HDU 6031 Innumerable Ancestors【LCA】

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题目链接

题意：n个点形成的一棵树，根节点为1，给两个点的集合A和B，从A和B中各取出一点，要求这两个点的LCA最大（深），输出最大深度。

先求一遍LCA，因为两个点的LCA不可能比这两个点都要深，所以按深度从大到小排序，剪枝：如果当前的最大深度比比较到的点要深则直接退出。

倍增LCA

#include <iostream>#include <cstring>#include <cstdio>#include <cmath>#include <algorithm>#include <map>using namespace std;const int maxn = 100000 + 10;int e;int n;int deep[maxn];int p[maxn][20];int head[maxn];int num1[maxn];int num2[maxn];struct Edge{    int v, next;} edge[maxn * 2];void addedge(int x, int y){    edge[e].v = y;    edge[e].next = head[x];    head[x] = e ++;}void Init(){    memset(head, -1, sizeof(head));    memset(p, -1, sizeof(p));    e = 0;}void init(){    int i, j;    //p[i][j]表示i结点的第2^j祖先    for (j = 1; (1 << j) <= n; j++)        for (i = 1; i <= n; i++)            if (p[i][j - 1] != -1)                p[i][j] = p[p[i][j - 1]][j - 1]; //i的第2^j祖先就是i的第2^(j-1)祖先的第2^(j-1)祖先}int lca(int a, int b) {    int i, j;    if (deep[a] < deep[b])  swap(a, b);    for ( i = 0; (1 << i) <= deep[a]; i++ );    i--;    for ( j = i; j >= 0; j-- )     {        if (deep[a] - (1 << j) >= deep[b])         {            a = p[a][j];        }    }    if (a == b) return a;    for (j = i; j >= 0; j--)     {        if (p[a][j] != -1 && p[a][j] != p[b][j])         {            a = p[a][j];            b = p[b][j];        }    }    return p[a][0];}void dfs(int u, int fa, int d){    deep[u] = d;    for (int i = head[u]; i != -1; i = edge[i].next)    {        int to = edge[i].v;        if (fa != to)        {            p[to][0] = u; //p[x][0]保存x的父节点为u;            dfs(to, u, d + 1);        }    }}bool cmp(int x, int y){    return deep[x] > deep[y];}int main(){    int m;    while (~scanf("%d %d", &n , &m))    {        int a, b;        Init();        for (int i = 0;  i < n - 1; i ++)        {            scanf("%d%d", &a, &b);            addedge(a, b);            addedge(b, a);        }        deep[1] = 1;        dfs(1, 1, 1);        init();        int k1, k2;        while (m --)        {            int max1 = 0;            scanf("%d", &k1);            for (int i = 0; i < k1; i ++)   scanf("%d", &num1[i]);            scanf("%d", &k2);            for (int i = 0; i < k2; i ++)   scanf("%d", &num2[i]);            sort(num1, num1 + k1, cmp);            sort(num2, num2 + k2, cmp);            for (int i = 0 ; i < k1; i ++)            {                if (max1 >= deep[num1[i]])  break;                for (int j = 0; j < k2; j ++)                    max1 = max(max1, deep[lca(num1[i], num2[j])]);            }            printf("%d\n", max1);        }    }    return 0;}

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