HDU 4638 Group (offline algorithm | discrete line segment tree query), hdu4638

HDU 4638 Group (offline algorithm | discrete line segment tree query), hdu4638

Address: HDU 4638
First, I wrote a letter to the MO team. It's very simple.
The Code is as follows:

#include <iostream>#include <string.h>#include <math.h>#include <queue>#include <algorithm>#include <stdlib.h>#include <map>#include <set>#include <stdio.h>#include <time.h>using namespace std;#define LL long long#define pi acos(-1.0)#pragma comment(linker, "/STACK:1024000000")const int mod=1e9+7;const int INF=0x3f3f3f3f;const double eqs=1e-9;const int MAXN=100000+10;int a[MAXN];LL ha[MAXN], ans[MAXN], res;struct node {        int l, r, id, pos;} fei[MAXN];bool cmp(node x, node y){        return x.pos<y.pos||x.pos==y.pos&&x.r<y.r;}void reduce(int tmp){        if(ha[tmp-1]&&ha[tmp+1]) res++;        else if(!ha[tmp-1]&&!ha[tmp+1]) res--;        ha[tmp]=0;}void add(int tmp){        if(ha[tmp-1]&&ha[tmp+1]) res--;        else if(!ha[tmp-1]&&!ha[tmp+1]) res++;        ha[tmp]=1;}int main(){        int t, n, m, i, j, k, tmp, l, r;        scanf("%d",&t);        while(t--) {                scanf("%d%d",&n,&m);                for(i=1; i<=n; i++) {                        scanf("%d",&a[i]);                }                k=sqrt(n*1.0);                for(i=0; i<m; i++) {                        scanf("%d%d",&fei[i].l,&fei[i].r);                        fei[i].id=i;                        fei[i].pos=fei[i].l/k;                }                sort(fei,fei+m,cmp);                memset(ha,0,sizeof(ha));                l=1;                r=0;                res=0;                for(i=0; i<m; i++) {                        while(r>fei[i].r) {                                tmp=a[r];                                reduce(tmp);                                r--;                        }                        while(r<fei[i].r) {                                r++;                                tmp=a[r];                                add(tmp);                        }                        while(l<fei[i].l) {                                tmp=a[l];                                reduce(tmp);                                l++;                        }                        while(l>fei[i].l) {                                l--;                                tmp=a[l];                                add(tmp);                        }                        ans[fei[i].id]=res;                }                for(i=0; i<m; i++) {                        printf("%I64d\n",ans[i]);                }        }        return 0;}

Then I read the question report and found that the line segment tree and offline query can also be used. And it's amazing ..
First, save it offline. Maintenance is performed from left to right to maintain the relative number of the previous values for the current enumerated values. For the current number of I, first update the value of this number to 1, representing an independent string, find the numbers a [I]-1 and a [I] + 1 that do not exist. If yes, the numbers above cannot be independent strings, compared with a [I], a [I] string can exist in the total, so we need to separate a [I]-1 and a [I] + 1-1, respectively. then, in the query where the R value is I, You can directly sum the line tree. The Code is as follows:

#include <iostream>#include <string.h>#include <math.h>#include <queue>#include <algorithm>#include <stdlib.h>#include <map>#include <set>#include <stdio.h>#include <time.h>using namespace std;#define LL long long#define pi acos(-1.0)#pragma comment(linker, "/STACK:1024000000")#define root 1, n, 1#define lson l, mid, rt<<1#define rson mid+1, r, rt<<1|1const int mod=1e9+7;const int INF=0x3f3f3f3f;const double eqs=1e-9;const int MAXN=100000+10;int a[MAXN], pos[MAXN], sum[MAXN<<2], ans[MAXN];struct node{        int l, r, id;}fei[MAXN];bool cmp(node x, node y){        return x.r<y.r;}void PushUp(int rt){        sum[rt]=sum[rt<<1]+sum[rt<<1|1];}void Update(int p, int x, int l, int r, int rt){        if(l==r){                sum[rt]+=x;                return ;        }        int mid=l+r>>1;        if(p<=mid) Update(p,x,lson);        else Update(p,x,rson);        PushUp(rt);}int Query(int ll, int rr, int l, int r, int rt){        if(ll<=l&&rr>=r){                return sum[rt];        }        int mid=l+r>>1, ans=0;        if(ll<=mid) ans+=Query(ll,rr,lson);        if(rr>mid) ans+=Query(ll,rr,rson);        return ans;}int main(){        int t, n, m, i, j;        scanf("%d",&t);        while(t--){                scanf("%d%d",&n,&m);                for(i=1;i<=n;i++){                        scanf("%d",&a[i]);                        pos[a[i]]=i;                }                for(i=0;i<m;i++){                        scanf("%d%d",&fei[i].l,&fei[i].r);                        fei[i].id=i;                }                memset(sum,0,sizeof(sum));                sort(fei,fei+m,cmp);                j=0;                for(i=1;i<=n;i++){                        Update(i,1,root);                        if(a[i]>1&&pos[a[i]-1]<i) Update(pos[a[i]-1],-1,root);                        if(a[i]<n&&pos[a[i]+1]<i) Update(pos[a[i]+1],-1,root);                        while(j<m&&fei[j].r<=i){                                ans[fei[j].id]=Query(fei[j].l,i,root);                                j++;                        }                }                for(i=0;i<m;i++){                        printf("%d\n",ans[i]);                }        }        return 0;}

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