POJ 2168 Joke with Turtles

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码代码的猿猿的AC之路

最新推荐文章于 2018-09-19 13:16:31 发布

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本文探讨了一种使用动态规划（DP）算法解决儿童智力谜题的方法，该谜题涉及三条乌龟在道路上爬行，并且部分乌龟在陈述时存在误导。文章详细介绍了算法实现步骤，并通过实例解释如何确定最少的撒谎者数量。

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脑子一抽用0初始化DP。。。TLE到爆了。。。。

Joke with Turtles

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 1203		Accepted: 319		Special Judge

Description

There is a famous joke-riddle for children:

Three turtles are crawling along a road. One turtle says: "There are two turtles ahead of me."
The other turtle says: "There are two turtles behind me." The third turtle says: "There are
two turtles ahead of me and two turtles behind me." How could this have happened?
The answer is -- the third turtle is lying!

Now in this problem you have n turtles crawling along a road. Some of them are crawling in a group, so that they do not see members of their group neither ahead nor behind them. Each turtle makes a statement of the form: "There are ai turtles crawling ahead of me and bi turtles crawling behind me." Your task is to find the minimal number of turtles that must be lying.
Let us formalize this task. Turtle i has xi coordinate. Some turtles may have the same coordinate. Turtle i tells the truth if and only if ai is the number of turtles such that xj > xi and bi is the number of turtles such that xj < xi. Otherwise, turtle i is lying.

Input

The first line of the input contains integer number n (1 <= n <= 1000). It is followed by n lines containing numbers ai and bi (0 <= ai, bi <= 1000) that describe statements of each turtle for i from 1 to n.

Output

On the first line of the output file write an integer number m -- the minimal number of turtles that must be lying, followed by m integers -- turtles that are lying. Turtles can be printed in any order. If there are different sets of m lying turtles, then print any of them.

Sample Input

Sample Output

2 1 4

Source

Northeastern Europe 2004

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn=1100;
    
int n,cnt[maxn][maxn],dp[maxn],s[maxn][3],pre[maxn],ans[maxn][2];

void init()
{
    memset(cnt,0,sizeof(cnt));
    memset(dp,-1,sizeof(dp));
}

int DP(int x)
{
    if(dp[x]!=-1) return dp[x]; 
    if(x==0) return dp[x]=0;

    int ret=0;
    for(int i=1;i<=x;i++)
    {
        if(ret<=DP(x-i)+min(cnt[x-i][n-x],i))
        {
            ret=dp[x-i]+min(cnt[x-i][n-x],i);
            pre[x]=x-i; 
            ans[x][0]=x-i;ans[x][1]=n-x;
        }
    }
    return dp[x]=ret;
}

int main()
{
while(scanf("%d",&n)!=EOF)
{
    init();
    for(int i=0;i<n;i++) 
    {
        scanf("%d%d",&s[i][0],&s[i][1]);
        cnt[s[i][0]][s[i][1]]++;
    }
    int res=DP(n);
    for(int i=n;i;i=pre[i])
    {
        for(int j=0,k=i-pre[i];k&&j<n;j++)
        {
            if(s[j][0]==ans[i][0]&&s[j][1]==ans[i][1])
            {
                k--;
                s[j][2]=1;
            }
        }
    }
    printf("%d",n-res);
    for(int i=0;i<n;i++) if(!s[i][2]) printf(" %d",i+1); 
    putchar(10);
}
    return 0;
}