Mario is world-famous plumber. His “burly” figure and amazing jumping ability reminded in our memory. Now the poor princess is in trouble again and Mario needs to save his lover. We regard the road to the boss’s castle as a line (the length is n), on every integer point i there is a brick on height hi. Now the question is how many bricks in [L, R] Mario can hit if the maximal height he can jump is H.

裸的模板題目，就是詢問區間第K小這里注意一下數組是從0開始的就行了

#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> using namespace std; const int MAXN = 1e5+10; struct Tree{int ch[2];int sum; }pool[MAXN*40+10]; int queries[MAXN+10][3], uniq[MAXN*2+10], val[MAXN+10], tot, n, m, Len, roots[MAXN+10]; void Insert(int &u, int l, int r, int v){++tot;pool[tot] = pool[u];u = tot;pool[u].sum ++;if(l == r) return ;int mid = (l + r) >> 1;if(v <= mid) Insert(pool[u].ch[0], l, mid, v);else Insert(pool[u].ch[1], mid+1, r, v); } int Query(int a, int b, int l, int r, int H){if(r <= H) return pool[b].sum - pool[a].sum;if(l > H) return 0;int mid = (l + r) >> 1;return Query(pool[a].ch[0], pool[b].ch[0], l, mid, H) + Query(pool[a].ch[1], pool[b].ch[1], mid+1, r, H); } void Init(){while(tot){pool[tot].ch[0] = pool[tot].ch[1] = pool[tot].sum = 0;tot--;} } void Read(){int tmp=0;scanf("%d%d", &n, &m);for(int i=1;i<=n;i++){scanf("%d", &val[i]);uniq[++tmp] = val[i];}for(int i=1;i<=m;i++){scanf("%d%d%d", &queries[i][0], &queries[i][1], &queries[i][2]);uniq[++tmp] = queries[i][2];}sort(uniq+1, uniq+1+tmp);Len = unique(uniq+1, uniq+1+tmp) - uniq - 1;for(int i=1;i<=n;i++)val[i] = lower_bound(uniq+1, uniq+1+Len, val[i]) - uniq;for(int i=1;i<=m;i++)queries[i][2] = lower_bound(uniq+1, uniq+1+Len, queries[i][2]) - uniq; } void Solve(){for(int i=1;i<=n;i++){roots[i] = roots[i-1];Insert(roots[i], 1, Len, val[i]);}for(int i=1;i<=m;i++)printf("%d\n", Query(roots[queries[i][0]], roots[queries[i][1]+1], 1, Len, queries[i][2])); } int main(){int T, t=0;scanf("%d", &T);while(t<T){printf("Case %d:\n", ++t);Init();Read();Solve();}return 0; }

轉載于:https://www.cnblogs.com/JeremyGJY/p/5921641.html

總結

以上是生活随笔為你收集整理的【可持久化线段树】【主席树】[HDU4417]Super Mario的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。