hackerrank Week of Code 34 Same Occurrence

之前也想到了利用前缀数组，记录每一位的0~i的值，但是这样效率并不高。

题解的方法是每出现一次这个数字，记录一下下标，然后通过求区间的交，这样优化就不会超时了。

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<cmath>
#include<set>
#include<stack>
#define ll long long
#define pb push_back
#define max(x,y) ((x)>(y)?(x):(y))
#define min(x,y) ((x)>(y)?(y):(x))
#define cls(name,x) memset(name,x,sizeof(name))//0或-1
#define fs first
#define sc second
#define mp make_pair
#define L(x) (1<<x)
using namespace std;
const int inf=1e9+10;
const ll llinf=1e16+10;
const int maxn=8e3+10;
const int maxm=2e5+10;
const int mod=1e9+7;
int n,q;
int ans[maxn][maxn];
vector<int> v[maxn];
map<int,int> mapp;
int main()
{
    //freopen("in.txt","r",stdin);
    cls(ans,-1);
    scanf("%d %d",&n,&q);
    for(int i=1;i<=n;i++)
        v[i].pb(0);
    for(int i=1;i<=n;i++)
    {
        int t;
        scanf("%d",&t);
        if(mapp[t]==0)
        mapp[t]=i;
        v[mapp[t]].pb(i);
    }
    map<int,int> c;
    while(q--)
    {
        int x,y;
        scanf("%d %d",&x,&y);
        if(ans[mapp[x]][mapp[y]]!=-1)
        {
            printf("%d
",ans[mapp[x]][mapp[y]]);
            continue;
        }
        if(x==y||(mapp[x]==0&&mapp[y]==0))
        {
            ans[mapp[x]][mapp[y]]=n*(n+1)/2;
        }
        else if(mapp[x]==0||mapp[y]==0)
        {
            int t=(mapp[x]!=0?mapp[x]:mapp[y]);
            int sum=0,len;
            for(int i=1;i<v[t].size();i++)
            {
                int len=v[t][i]-v[t][i-1]-1;
                sum+=len*(len+1)/2;
            }
            len=n-v[t][v[t].size()-1];
            sum+=len*(len+1)/2;
            ans[mapp[x]][mapp[y]]=sum;
        }
        else
        {
            c.clear();
            int sum=0;
            int t1=mapp[x],t2=mapp[y];
            /*for(int i=0;i<v[t1].size();i++)
                printf("%d ",v[t1][i]);
            printf("
");
            for(int i=0;i<v[t2].size();i++)
                printf("%d ",v[t2][i]);
            printf("
");*/
            int xl,xr,idx1=0;
            int yl,yr,idx2=0;
            while(idx1<v[t1].size()&&idx2<v[t2].size())
            {
                xl=v[t1][idx1];
                if(idx1+1>=v[t1].size())
                    xr=n;
                else xr=v[t1][idx1+1]-1;

                yl=v[t2][idx2];
                if(idx2+1>=v[t2].size())
                    yr=n;
                else yr=v[t2][idx2+1]-1;

                //printf("[%d %d:%d] [%d %d:%d]
",xl,xr,idx1,yl,yr,idx2);
                int len=(min(xr,yr)-max(xl,yl)+1);
                sum+=c[idx1-idx2]*len+len*(len-1)/2;
                c[idx1-idx2]+=len;
                if(xr<yr) idx1++;
                else if(xr>yr) idx2++;
                else {idx1++;idx2++;}
            }
            ans[mapp[x]][mapp[y]]=sum;
        }
        printf("%d
",ans[mapp[x]][mapp[y]]);
    }
    return 0;
}