POJ 1330 Nearest Common Ancestors
题意:求LCA……这棵树是单向边,入度为0的为根,只有一组查询。
解法:st求LCA(只会这个)。dp[i][j]表示第i个点的第2j个祖先是谁,转移方程dp[i][j] = dp[dp[i][j - 1]][j - 1]。先dfs一边记录dp[i][0],和点的深度,求LCA的时候先把深度大的点升到同一深度,如果此时两个点重合则说明LCA就是当前点,否则再同时上升,他俩的父亲就是LCA。
代码:
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<string.h>
#include<math.h>
#include<limits.h>
#include<time.h>
#include<stdlib.h>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#define LL long long
using namespace std;
int n;
vector <int> edge[10005];
int indegree[10005];
int dp[10005][16];
int deep[10005];
void dfs(int root, int d)
{
deep[root] = d;
int len = edge[root].size();
for(int i = 0; i < len; i++)
{
dp[edge[root][i]][0] = root;
dfs(edge[root][i], d + 1);
}
}
void build(int root)
{
memset(dp, 0, sizeof dp);
memset(deep, 0, sizeof deep);
dfs(root, 0);
dp[root][0] = root;
for(int j = 1; j < 16; j++)
for(int i = 1; i <= n; i++)
dp[i][j] = dp[dp[i][j - 1]][j - 1];
}
int lca(int a, int b)
{
if(deep[a] > deep[b]) swap(a, b);
for(int i = 15; i >= 0; i--)
if(deep[dp[b][i]] >= deep[a]) b = dp[b][i];
if(a == b) return a;
for(int i = 15; i >= 0; i--)
{
if(dp[a][i] != dp[b][i])
{
a = dp[a][i];
b = dp[b][i];
}
}
return dp[a][0];
}
int main()
{
int T;
while(~scanf("%d", &T))
{
while(T--)
{
memset(indegree, 0, sizeof indegree);
for(int i = 0; i < 10005; i++)
edge[i].clear();
scanf("%d", &n);
for(int i = 0; i < n - 1; i++)
{
int a, b;
scanf("%d%d", &a, &b);
edge[a].push_back(b);
indegree[b]++;
}
int root;
for(int i = 1; i <= n; i++)
if(indegree[i] == 0)
{
root = i;
break;
}
build(root);
int a, b;
scanf("%d%d", &a, &b);
printf("%d\n", lca(a, b));
}
}
return 0;
}

浙公网安备 33010602011771号