URAL 1003 Parity

URAL_1003

    这个题目可以用并查集做，类似“食物链”的题目。只不过由于N的范围比较大，一开始可以先离散化一下。

#include<stdio.h>

#include<string.h>

#include<stdlib.h>

#define MAXD 100010

int N, Q, a[MAXD], p[MAXD], d[MAXD];

struct Que

{

    int x, y;

    char b[5];

}que[MAXD];

int cmp(const void *_p, const void *_q)

{

    int *p = (int *)_p, *q = (int *)_q;

    return *p < *q ? -1 : 1;

}

int find(int x)

{

    int fa;

    if(p[x] == x)

        return x;

    fa = find(p[x]);

    d[x] = (d[x] + d[p[x]]) % 2, p[x] = fa;

    return fa;

}

void init()

{

    int i, j, k;

    scanf("%d", &Q);

    for(i = 0; i < Q; i ++)

    {

        scanf("%d%d%s", &que[i].x, &que[i].y, que[i].b);

        a[i * 2 + 1] = que[i].x, a[i * 2 + 2] = que[i].y;

    }

    qsort(a, 2 * Q, sizeof(a[0]), cmp);

    a[0] = N = 0;

    for(i = 1; i <= 2 * Q; i ++)

        if(a[i] != a[i - 1])

            a[++ N] = a[i];

}

int BS(int x)

{

    int min = 1, max = N + 1, mid;

    for(;;)

    {

        mid = (min + max) >> 1;

        if(min == mid)

            break;

        if(a[mid] <= x)

            min = mid;

        else

            max = mid;

    }

    return mid;

}

void solve()

{

    int i, j, k, x, y, tx, ty, delta;

    for(i = 0; i <= N; i ++)

        p[i] = i, d[i] = 0;

    for(i = 0; i < Q; i ++)

    {

        x = BS(que[i].x) - 1, y = BS(que[i].y);

        tx = find(x), ty = find(y), delta = que[i].b[0] == 'e' ? 0 : 1;

        if(tx != ty)

            p[ty] = tx, d[ty] = (delta + d[x] - d[y] + 2) % 2;

        else

        {

            if(d[y] != (d[x] + delta) % 2)

                break;

        }

    }

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

}

int main()

{

    for(;;)

    {

        scanf("%d", &N);

        if(N == -1)

            break;

        init();

        solve();

    }

    return 0;

}