hdu 4293 poj 2168 dp

Groups

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 153    Accepted Submission(s): 66

Problem Description

　　After the regional contest, all the ACMers are walking alone a very long avenue to the dining hall in groups. Groups can vary in size for kinds of reasons, which means, several players could walk together, forming a group.
　　As the leader of the volunteers, you want to know where each player is. So you call every player on the road, and get the reply like “Well, there are A_i players in front of our group, as well as B_i players are following us.” from the i^th player.
　　You may assume that only N players walk in their way, and you get N information, one from each player.
　　When you collected all the information, you found that you’re provided with wrong information. You would like to figure out, in the best situation, the number of people who provide correct information. By saying “the best situation” we mean as many people as possible are providing correct information.

 

Input

　　There’re several test cases.
　　In each test case, the first line contains a single integer N (1 <= N <= 500) denoting the number of players along the avenue. The following N lines specify the players. Each of them contains two integers A_i and B_i (0 <= A_i,B_i < N) separated by single spaces.
　　Please process until EOF (End Of File).

 

Output

　　For each test case your program should output a single integer M, the maximum number of players providing correct information.

 

Sample Input

32 00 22 232 00 22 2

 

Sample Output

22
Hint
The third player must be making a mistake, since only 3 plays exist.
 

 

Source

///和 poj 2168 非常类似，坑爹的网络赛///dp[i][j]为前j个人中，i+1到j为一组时，说真话的最大人数。#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>using namespace std;const int maxn=502;int p[maxn],s[maxn][maxn],dp[maxn][maxn],n;int main(){    while(~scanf("%d",&n))    {        memset(dp,0,sizeof(dp));        memset(p,0,sizeof(p));        memset(s,0,sizeof(s));        int i,j,pr,next;        for(i=1;i<=n;i++)        {            scanf("%d%d",&pr,&next);            if(pr+next<n) s[pr][next]++;            if(s[pr][next]>n-pr-next)///预判同一组人数是否符合要求            {                s[pr][next]=n-pr-next;            }        }        for(j=1;j<=n;j++)        {            for(i=0;i<j;i++)            {                dp[i][j]=p[i];                dp[i][j]+=s[i][n-j];                if(p[j]<dp[i][j]) p[j]=dp[i][j];            }        }        int ans=0;        for(i=0;i<n;i++)        ans=max(ans,dp[i][n]);        printf("%d\n",ans);    }    return 0;}