hdu 4638 Group（莫队算法+分块）

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

Total Submission(s): 1966 Accepted Submission(s): 1012



[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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].

[align=left]Output[/align]
For every query output a number indicate there should be how many group so that the sum of value is max.

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

1
2

[align=left]Source[/align]
2013 Multi-University Training Contest 4

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题解：莫队算法+分块

脑残错误害死人啊。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define N 100003
using namespace std;
int n,m,t;
int ans1
,num
,b
,p
,vis
,belong
,a
;
struct data
{
int l,r,num;
}; data a1
;
int cmp1(data a,data b)
{
if (belong[a.l]==belong[b.l])
return a.r<b.r;
return belong[a.l]<belong[b.l];
}
int main()
{
scanf("%d",&t);
for (int k=1;k<=t;k++)
{
scanf("%d%d",&n,&m);
for (int i=1;i<=n;i++)
scanf("%d",&a[i]);
for (int i=1;i<=m;i++)
scanf("%d%d",&a1[i].l,&a1[i].r),a1[i].num=i;
int size=ceil(sqrt(n));
for (int i=1;i<=n;i++)
belong[i]=(i-1)/size+1;
sort(a1+1,a1+m+1,cmp1);
memset(vis,0,sizeof(vis));
memset(ans1,0,sizeof(ans1));
int l=1;  int r=0; int ans=0;
for (int i=1;i<=m;i++)
{
while (r<a1[i].r)
{
r++;
vis[a[r]]=1;
if (vis[a[r]-1]&&vis[a[r]+1])
ans--;
if (!vis[a[r]-1]&&!vis[a[r]+1])
ans++;
}
while (r>a1[i].r)
{
vis[a[r]]=0;
if (vis[a[r]-1]&&vis[a[r]+1])
ans++;
if (!vis[a[r]-1]&&!vis[a[r]+1])
ans--;
r--;
}
while (l>a1[i].l)
{
l--;
if (vis[a[l]-1]&&vis[a[l]+1])
ans--;
if (!vis[a[l]-1]&&!vis[a[l]+1])
ans++;
vis[a[l]]=1;
}
while (l<a1[i].l)
{
vis[a[l]]=0;
if (vis[a[l]-1]&&vis[a[l]+1])
ans++;
if (!vis[a[l]-1]&&!vis[a[l]+1])
ans--;
l++;
}
ans1[a1[i].num]=ans;
}
for (int i=1;i<=m;i++)
printf("%d\n",ans1[i]);
}
}