题目链接
https://www.patest.cn/contests/gplt/L2-010
思路
因为 题意中 朋友的朋友 就是朋友 那么 朋友的关系 用 并查集 保存
但是 敌对关系 只有直接的敌对关系才是具有敌对关系 所以直接用结构体保存就好
AC代码
#include <cstdio>
#include <cstring>
#include <ctype.h>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <map>
#include <stack>
#include <set>
#include <numeric>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long LL;
const double PI = 3.14159265358979323846264338327;
const double E = 2.718281828459;
const double eps = 1e-6;
const int MAXN = 0x3f3f3f3f;
const int MINN = 0xc0c0c0c0;
const int maxn = 1e2 + 5;
const int MOD = 1e9 + 7;
int pre[maxn];
struct Node
{
vector <int> v;
};
map <int, Node> M;
int find(int x)
{
int r = x;
while (pre[r] != r)
r = pre[r];
pre[x] = r;
return r;
}
void join(int x, int y)
{
int fx = find(x), fy = find(y);
if (x != fy)
pre[fx] = fy;
}
bool Hos(int x, int y)
{
vector <int>::iterator it;
for (it = M[x].v.begin(); it != M[x].v.end(); it++)
{
if ((*it) == y)
return true;
}
return false;
}
int main()
{
int n, m, k;
scanf("%d%d%d", &n, &m, &k);
for (int i = 0; i < 2; i++)
{
for (int j = 0; j <= n; j++)
pre[j] = j;
}
int x, y, flag;
for (int i = 0; i < m; i++)
{
scanf("%d%d%d", &x, &y, &flag);
if (flag == 1)
join(x, y);
else
{
M[x].v.push_back(y);
M[y].v.push_back(x);
}
}
for (int i = 0; i < k; i++)
{
scanf("%d%d", &x, &y);
if (find(x) == find(y))
{
if (Hos(x, y) == true)
printf("OK but...
");
else
printf("No problem
");
}
else
{
if (Hos(x, y) == true)
printf("No way
");
else
printf("OK
");
}
}
}