• SPOJ


    SPOJ - 3267

    主席树的又一种写法。 从后端点开始添加主席树, 然后如果遇到出现过的元素先把那个点删除, 再更新树, 最后查询区间就好了。

     1 #include<bits/stdc++.h>
     2 using namespace std;
     3 #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
     4 #define LL long long
     5 #define ULL unsigned LL
     6 #define fi first
     7 #define se second
     8 #define pb push_back
     9 //#define lson l,m,rt<<1
    10 //#define rson m+1,r,rt<<1|1
    11 #define max3(a,b,c) max(a,max(b,c))
    12 #define min3(a,b,c) min(a,min(b,c))
    13 typedef pair<int,int> pll;
    14 const int INF = 0x3f3f3f3f;
    15 const LL mod =  (int)1e9+7;
    16 const int N = 3e4 + 100;
    17 const int M = 1e6 + 100;
    18 int root[N], lson[M], rson[M], cnt[M];
    19 int a[N];
    20 map<int,int> mp;
    21 int tot;
    22 int Build(int l, int r){
    23     int now = ++tot;
    24     cnt[now] = 0;
    25     if(l < r){
    26         int m = l+r >> 1;
    27         lson[now] = Build(l, m);
    28         rson[now] = Build(m+1, r);
    29     }
    30     return now;
    31 }
    32 int Update(int l, int r, int pre, int c, int v){
    33     int now = ++tot;
    34     cnt[now] = cnt[pre] + v;
    35     if(l < r){
    36         int m = l+r >> 1;
    37         if(c <= m){
    38             rson[now] = rson[pre];
    39             lson[now] = Update(l, m, lson[pre], c, v);
    40         }
    41         else {
    42             lson[now] = lson[pre];
    43             rson[now] = Update(m+1, r, rson[pre], c, v);
    44         }
    45     }
    46     return now;
    47 }
    48 int Query(int l, int r, int p, int R){
    49     if(l == r) return cnt[p];
    50     int m = l+r >> 1;
    51     if(R <= m){
    52         return Query(l, m, lson[p], R);
    53     }
    54     else {
    55         return Query(m+1,r,rson[p],R) + cnt[lson[p]];
    56     }
    57 }
    58 int main(){
    59     int n;
    60     scanf("%d", &n);
    61     for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
    62     root[n+1] = Build(1, n);
    63     for(int i = n; i >= 1; i--){
    64         if(mp[a[i]] == 0){
    65             root[i] = Update(1, n, root[i+1], i, 1);
    66         }
    67         else {
    68             int tmp = Update(1, n, root[i+1], mp[a[i]], -1);
    69             root[i] = Update(1, n, tmp, i, 1);
    70         }
    71         mp[a[i]] = i;
    72     }
    73     int m, l, r;
    74     scanf("%d", &m);
    75     for(int i = 1; i <= m; i++){
    76         scanf("%d%d", &l, &r);
    77         printf("%d
    ", Query(1,n,root[l],r));
    78     }
    79     return 0;
    80     
    81 }
    SPOJ-3267
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  • 原文地址:https://www.cnblogs.com/MingSD/p/9094322.html
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