UVALive 4764 Bing it（dp）

题目链接：https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2765点击打开链接

4764 - Bing it

Time limit: 3.000 seconds

I guess most of you played cards on the trip to Harbin, but I’m sure you have never played thefollowing card game. This card game hasN rounds and 100000 types of cards numbered from 1 to100000. A new card will be opened when each round begins. You can “bing” this new card. And if thecard you last “bing” is the same with this new one, you will get 1 point. You can ”bing” only one card,but you can change it into a new one. For example, the order of the 4 cards is 1 3 4 3. You can “bing”1 in the rst round and change it into 3 in the second round. You get no point in the third round, butget 1 point in the last round. Additionally, there is a special card 999. If you “bing” it and it is openedagain, you will get 3 point.

Given the order ofN cards, tell me the maximum points you can get.Input

The input le will contain multiple test cases. Each test case will consist of two lines. The rst line ofeach test case contains one integerN (2≤ N≤ 100000). The second line of each test case contains asequence of nintegers, indicating the order of cards. A single line with the number ‘0’ marks the endof input; do not process this case.

Output

For each input test case, print the maximum points you can get.

Sample Input

2
11
5
1 999 3 3 9990

Sample Output

13 

#include <iostream>#include <queue>#include <stdio.h>#include <stdlib.h>#include <stack>#include <limits>#include <string>#include <string.h>#include <vector>#include <set>#include <map>#include <algorithm>#include <math.h>#define maxn 100010using namespace std;set < int > point[100010];set < int > :: iterator it;int dp[maxn],a[maxn];int judge(int x,int y){    if(x==y&&x==999)        return 3;    else if(x==y)        return 1;    else        return 0;}int main(){    int n=0;    while(~scanf("%d",&n)&&n)    {        for(int i=0;i<=100000;i++)            {                point[i].clear();                dp[i]=0;                a[i]=0;            }        for(int i=0;i<n;i++)        {            scanf("%d",&a[i]);            point[a[i]].insert(i);        }        dp[0]=0;        for(int i=1;i<n;i++)        {            dp[i]=max(dp[i-1]+judge(a[i],a[i-1]),dp[i]);            for(it=point[a[i]].begin();it!=point[a[i]].end();it++)            {                if((*it)>=i-1)                    break;                dp[i]=max(dp[(*it)]+judge(a[i],a[(*it)]),dp[i]);            }        }        printf("%d\n",dp[n-1]);    }}