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;

}