【离散化+并查集】POJ_1733_Parity game

最新推荐文章于 2020-05-02 15:41:10 发布

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最新推荐文章于 2020-05-02 15:41:10 发布

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本文介绍了一个名为Paritygame的博弈问题，玩家通过询问子序列中1的数量为奇数还是偶数来猜测整个由0和1组成的序列。文章详细阐述了问题背景、输入输出格式及示例，并提供了一段C++代码实现，用于验证朋友的回答是否存在矛盾。

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Parity game

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9578		Accepted: 3713

Description

Now and then you play the following game with your friend. Your friend writes down a sequence consisting of zeroes and ones. You choose a continuous subsequence (for example the subsequence from the third to the fifth digit inclusively) and ask him, whether this subsequence contains even or odd number of ones. Your friend answers your question and you can ask him about another subsequence and so on. Your task is to guess the entire sequence of numbers.

You suspect some of your friend's answers may not be correct and you want to convict him of falsehood. Thus you have decided to write a program to help you in this matter. The program will receive a series of your questions together with the answers you have received from your friend. The aim of this program is to find the first answer which is provably wrong, i.e. that there exists a sequence satisfying answers to all the previous questions, but no such sequence satisfies this answer.

Input

The first line of input contains one number, which is the length of the sequence of zeroes and ones. This length is less or equal to 1000000000. In the second line, there is one positive integer which is the number of questions asked and answers to them. The number of questions and answers is less or equal to 5000. The remaining lines specify questions and answers. Each line contains one question and the answer to this question: two integers (the position of the first and last digit in the chosen subsequence) and one word which is either `even' or `odd' (the answer, i.e. the parity of the number of ones in the chosen subsequence, where `even' means an even number of ones and `odd' means an odd number).

Output

There is only one line in output containing one integer X. Number X says that there exists a sequence of zeroes and ones satisfying first X parity conditions, but there exists none satisfying X+1 conditions. If there exists a sequence of zeroes and ones satisfying all the given conditions, then number X should be the number of all the questions asked.

Sample Input

10
5
1 2 even
3 4 odd
5 6 even
1 6 even
7 10 odd

Sample Output

Source

CEOI 1999

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=1e5+10;
int n,m,f[maxn],val[maxn],b[maxn],d[maxn],x[maxn],y[maxn];
void Init(){
    for(int i=0;i<=(int)(1e5);i++){
        f[i]=i;
        val[i]=0;
    }
}
int Find(int x){
    if(f[x]!=x){
        int temp=Find(f[x]);
        val[x]^=val[f[x]];
        f[x]=temp;
    }
    return f[x];
}
int main()
{
    char s[11];
    scanf("%d%d",&n,&m);
    Init();
    int idx=m,top=0;
    for(int i=0;i<m;i++){
        scanf("%d%d%s",&x[i],&y[i],s);
        if(s[0]=='e') d[i]=0;
        else d[i]=1;
        b[top++]=x[i];b[top++]=y[i];
        b[top++]=x[i]+1;//离散化防止原来不相邻的数现在相邻
        b[top++]=y[i]+1;
    }
    sort(b,b+top);
    int index=unique(b,b+top)-b;
    for(int i=0;i<m;i++){
        int xx=lower_bound(b,b+index,x[i])-b+1;
        int yy=lower_bound(b,b+index,y[i])-b+1;
        xx--;
        int fx=Find(xx);
        int fy=Find(yy);
        if(fx!=fy){
            f[fy]=fx;
          //  printf("^ = %d %d\n",val[xx]^val[yy],~(0));
            if(d[i])
                val[fy]=!(val[xx]^val[yy]);
            else val[fy]=(val[xx]^val[yy]);
        }
        else{
            int ans=(val[xx]^val[yy]);
            if(ans!=d[i]&&idx==m){
               idx=i;
            }
        }
       // printf("val[%d] = %d val[%d] = %d\n",xx,val[xx],yy,val[yy]);
    }
    printf("%d\n",idx);
    return 0;
}


/*
10
5
1 2 odd
4 6 even
1 6 odd
3 3 odd
1 2 odd

3
*/