POJ1733 Parity game(带权并查集+离散化)

2018 年 04 月 20 日 • 阅读: 1745 • 数据结构阅读设置

Parity game

Time Limit: 1000MSMemory Limit: 65536K
Total Submissions: 10777Accepted: 4139


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.


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).


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

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

Sample Output



CEOI 1999




使用并查集来维护。序列中1的数量可以理解为和,偶数用0表示,奇数用1表示。x和y之间1的数量s转换为y比x-1大s,具体可以参考HDOJ3038 How Many Answers Are Wrong这道题,还有就是数据太大,无法用数组保存,需要进行离散化处理。


#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 50010;
int F[maxn], sum[maxn], a[maxn], tmp[maxn];
struct Node
    int l, r;

int find(int x)
    if (x == F[x])
        return x;
    int t = find(F[x]);
    sum[x] = (sum[x] + sum[F[x]]) % 2; // 更新
    F[x] = t;
    return F[x];

int main()
    int n, m, x, y;
    char s[10];
    scanf("%d%d", &n, &m);
    int ans = m;
    for (int i = 1; i <= (m << 1); ++i)
        F[i] = i;
        sum[i] = tmp[i]= 0;
    bool flag = true;
    int k = 0;
    for (int i = 1; i <= m; ++i)
        scanf("%d%d%s", &node[i].l, &node[i].r, s);
        tmp[i] = ((s[0] == 'e') ? 0 : 1);
        a[k++] = node[i].l;
        a[k++] = node[i].r;
    sort(a, a + k); // 离散化
    int size = unique(a, a + k) - a;
    for (int i = 1; i <= m; ++i)
        x = lower_bound(a, a + size, node[i].l) - a + 1;
        y = lower_bound(a, a + size, node[i].r) - a + 1;
        x -= 1;
        int rootx = find(x), rooty = find(y);
        if (rootx != rooty) // 合并
            F[rooty] = rootx; // 让rootx为父节点
            sum[rooty] = (2 + sum[x] - sum[y] + tmp[i]) % 2; // 更新值
            if ((2 + sum[y] - sum[x]) % 2 != tmp[i]) // 判断
                if (flag)
                    ans = i - 1;
                    flag = false;
    printf("%d\n", ans);
    return 0;


The end.
2018-04-20 星期五
最后编辑于: 2018 年 05 月 06 日