[(pos<<1)+1].left+1); node[pos].mark = 0; } } //建立线段树 void buildTree(int left , int right , int pos){ node[pos].left = left; node[pos].right = right; node[pos].mark = 0; if(left == right){ node[pos].sum = 0; return; } int mid = (left+right)>>1; buildTree(left , mid , pos<<1); buildTree(mid+1 , right , (pos<<1)+1); push_up(pos); } //更新 void update(int left , int right , int value , int pos){ if(left <= node[pos].left && right >= node[pos].right){ node[pos].mark += value; node[pos].sum += value*(node[pos].right-node[pos].left+1); return; } push_down(pos); int mid = (node[pos].left + node[pos].right)>>1; if(right <= mid) update(left , right , value , pos<<1); else if(left > mid) update(left , right , value , (pos<<1)+1); else{ update(left , mid , value , pos<<1); update(mid+1 , right , value , (pos<<1)+1); } push_up(pos); } //查询 int query(int left , int right , int pos){ if(node[pos].left == left && node[pos].right == right) return node[pos].sum; int mid = (node[pos].left+node[pos].right)>>1; push_down(pos);//继续向下更新,这样才能完全更新完毕 if(right <= mid) return query(left , right , pos<<1); else if(left > mid) return query(left , right , (pos<<1)+1); else return query(left , mid , pos<<1) + query(mid+1 , right , (pos<<1)+1); } int main(){ int x , y; while(scanf("%d" , &N) && N){ buildTree(1 , N , 1); int cnt = N; while(cnt--){ scanf("%d%d" , &x , &y); update(x , y , 1 , 1); } for(int i = 1 ; i <= N ; i++){ if(i == 1) printf("%d" , query(i , i , 1)); else printf(" %d" , query(i , i , 1)); } printf("\n"); } return 0; }