题意:数列区间
思路:还算是简单的线段树操作,关键两点:
1 ,更新过程;
2,最后的数列标记,记住是最长连续区间长的。给组测试数据: 3 2 1 3 2 2 答案:1
代码:
#include<iostream>
#include<stdio.h>
#include<string>
#include<string.h>
using namespace std;
const int maxnode= 1000005;
class node
{
public:
int lt,rt;
int color;
int num;
} ;
class data
{
public:
int a,b;
};
data st[200003];
node tree[maxnode*4];
int col[maxnode],vis[maxnode];
int n,m;
void create(int lt,int rt,int index)
{
tree[index].lt =lt; tree[index].rt =rt;
tree[index].color =0;
if(lt==rt) return ;
int mid=(lt+rt)>>1;
create(lt,mid,index*2);
create(mid+1,rt,index*2+1);
}
void add(int lt,int rt,int index,int color)
{
if(tree[index].lt ==lt&&tree[index].rt ==rt){
tree[index].color =color;
return ;
}
if(tree[index].color >0) { //更新 至关重要
tree[index*2].color =tree[index].color ;
tree[index*2+1].color =tree[index].color ;
tree[index].color =-1;
}
tree[index].color =-1;
int mid;
mid=(tree[index].lt +tree[index].rt )>>1;
if(lt>mid)
add(lt,rt,index*2+1,color);
else if(rt<=mid)
add(lt,rt,index*2,color);
else {
add(lt,mid,index*2,color);
add(mid+1,rt,index*2+1,color);
}
}
void query(int lt,int rt,int index)
{
/*cout<<index<<endl;*/
if(tree[index].color >0) {
for(int i=tree[index].lt ;i<=tree[index].rt ;i++)
col[i]=tree[index].color ;
return ;
}
if(tree[index].color ==0) return ;
if(tree[index].color ==-1) {
int m=0,n=0,mid=(tree[index].lt +tree[index].rt )>>1;
if(lt>mid) query(lt,rt,index*2+1);
else if(rt<=mid) query(lt,rt,index*2);
else {
query(lt,mid,index*2);
query(mid+1,rt,index*2+1);
}
}
}
int main()
{
while(scanf("%d%d",&n,&m)!=EOF){
int i,a,b;
create(1,n,1);
for(i=0;i<m;i++){
scanf("%d%d",&a,&b);
add(a,b,1,i+1);
st[i].a =a;
st[i].b =b;
}
int maxsum=0;
memset(col,0,sizeof(col));
query(1,n,1);
/*for(i=0;i<m;i++){
query(st[i].a,st[i].b,1,i+1);
}*/
memset(vis,0,sizeof(vis));
/*for(i=1;i<=n;i++) cout<<col[i]<<" ";
cout<<endl;*/
for(i=1;i<=n;i++) {
if(col[i]&&!vis[i]) {
int t=1;
for(int j=i+1;j<=n;j++){
if(col[i]==col[j]) {
t++;
vis[j]=1;
}
else break;
}
maxsum=maxsum>t?maxsum:t;
}
}
printf("%d\n",maxsum);
}
return 0;
}