POJ 1733 Parity game

时间：2015-10-04 15:54:12 阅读：184 评论：0 收藏：0 [点我收藏+]

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：

3

题意：有一个只包含0和1的序列，现在有些问题及其答案（问题是从a位置到b位置有偶数个1还是奇数个1），判断最早的那个不对的答案的位置，输出即可，但是由于这道题的取值范围比较大，所以需要离散化处理，（算是再次让我复习了一下）。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

const int N=100100;

struct node
{
    int a, b, c;
}no[N];
int f[N], Hash[N], r[N]; ///Hash数组存放离散化处理后的数据

int Find(int x)
{
    int k = f[x];

    if (f[x] != x)
    {
        f[x] = Find(f[x]);
        r[x] = (r[x]+r[k])%2;
    }

    return f[x];
}

int main ()
{
    int n, m, a, b, k, na, nb, i;
    char s[10];

    while (scanf("%d", &n) != EOF)
    {
        k = 0;

        scanf("%d", &m);
        for (i = 0; i < m; i++)
        {
            scanf("%d %d %s", &a, &b, s);

            no[i].a = a; no[i].b = b;
            if (s[0] == ‘e‘) no[i].c = 0;
            else no[i].c = 1;

            Hash[k++] = a; Hash[k++] = a-1;
            Hash[k++] = b; Hash[k++] = b-1;
        }

        sort(Hash, Hash+k); ///排序
        k = unique(Hash, Hash+k)-Hash; ///去重

        for (i = 0; i < k; i++)
        {
            f[i] = i;
            r[i] = 0;
        }

        for (i = 0; i < m; i++)
        {
            a = lower_bound(Hash, Hash+k, no[i].a-1)-Hash; ///二分查找no[i].a-1在Hash数组中的位置
            b = lower_bound(Hash, Hash+k, no[i].b)-Hash;

            na = Find(a); nb = Find(b);

            if (na == nb && (no[i].c+r[b])%2 != r[a])
                break;
            else if (na < nb)
            {
                f[na] = nb;
                r[na] = ((no[i].c-r[a])+r[b]+2)%2;
            }
            else if (na > nb)
            {
                f[nb] = na;
                r[nb] = ((r[a]-no[i].c)-r[b]+2)%2;
            }
        }

        printf("%d\n", i);
    }

    return 0;
}

POJ 1733 Parity game

原文：http://www.cnblogs.com/syhandll/p/4854543.html

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