hdu 4747 Mex (线段树)
不错的思维题，犀利的线段树。解题思路百度很多。。我那蹩脚的表达能力，就不误导大家了。
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<map>
#define ll long long
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
using namespace std ;
const int maxn = 222222 ;
map<int,int> mp ;
int col[maxn<<2] , b[maxn] , a[maxn] , mx[maxn<<2] ;
ll ans = 0 , sum[maxn<<2] ;
int nxt[maxn] ;
inline void push_up ( int rt ) {
sum[rt] = sum[rt<<1] + sum[rt<<1|1] ;
mx[rt] = mx[rt<<1|1] ;
}
void push_down ( int rt , int m ) {
if ( col[rt] != -1 ) {
int ls = rt << 1 , rs = rt << 1 | 1 ;
mx[ls] = mx[rs] = col[ls] = col[rs] = col[rt] ;
sum[ls] = (ll) col[ls] * ( m - ( m >> 1 ) ) ;
sum[rs] = (ll) col[rs] * ( m >> 1 ) ;
col[rt] = -1 ;
}
}
void build ( int l , int r , int rt ) {
col[rt] = -1 ;
if ( l == r ) {
mx[rt] = sum[rt] = b[l] ;
return ;
}
int m = ( l + r ) >> 1 ;
build ( lson ) ;
build ( rson ) ;
push_up ( rt ) ;
}
int find ( int l , int r , int rt , int v ) {
if ( l == r ) return l ;
push_down ( rt , r - l + 1 ) ;
int m = ( l + r ) >> 1 ;
if ( mx[rt<<1] >= v ) return find ( lson , v ) ;
else return find ( rson , v ) ;
}
void update ( int a , int b , int c , int l , int r , int rt ) {
if ( a <= l && r <= b ) {
col[rt] = c ;
sum[rt] = (ll) c * ( r - l + 1 ) ;
mx[rt] = c ;
return ;
}
push_down ( rt , r - l + 1 ) ;
int m = ( l + r ) >> 1 ;
if ( a <= m ) update ( a , b , c , lson ) ;
if ( m < b ) update ( a , b , c , rson ) ;
push_up ( rt ) ;
}
void update ( int a , int l , int r , int rt ) {
if ( l == r ) {
mx[rt] = -1 ;
sum[rt] = 0 ;
return ;
}
push_down ( rt , r - l + 1 ) ;
int m = ( l + r ) >> 1 ;
if ( a <= m ) update ( a , lson ) ;
else update ( a , rson ) ;
push_up ( rt ) ;
}
void solve ( int n ) {
int i , j , k , l , r ;
for ( i = 1 ; i <= n ; i ++ ) {
ans += sum[1] ;
update ( i , 1 , n , 1 ) ;
r = nxt[i] ;
if ( mx[1] < a[i] ) continue ;
l = find ( 1 , n , 1 , a[i] ) ;
if ( l <= r ) update ( l , r , a[i] , 1 , n , 1 ) ;
}
}
int main () {
int n , i , j , k ;
while ( scanf ( "%d" , &n ) != EOF ) {
if ( n == 0 ) break ;
ans = 0 ;
mp.clear () ;
for ( i = 1 ; i <= n ; i ++ ) scanf ( "%d" , &a[i] ) ;
mp[a[1]] = 1 ;
if ( a[1] == 0 ) k = b[1] = 1 ;
else k = b[1] = 0 ;
for ( i = 2 ; i <= n ; i ++ ) {
mp[a[i]] = 1 ;
while ( mp[k] ) k ++ ;
b[i] = k ;
}
mp.clear () ;
for ( i = n ; i>= 1 ; i -- ) {
if ( !mp[a[i]] ) nxt[i] = n ;
else nxt[i] = mp[a[i]] - 1 ;
mp[a[i]] = i ;
}
build ( 1 , n , 1 ) ;
solve ( n ) ;
printf ( "%I64d\n" , ans ) ;
}
return 0 ;
}