Parity game

Parity game

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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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

 并查集+离散化：

　　　　首先了解怎么用并查集，数列是由0和1组成的，0和1总是有很多特别的解法，例如此题，ran[x]表示总和sum[0~x]，同时ta也是1的个数num[0~x]，而题中只有两种表达，odd和even所以就更方便了。并且num[l~r] = ran[r] - ran[l-1];所以输入num[l~r] = odd or even -> ran[r] - ran[l-1] = 1 or 0;并查集内ran值的关系搞出来了，就so easy 了。

　　　　然后是离散化，第一次打离散代码，忘记此题数组范围需要扩大2倍了，注意！先搞个数组将输入值记录下来 -> sort一下 -> unique函数去重  -> 在运用数据时二分搜索找到相对应的序号值;就这样了。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int pre[5005*2];
int ran[5005*2];
int num[5005*2];
int a[5005],b[5005];
char ch[5005][10];
int n,cnt;
void init(){
    for(int i = 0; i < cnt; i++){
        pre[i] = i;
        ran[i] = 0;
    }
}
int findl(int x){
    int l = 0,r = cnt-1;
    while(l <= r){
        int mid = (l+r)>>1;
        if(num[mid] == x)return mid;
        else if(num[mid] > x){
            r = mid-1;
        }
        else l = mid+1;
    }
}
int find(int x){
    if(pre[x] == x)
        return pre[x];
    int fx = pre[x];
    pre[x] = find(pre[x]);
    ran[x] += ran[fx];
    ran[x] %= 2;
    return pre[x];
}
int main(){
    int T,x,y,d;
    bool flag = 0;
    scanf("%d%d",&n,&T);
    int ans = T;
    cnt = 0;
    for(int i = 0; i < T; i++){
        scanf("%d%d%s",&a[i],&b[i],ch[i]);
        num[cnt++] = --a[i];
        num[cnt++] = b[i];
    }
    sort(num,num+cnt);
    cnt = unique(num,num+cnt) - num;
    init();
    for(int i = 0; i < T; i++){
        x = findl(a[i]);
        y = findl(b[i]);
        if(ch[i][0] == 'e')
            d = 0;
        else d = 1;
        int fx = find(x);
        int fy = find(y);
        if(fx == fy){
            if((ran[x] - ran[y] + 2) % 2 != d && !flag){
                flag = 1;ans = i;
            }
        }
        else {
            pre[fx] = fy;
            ran[fx] = (ran[y] - ran[x] + 2 + d) % 2;
        }
    }
    printf("%d
",ans);
    return 0;
}

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int pre[5005*2];
int ran[5005*2];
int num[5005*2];
int a[5005],b[5005];
char ch[5005][10];
int n,cnt;
void init(){
    for(int i = 0; i < cnt; i++){
        pre[i] = i;
        ran[i] = 0;
    }
}
int findl(int x){
    int l = 0,r = cnt-1;
    while(l <= r){
        int mid = (l+r)>>1;
        if(num[mid] == x)return mid;
        else if(num[mid] > x){
            r = mid-1;
        }
        else l = mid+1;
    }
}
int find(int x){
    if(pre[x] == x)
        return pre[x];
    int fx = pre[x];
    pre[x] = find(pre[x]);
    ran[x] ^= ran[fx];
    return pre[x];
}
int main(){
    int T,x,y,d;
    bool flag = 0;
    scanf("%d%d",&n,&T);
    int ans = T;
    cnt = 0;
    for(int i = 0; i < T; i++){
        scanf("%d%d%s",&a[i],&b[i],ch[i]);
        num[cnt++] = --a[i];
        num[cnt++] = b[i];
    }
    sort(num,num+cnt);
    cnt = unique(num,num+cnt) - num;
    init();
    for(int i = 0; i < T; i++){
        x = findl(a[i]);
        y = findl(b[i]);
        if(ch[i][0] == 'e')
            d = 0;
        else d = 1;
        int fx = find(x);
        int fy = find(y);
        if(fx == fy){
            if(ran[x]^ran[y] != d && !flag){
                flag = 1;ans = i;
            }
        }
        else {
            pre[fx] = fy;
            ran[fx] = ran[y]^ran[x]^d;
        }
    }
    printf("%d
",ans);
    return 0;
}