HDU 4699 Editor（双向链表）

双向链表直接模拟。

用一个辅助数组maxSum来维护一下前k项中[1,k]的最大和。

因为光标是一格一格的移动，所以每次光标右移的时候动态更新一下即可。

时间复杂度O(n)。

 

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>

using namespace std;

const int MAXN = 1000100;
const int INF = 1 << 30;

struct node
{
    int val;
    node *next;
    node *pre;
};

int sum[MAXN];
int maxSum[MAXN];

int main()
{
    //freopen( "1004.in", "r", stdin );
    //freopen( "test.txt", "w", stdout );
    int Q;
    node *head = ( node *)malloc(sizeof(node));

    while ( scanf( "%d", &Q ) == 1 )
    {
        head->pre = NULL;
        head->next = NULL;

        node *cur = head;
        maxSum[0] = -INF;
        sum[0] = 0;
        int pos = 0;

        char op[4];
        int a;
        for ( int i = 0; i < Q; ++i )
        {
            scanf( "%s", op );
            if ( op[0] == 'I' )
            {
                scanf( "%d", &a );

                ++pos;
                sum[pos] = sum[pos-1] + a;
                maxSum[pos] = max( sum[pos], maxSum[pos-1] );

                node *p = (node *)malloc( sizeof(node) );
                p->val = a;
                p->next = NULL;
                p->pre = cur;

                if ( cur->next )
                {
                    p->next = cur->next;
                    cur->next->pre = p;
                }

                cur->next = p;
                cur = cur->next;
            }
            else if ( op[0] == 'Q' )
            {
                scanf( "%d", &a );
                if ( pos < a ) a = pos;
                printf( "%d\n", maxSum[a] );
            }
            else if ( op[0] == 'L' )
            {
                if ( cur->pre != NULL )
                {
                    cur = cur->pre;
                    --pos;
                }
            }
            else if ( op[0] == 'R' )
            {
                //printf( "cur->next=%d %d\n", cur->next, NULL );
                if ( cur->next != NULL )
                {
                    cur = cur->next;
                    ++pos;
                    sum[pos] = sum[pos - 1] + cur->val;
                    maxSum[pos] = max( sum[pos], maxSum[pos-1] );
                }
            }
            else if ( op[0] == 'D' )
            {
                --pos;

                node *p = cur;
                cur = p->pre;

                cur->next = p->next;

                if ( p->next )
                    p->next->pre = cur;

                free(p);
            }
/*
            node *pp = head->next;
            while ( pp )
            {
                printf( "%d ", pp->val );
                pp = pp->next;
            }
            puts("");
*/
        }
    }
    return 0;
}