LCA问题,用了离线的tarjan算法。输入输出参考了博客http://www.cnblogs.com/rainydays/archive/2011/06/20/2085503.html
tarjan算法是用了dfs+并查集的方式做的。这里输入输出有个不错的地方,就是用scanf("%[^0-9]", st);跳过非数字。
里面用数组g来表示多维的树,还用并查集的id数组的-1来表示是否访问。
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std; #define MAXN 909 /*
g for the edges
hasroot for whether the node is under root, it helps to identify the root
id for disjoint-set
lca for LCA of two nodes
sum for the count for ancestors in result
*/
int n, root;
bool g[MAXN][MAXN], hasroot[MAXN];
int id[MAXN], lca[MAXN][MAXN];
int sum[MAXN]; void input()
{
int a, b, m;
char str[100];
memset(g, 0, sizeof(g));
memset(hasroot, 0, sizeof(hasroot));
for (int i = 0; i < n; i++)
{
scanf("%d", &a);
a--;
scanf("%[^0-9]", str);
scanf("%d", &m);
scanf("%[^0-9]", str);
for (int i = 0; i < m; i++)
{
scanf("%d", &b);
b--;
hasroot[b] =true;
g[a][b] = g[b][a] =true;
}
}
for (int i = 0; i < n; i++)
if (!hasroot[i])
{
root = i;
break;
}
} // for disjoint-set
int find(int i)
{
if (id[i] == i)
return i;
return id[i] = find(id[i]);;
} void merge(int i, int j)
{
id[find(i)] = find(j);
} // do the tarjan algo and update lca table
void tarjan(int rt)
{
id[rt] = rt;
// id[k] != -1 means visited
for (int i = 0; i < n; i++)
if (g[rt][i] && id[i] == -1)
{
tarjan(i);
merge(i, rt); // the order matters, because of the implementaion of merge
}
for (int i = 0; i < n; i++)
if (id[i] != -1)
lca[rt][i] = lca[i][rt] = find(i);
} void solve()
{
int m;
char str[100];
scanf("%d", &m);
for (int i =0; i < m; i++)
{
int a, b;
scanf("%[^0-9]", str);
scanf("%d", &a);
scanf("%[^0-9]", str);
scanf("%d", &b);
a--;
b--;
sum[lca[a][b]]++;
}
for (int i =0; i < n; i++)
if (sum[i])
printf("%d:%d\n", i + 1, sum[i]);
} int main()
{
//freopen("d:\\\\t.txt", "r", stdin);
while (scanf("%d", &n) != EOF)
{
char str[100];
input();
memset(id, -1, sizeof(id));
memset(sum, 0, sizeof(sum));
tarjan(root);
solve();
scanf("%[^0-9]", str);
}
return 0;
}