codeforces 1167E

题意：删除原数组中值域为[ l,r ]的数字，使数组中剩下的数单调递增

思路：预处理每个数值的左边界和右边界，那么结果中的形式一定是这样的

           [_____1________]              [______5_______]                  [_____10________]

           3                            7             10                      15                 20                           23

           又因为只能删除一段值域，所以一定有一边一开始就是排好序的，所以我们预处理前缀为上图最长的序列和后缀为上图最长的序列。这两边具有单调性，所以左指针右移时右指针一定右移或不动。有单调性质的序列可以用指针移动来做。

#include 
using namespace std;
typedef long long LL;
LL a[1000005],vis[1000005],l[1000005],r[1000005];
vector ve;
const LL inf = 0x3f3f3f3f3f3f3f3f;
void init( LL n ){
    for( LL i =0;i <= n;i++ ) vis[i]= 0;
}
int main()
{
    LL n,x,ans = 0,mn = inf;
    scanf("%I64d%I64d",&n,&x);
    for( LL i = 1;i <= n;i++ ){
        scanf("%I64d",&a[i]);
        ve.push_back( a[i] );
        mn = min( mn,a[i] );
    }
    sort( ve.begin(),ve.end() );
    init(x);
    for( LL i  = 1;i <= n;i++ ){
        if( !vis[ a[i] ] ){
            vis[ a[i] ] = 1;
            l [ a[i] ] = i;
        }
    }
    init(x);
    for( LL i = n;i >= 1;i-- ){
        if( !vis[ a[i]  ] ){
            vis[ a[i] ] = 1;
            r[ a[i] ] = i;
        }
    }
    auto t = unique( ve.begin(),ve.end() );
    ve.erase( t,ve.end() );
    LL le = 0,ri = ve.size() - 1;
    while( le +1 < ve.size() && l[ ve[le+1] ] > r[ ve[le] ] ) le++;    //指针移动写法                        
    while( ri-1 >= 0 && r[ ve[ri-1] ] < l[ ve[ri] ]  ) ri--;
    if( le >= ri ){
        printf("%I64d",x*( x-1 )/2 + x);
        return 0;
    }
    ans += (ve[0] - 1 - 1 + 2) *( x - (ve[ri-1]+1)+2 );
    LL L = 0,R = ri;
    for( ; L < ve.size() && L <= le;L++ ){
        while( R < ve.size() && l[ ve[R] ] <= r[ ve[L] ] ) R++;   //指针移动写法
        ans += (x - (ve[R-1] +1) + 2) *( ve[L+1] - 1 - (ve[L] + 1) + 2 );
    }
    printf("%I64d\n",ans);
    return 0;
}
/* 10 100
1 5 6 6 9 5 5 3 2 6
*/