hdu 1556 Color the ball(线段树的另一种应用)

/*线段树这道题做完以后，收获很大，学到了线段树的另一种应用：如果需要对某个区间内所有元素进行同样操作，则只需要对线段树上对应区间进行操作即可。统计每个气球被涂次数，只需要匹配到线段树的对应的区间即可，不需要统计到叶子节点。否则会超时，在最后统计结果的时候则需要搜索到叶子节点。即：①处的作用*/#include <iostream>using namespace std;const int nMax = 100010;//int A[nMax];struct Node{int l, r;int num;Node(){}Node(int l,int r,int num) : l(l), r(r), num(num){}}node[nMax * 4];int ans[nMax];int N;void build(int rt, int l, int r){if(l < r){int mid = (l + r) / 2;build(rt * 2, l, mid);build(rt * 2 + 1, mid + 1, r);node[rt] = Node(l, r, 0);}else if(l == r)node[rt] = Node(l, r, 0);}void update(int rt, int l, int r){if(node[rt].l ==l && node[rt].r == r)//①node[rt].num ++;else{int mid = (node[rt].l + node[rt].r) / 2;if(mid >= r)update(rt * 2, l, r);else if(mid + 1 <= l)update(rt * 2 + 1, l, r);else{update(rt * 2, l, mid);update(rt * 2 + 1, mid + 1, r);}}}void solve(int rt){int i;if(node[rt].num > 0){for(i = node[rt].l; i <= node[rt].r; ++ i)ans[i] += node[rt].num;}if(node[rt].l < node[rt].r){solve(rt * 2);solve(rt * 2 + 1);}}int main(){//freopen("e://data.in", "r", stdin);while(scanf("%d", &N) != EOF && N){int a, b;int i;build(1, 1, N);for(i = 1; i <= N; ++ i){scanf("%d%d", &a, &b);update(1, a, b);}memset(ans, 0, sizeof(ans));solve(1);for(i = 1; i <= N; ++ i){if(i != 1) printf(" ");printf("%d", ans[i]);}printf("\n");}return 0;}