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  • hdu 3333 Turing Tree

    Problem Description:
    After inventing Turing Tree, 3xian always felt boring when solving problems about intervals, because Turing Tree could easily have the solution. As well, wily 3xian made lots of new problems about intervals. So, today, this sick thing happens again...
    Now given a sequence of N numbers A1, A2, ..., AN and a number of Queries(i, j) (1≤i≤j≤N). For each Query(i, j), you are to caculate the sum of distinct values in the subsequence Ai, Ai+1, ..., Aj.
    View Code
      1 #include <cstdio>
      2 #include <cstring>
      3 #include <algorithm>
      4 using namespace std;
      5 #define lson l,m,rt<<1
      6 #define rson m+1,r,rt<<1|1
      7 #define maxn 30005
      8 struct node
      9 {
     10     __int64 sum;
     11 }setree[maxn<<2];
     12 int pre[maxn];
     13 __int64 sorted[maxn];
     14 __int64 num[maxn],ans[100005];
     15 struct op
     16 {
     17     int l,r,id;
     18 }mes[200005];
     19 void build(int l,int r,int rt)
     20 {
     21     setree[rt].sum=0;
     22     if(l==r)
     23     return;
     24     int m=(l+r)>>1;
     25     build(lson);
     26     build(rson);
     27 }
     28 void pushup(int rt)
     29 {
     30     setree[rt].sum=setree[rt<<1].sum+setree[rt<<1|1].sum;
     31 }
     32 void update(int l,int r,int rt,int num,__int64 val)
     33 {
     34     if(l==r){
     35         setree[rt].sum=val;
     36         return;
     37     }
     38     int m=(l+r)>>1;
     39     if(num<=m)
     40     update(lson,num,val);
     41     else
     42     update(rson,num,val);
     43     pushup(rt);
     44 } 
     45 __int64 query(int l,int r,int rt,int L,int R)
     46 {
     47     if(L<=l&&r<=R)
     48     return setree[rt].sum;
     49     int m=(l+r)>>1;
     50     __int64 ans=0;
     51     if(L<=m)
     52     ans+=query(lson,L,R);
     53     if(R>m)
     54     ans+=query(rson,L,R);
     55     return ans;
     56 }
     57 bool cmp(struct op a,struct op b)
     58 {
     59     if(a.r==b.r)
     60     return a.l>b.l;
     61     return a.r<b.r;
     62 }
     63 int binsearch(int l,int r,__int64 key)
     64 {
     65     int m=(l+r)>>1;
     66     if(sorted[m]==key)
     67     return m;
     68     if(key<sorted[m])
     69     return binsearch(l,m-1,key);
     70     return binsearch(m+1,r,key);
     71 }
     72 int main()
     73 {
     74     int t;
     75     scanf("%d",&t);
     76     while(t--){
     77         int n,m,cnt=1;
     78         scanf("%d",&n);
     79         for(int i=1;i<=n;i++){
     80         scanf("%I64d",num+i);
     81         sorted[i]=num[i];
     82         }        
     83         sort(sorted+1,sorted+n+1);
     84         for(int i=2;i<=n;i++)
     85         if(sorted[i]!=sorted[i-1])
     86         sorted[++cnt]=sorted[i];
     87         build(1,n,1);
     88         memset(pre,-1,sizeof(pre));
     89         scanf("%d",&m);
     90         for(int i=0;i<m;i++){
     91             scanf("%d%d",&mes[i].l,&mes[i].r);
     92             mes[i].id=i;
     93         }
     94         sort(mes,mes+m,cmp);
     95         int k=1;
     96         for(int i=0;i<m;i++){
     97             while(k<=mes[i].r){
     98                 int pos=binsearch(1,cnt,num[k]);
     99                 if(pre[pos]==-1){
    100                     pre[pos]=k;
    101                     update(1,n,1,k,num[k]);
    102                 }
    103                 else{
    104                     int pos1=pre[pos];
    105                     update(1,n,1,pos1,0);
    106                     pre[pos]=k;
    107                     update(1,n,1,k,num[k]);
    108                 }
    109                 k++;
    110             }
    111             ans[mes[i].id]=query(1,n,1,mes[i].l,mes[i].r);
    112         }
    113         for(int i=0;i<m;i++)
    114         printf("%I64d\n",ans[i]);
    115     }
    116     return 0;
    117 }
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  • 原文地址:https://www.cnblogs.com/kim888168/p/2879218.html
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