hdu-4638

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.

InputFirst 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]. 
OutputFor 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

题意：给你N个数，然后给出M个查询；对于每一个查询，该区间内最少可以有多少个连续的序列；

题解：莫队（也可以用树，单位现在还不会，以后补上~~）；

AC代码为：

 #include<bits/stdc++.h>
 using namespace std;

 typedef long long ll;
 const int M =  + ;

 int a[M], vis[M], ans, ret[M];
 struct node
 {
     int id, l, r, pos;
 }v[M];

 int cmp(node x, node y)
 {
     if (x.pos<y.pos)
         return ;
     else if (x.pos == y.pos&&x.r<y.r)
         return ;
     return ;
 }

 inline void insert(int x)
 {
     x = a[x];
     vis[x] = ;
     if (vis[x - ] && vis[x + ])
     {
         ans--;
     }
     else if (!vis[x - ] && !vis[x + ])
     {
         ans++;
     }
 }

 inline void erase(int x)
 {
     x = a[x];
     vis[x] = ;
     if (vis[x - ] && vis[x + ])
     {
         ans++;
     }
     else if (!vis[x - ] && !vis[x + ])
     {
         ans--;
     }
 }

 int main()
 {
     int t, n, m, i, j, k;
     int l, r;
     scanf("%d", &t);
     while (t--)
     {
         scanf("%d%d", &n, &m);
         for (i = ; i <= n; i++)
             scanf("%d", &a[i]);
         int x = sqrt(n);
         for (i = ; i <= m; i++)
         {
             scanf("%d%d", &v[i].l, &v[i].r);
             v[i].id = i;
             v[i].pos = v[i].l / x;
         }
         sort(v + , v + m + , cmp);
         memset(vis, , sizeof(vis));
         ans = ;
         insert();
         int p = , q = ;
         for (i = ; i <= m; i++)
         {
             l = v[i].l;
             r = v[i].r;
             while (q<r) insert(++q);
             while (q>r) erase(q--);
             while (p<l) erase(p++);
             while (p>l) insert(--p);
             ret[v[i].id] = ans;
         }
         for (i = ; i <= m; i++)
             printf("%d\n", ret[i]);
     }
     return ;