BZOJ4636 蒟蒻的数列

水题，造个线段树，把标记打线段树上，然后dfs一遍就行了

a可能等于b，判掉就好

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<iomanip>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<bitset>
using namespace std;
#define MAXN 1010
#define MAXM 10000010
#define ll long long
#define INF 1000000000
#define MOD 1000000007
#define eps 1e-8
int v[MAXM],son[MAXM][2];
int tot;
int n;
ll ans;
int rt;
void change(int &x,int y,int z,int l,int r,int cv){
if(l>r){
return ;
}
if(!x){
x=++tot;
}
if(y==l&&z==r){
v[x]=max(v[x],cv);
return ;
}
int mid=y+z>>1;
if(r<=mid){
change(son[x][0],y,mid,l,r,cv);
}else if(l>mid){
change(son[x][1],mid+1,z,l,r,cv);
}else{
change(son[x][0],y,mid,l,mid,cv);
change(son[x][1],mid+1,z,mid+1,r,cv);
}
}
void dfs(int x,int l,int r,int now){
if(!x){
return ;
}
now=max(now,v[x]);
if(!son[x][0]&&!son[x][1]){
ans+=(ll)(r-l+1)*now;
return ;
}
int mid=l+r>>1;
dfs(son[x][0],l,mid,now);
dfs(son[x][1],mid+1,r,now);
if(!son[x][0]){
ans+=(ll)(mid-l+1)*now;
}
if(!son[x][1]){
ans+=(ll)(r-mid)*now;
}
}
int main(){
int l,r,x;
scanf("%d",&n);
while(n--){
scanf("%d%d%d",&l,&r,&x);
change(rt,0,INF,l,r-1,x);
}
dfs(rt,0,INF,0);
printf("%lld\n",ans);
return 0;
}

/*

*/