hdu 5997 rausen loves cakes

Problem Description

Rausen loves cakes. One day, he bought n cakes
and the color of each cake is described as an integer in [1,1000000].
Rausen lines the cakes from left to right.

Before eating, rausen proceeds q operations
on cakes.

At one time point, rausen would replace all cakes of x color
with those of color y.

At another time point, rausen would calculate the number of segment colors in the interval[x,y].
A color segment is defined as an interval of one single color. For example,'1 4 4 1 1' involves 3 color segments.

Nevertheless, rausen finds that he cannot compile the statistics of color segments in the interval, which makes him weep like a helpless crybaby (bazinga). Please help rausen resolve the problem to placate him.

 

Input

There are multiple test cases. The first line of input contains an integer T indicating
the number of test cases. For each test case:

The first line contains 2 integers n,q.

In the following q lines,each
line contains 3 integers: op(1≤op≤2), x and y which
describe one operation:

If op=1,
then rausen is to proceed a substitution operation and this is when you replace cakes of color x with
those of color y.x and y satisfy (1≤x,y≤1000000).

if op=2,
then rausen is to proceed a counting operation and this is when you are required to input the color segments in the interval [x,y].x and y satisfy (1≤x≤y≤n)

(1≤T≤5),(1≤n≤105),(1≤q≤105)

 

Output

For every counting operation of each case, a single line contains one number as the answer.

 

Sample Input

1
5 3
1 4 4 10 1
2 1 5
1 4 10
2 3 5

 

Sample Output

4
2

 

Source

BestCoder Round #90

给你一个数列，有两种操作。

1、把所有值为a的改成b

2、询问[l,r]中一共有多少段

首先可以发现，段数肯定是越来越少。

考虑合并a,b。每次把小的并到大的上面，那么复杂度最终是log级别的

对于询问，我们把每段第一个位置赋值为1，用树状数组维护区间和。

暴力修改的时候顺便修改当前点的权值即可

#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
using namespace std;
struct line
{
int s,t;
int next;
}a[2000011];
int head[1000011];
int edge;
inline void addedge(int s,int t)
{
a[edge].next=head[s];
head[s]=edge;
a[edge].s=s;
a[edge].t=t;
}
int tr[100011];
int n;
inline int lowbit(int x)
{
return x&(-x);
}
inline void add(int x,int xx)
{
int i;
for(i=x;i<=n;i+=lowbit(i))
tr[i]+=xx;
}
inline int ask(int x)
{
int i,sum=0;
for(i=x;i>=1;i-=lowbit(i))
sum+=tr[i];
return sum;
}
int col[100011],sz[1000011];
int fx[1000011];
int main()
{
int T;
scanf("%d",&T);
while(T>0)
{
T--;
int q;
edge=0;
memset(head,0,sizeof(head));
memset(sz,0,sizeof(sz));
memset(tr,0,sizeof(tr));
scanf("%d%d",&n,&q);
int i,j;
for(i=1;i<=n;i++)
{
scanf("%d",&col[i]);
edge++;
addedge(col[i],i);
sz[col[i]]++;
if(col[i]!=col[i-1])
add(i,1);
}
for(i=1;i<=1000000;i++)
fx[i]=i;
int xx,x,y;
for(i=1;i<=q;i++)
{
scanf("%d%d%d",&xx,&x,&y);
if(xx==1)
{
int s=x,t=y;
x=fx[x];
y=fx[y];
if(sz[x]>sz[y])
{
int tt=x;
x=y;
y=tt;
tt=fx[s];
fx[s]=fx[t];
fx[t]=tt;
}
if(sz[x]==0||x==y)
continue;
int la=0;
for(j=head[x];j!=0;j=a[j].next)
{
int t=a[j].t;
if(col[t]!=col[t-1])
add(t,-1);
if(t<=n-1&&col[t+1]!=col[t])
add(t+1,-1);
col[t]=y;
if(col[t]!=col[t-1])
add(t,1);
if(t<=n-1&&col[t+1]!=col[t])
add(t+1,1);
la=j;
}
a[la].next=head[y];
head[y]=head[x];
head[x]=0;
sz[y]+=sz[x];
sz[x]=0;
}
else
printf("%d\n",ask(y)-ask(x-1)+(col[x]==col[x-1]));
}
}
return 0;
}