[HDU4417]Super Mario（主席树+离散化）

传送门

又是一道主席树模板题，注意数组从0开始，还有主席树耗费空间很大，数组开大点，之前开小了莫名其妙TLE。QAQ

——代码

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #define ls son[now][0], l, mid
 #define rs son[now][1], mid + 1, r

 using namespace std;

 const int N = ;

 int T, n, m, sz, tot, ans;
 int a[N], b[N], rt[N], son[ * N][], sum[ * N];

 inline void build(int &now, int l, int r)
 {
     now = ++tot;
     sum[now] = ;
     if(l == r) return;
     int mid = (l + r) >> ;
     build(ls);
     build(rs);
 }

 inline void update(int &now, int l, int r, int last, int x)
 {
     now = ++tot;
     son[now][] = son[last][];
     son[now][] = son[last][];
     sum[now] = sum[last] + ;
     if(l == r) return;
     int mid = (l + r) >> ;
     if(x <= mid) update(ls, son[now][], x);
     else update(rs, son[now][], x);
 }

 inline void query(int s, int t, int l, int r, int x)
 {
     if(l == r)
     {
         ans += sum[t] - sum[s];
         return;
     }
     int mid = (l + r) >> ;
     if(x <= mid) query(son[s][], son[t][], l, mid, x);
     else
     {
         ans += sum[son[t][]] - sum[son[s][]];
         query(son[s][], son[t][], mid + , r, x);
     }
 }

 inline void clear()
 {
     n = m = sz = tot = ;
     memset(a, , sizeof(a));
     memset(b, , sizeof(b));
     memset(sum, , sizeof(sum));
     memset(son, , sizeof(son));
     memset(rt, , sizeof(rt));
 }

 int main()
 {
     int i, j = , k, x, y, z;
     scanf("%d", &T);
     while(T--)
     {
         scanf("%d %d", &n, &m);
         for(i = ; i <= n; i++) scanf("%d", &a[i]), b[i] = a[i];
         sort(b, b + n + );
         sz = unique(b + , b + n + ) - (b + );
         tot = ;
         build(rt[], , sz);
         for(i = ; i <= n; i++)
         {
             k = lower_bound(b + , b + sz + , a[i]) - b;
             update(rt[i], , sz, rt[i - ], k);
         }
         printf("Case %d:\n", ++j);
         for(i = ; i <= m; i++)
         {
             scanf("%d %d %d", &x, &y, &z);
             k = upper_bound(b + , b + sz + , z) - b - ;
             if(k)
             {
                 ans = ;
                 query(rt[x], rt[y + ], , sz, k);
                 printf("%d\n", ans);
             }
             else printf("0\n");
         }
     }
     return ;
 }