gplt L2-010. 排座位（普通並查集）

https://www.patest.cn/contests/gplt/L2-010

題意：中文題。

思路：剛開始一看兩者之間關係有多種，還以爲是帶權並查集，後來想想天梯賽不都是裸題麼。。只需要將是朋友關係的合併成一個集合，非朋友關係的用矩陣表達，兩判斷條件相與即可。

ps：暈着頭敲的，起太早了。。

#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <iostream>

using namespace std;

typedef long long ll;
const int N = 105;
const int INF = 0x3f3f3f3f;

int pre[N], relation[N][N];

int Find(int x)
{
    int r = x;
    while(r != pre[r])
        r = pre[r];
    int i = x;
    while(pre[i] != r)
    {
        int tmp = pre[i];
        pre[i] = r;
        i = tmp;
    }
    return r;
}

void Union(int p1, int p2)
{
    int x = Find(p1);
    int y = Find(p2);
    if(x != y)
    {
        pre[x] = y;
    }
}

void init()
{
    memset(relation, 0, sizeof(relation));
    for(int i = 1; i <= N; i++)
        pre[i] = i;
}

int main()
{
  //  freopen("in.txt", "r", stdin);
    int n, m, k, x, y, val;
    while(~scanf("%d%d%d", &n, &m, &k))
    {
        init();
        for(int i = 1; i <= m; i++)
        {
            scanf("%d%d%d", &x, &y, &val);
            relation[x][y] = relation[y][x] = val;
            if(val == 1) Union(x, y);
        }
        for(int i = 1; i <= k; i++)
        {
            scanf("%d%d", &x, &y);
            if(Find(x)==Find(y) && relation[x][y]!=-1 && relation[y][x]!=-1) printf("No problem\n");
            else if(Find(x)!=Find(y) && relation[x][y]!=-1 && relation[y][x]!=-1) printf("OK\n");
            else if(Find(x)==Find(y) && (relation[x][y]==-1 || relation[y][x]==-1)) printf("OK but...\n");
            else if(Find(x)!=Find(y) && relation[x][y]==-1 && relation[y][x]==-1) printf("No way\n");
        }
    }
    return 0;
}