Group

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 208    Accepted Submission(s): 122

Problem Description

There are n men ,every man has an ID(1..n).their ID is unique. Whose ID is i and i-1 are friends, Whose ID is i and i+1 are friends. These n men stand in line. Now we select an interval of men to make some group. K men in a group can create K*K value. The value of an interval is sum of these value of groups.

 The people of same group's id must be continuous. Now we chose an interval of men and want to know there should be how many groups so the value of interval is max.

 

 

Input

First line is T indicate the case number.

For each case first line is n, m(1<=n ,m<=100000) indicate there are n men and m query.

Then a line have n number indicate the ID of men from left to right.

Next m line each line has two number L,R(1<=L<=R<=n),mean we want to know the answer of [L,R].

 

 

Output

For every query output a number indicate there should be how many group so that the sum of value is max.

 

 

Sample Input

1 5 2 3 1 2 5 4 1 5 2 4

 

 

Sample Output

1 2

 

 

Source

 

 

Recommend

zhuyuanchen520

 

题意：找到区间能，有多少组连续数字串 

 （转）思路：显然，我们要使得value最大，就要尽量将连续的ID分在一组，所以问题转化为求一个区间中连续ID区间的个数。我们从左往右扫描，依次考虑右端点为i的询问，设dp[l]为区间[l，i]的连续区间个数，po[i]为i出现的位置，若还未出现，则为0，设我们当前考虑的右端点为a[i]，首先我们假设a[i]不能和区间[1,i-1]中的任何一个数分到一组，则我们要将dp[1]到dp[i-1]全部加1，然后考虑po[a[i]+1]是否不为0，若不为0则说明a[i]-1已经在前面出现，则我们需要将dp[1]到dp[po[a[i]+1]]全部减一个1，因为a[i]可以和a[i]+1分为一组，则我们之前加的1是多余的。对于a[i]-1的情况同理。以上操作可以由线段树或者树状数组什么的实现，然后再将询问按照右端点从小到大排序，离线处理即可，以下是代码实现

线段树：

#include
     
      #include
      
       #include
       
        #include
        
         using namespace std;const int N=100010;#define L(rt) (rt<<1)#define R(rt) (rt<<1|1)struct Tree{    int l,r;    int num;    //num记录(l,r)区间内的组数  }tree[N<<2];struct node{    int l,r;    int id;}f[N];int cmp(node a,node b){    return a.r
         
          >1;    build(L,mid,L(rt));    build(mid+1,R,R(rt));}void update(int L,int R,int val,int rt){    if(tree[rt].l==L && tree[rt].r==R){        tree[rt].num+=val;        return ;    }    int mid=(tree[rt].l+tree[rt].r)>>1;    if(R<=mid)        update(L,R,val,L(rt));    else if(L>=mid+1)        update(L,R,val,R(rt));    else{        update(L,mid,val,L(rt));        update(mid+1,R,val,R(rt));    }    tree[rt].num=tree[L(rt)].num+tree[R(rt)].num;}int query(int L,int R,int rt){    if(tree[rt].l==L && tree[rt].r==R)        return tree[rt].num;    int mid=(tree[rt].l+tree[rt].r)>>1;    if(R<=mid)        return query(L,R,L(rt));    else if(L>=mid+1)        return query(L,R,R(rt));    else{        int a=query(L,mid,L(rt));        int b=query(mid+1,R,R(rt));        return a+b;    }}int main(){    //freopen("input.txt","r",stdin);    int t,n,m;    scanf("%d",&t);    while(t--){        scanf("%d%d",&n,&m);        for(int i=1;i<=n;i++){            scanf("%d",&a[i]);            loc[a[i]]=i;        }        for(int i=1;i<=m;i++){            scanf("%d%d",&f[i].l,&f[i].r);            f[i].id=i;        }        sort(f+1,f+m+1,cmp);    //将查找区间以右右区间递增排序，方便在更改线段树的时候查找，不会漏掉        build(1,n,1);        int i,j=1;        for(i=1;i<=n;i++){            update(i,i,1,1);    //每次新进的一位数，假设独立，所有已i结尾的区间都+1             if(a[i]
          
           1 && loc[a[i]-1]
          
         
        
       
      
     

 

#include
     
      #include
      
       #include
       
        #include
        
         using namespace std;const int N=100010;int n,m;int arr[N],a[N],loc[N],res[N];struct node{    int l,r;    int id;}f[N];int cmp(node a,node b){    return a.r
         
          0){        ans+=arr[i];        i-=lowbit(i);    }    return ans;}int main(){    //freopen("input.txt","r",stdin);    int t;  //刚开在这里使用 int t,n,m; 这里的局部变量影响了在main函数外面定义的全局变量n（默认值为0），导致我无奈了。。。。。    scanf("%d",&t);    while(t--){        scanf("%d%d",&n,&m);        for(int i=1;i<=n;i++){            scanf("%d",&a[i]);            loc[a[i]]=i;        }        for(int i=1;i<=m;i++){            scanf("%d%d",&f[i].l,&f[i].r);            f[i].id=i;        }        sort(f+1,f+m+1,cmp);    //将查找区间以右右区间递增排序，方便在更改线段树的时候查找，不会漏掉        memset(arr,0,sizeof(arr));        int i,j=1;        for(i=1;i<=n;i++){            update(i,1);    //每次新进的一位数，假设独立，所有已i结尾的区间都+1            if(a[i]
          
           1 && loc[a[i]-1]