不算难题,读懂就可以做。这题unique去重和lower_bound二分算是对这道题目一个比较好的运用,其实用set也可以做。自带去重了。
int n, m;
int x, y;
cin >> n;
for (int i = 1; i <= n; i++)
{
cin >> m;
node[i].len = m;
for (int j = 1; j <= m; j++)
cin >> node[i].a[j];
}
cin >> m;
for (int i = 0; i < m; i++)
{
cin >> x >> y;
sort(node[x].a + 1, node[x].a + node[x].len + 1);
sort(node[y].a + 1, node[y].a + node[y].len + 1);
int p1 = unique(node[x].a + 1, node[x].a + node[x].len + 1) - node[x].a - 1;
int p2 = unique(node[y].a + 1, node[y].a + node[y].len + 1) - node[y].a - 1;
int cnt = 0;
for (int j = 1; j <= p1; j++)
{
int t = node[x].a[j];
int p = lower_bound(node[y].a + 1, node[y].a + p2 + 1, t) - node[y].a;
if (t == node[y].a[p])
++cnt;
}
printf("%.2lf%%\n", 100 * (double(cnt) / double(p1 + p2 - cnt)));
}