HDU-6031 Innumerable Ancestors(二分+树上倍增)

题意

给一棵树，$m$次询问，每次询问给两个点集问从两个点集中各取一个点的$LCA$的最大深度。

思路

二分答案。对于某个二分过程中得到的$Mid$，如果可行则两个点集在$Mid$所在的深度存在公共的祖先。枚举点集内的点，倍增找到他在这个深度的祖先就行。

代码

#include 
#define DBG(x) cerr << #x << " = " << x << endl

using namespace std;

const int N = 100000 + 5;
const int M = 200000 + 5;

int n, m, k1, k2;
int tot, head[N];
int A[N], B[N];
int d[N], f[N][25], maxd, t;
struct edgenode {
    int to, next;
}edge[M];
set st;
queue Q;

void addedge(int u, int v) {
    edge[tot].to = v;
    edge[tot].next = head[u];
    head[u] = tot++;
}

void bfs() {
    while(!Q.empty()) Q.pop();
    Q.push(1); d[1] = 1;
    while(!Q.empty()) {
        int tmp = Q.front(); Q.pop();
        for(int i = head[tmp]; i != -1; i = edge[i].next) {
            int v = edge[i].to;
            if(d[v] != 0) continue;
            d[v] = d[tmp] + 1;
            f[v][0] = tmp;
            for(int j = 1; j <= t; j++) f[v][j] = f[f[v][j - 1]][j - 1];
            Q.push(v);
        }
    }
}

int Find(int x, int dep) {
    if(d[x] < dep) return -1;
    if(d[x] == dep) return x;
    for(int i = t; i >= 0; i--) if(d[f[x][i]] >= dep) x = f[x][i];
    return x;
}

bool judge(int x) {
    st.clear();
    for(int i = 1; i <= k1; i++) {
        int fa = Find(A[i], x);
        if(fa != -1) st.insert(fa);
    }
    for(int i = 1; i <= k2; i++) {
        int fa = Find(B[i], x);
        if(st.find(fa) != st.end()) return true;
    }
    return false;
}

int calc(int L, int R) {
    int res = L;
    while(L <= R) {
        int Mid = (L + R) / 2;
        if(judge(Mid)) res = max(res, Mid), L = Mid + 1;
        else R = Mid - 1;
    }
    return res;
}

int main() {
    while(scanf("%d%d", &n, &m) != EOF){
        tot = 0;
        memset(head, -1, sizeof head);
        memset(f, 0, sizeof f);
        memset(d, 0, sizeof d);
        for(int i = 1, x, y; i <= n - 1; i++) {
            scanf("%d%d", &x, &y);
            addedge(x, y);
            addedge(y, x);
        }
        t = (int)(log(n) / log(2)) + 1;
        bfs();
        while(m--) {
            maxd = 1;
            scanf("%d", &k1);
            for(int i = 1; i <= k1; i++) scanf("%d", &A[i]), maxd = max(maxd, d[A[i]]);
            scanf("%d", &k2);
            for(int i = 1; i <= k2; i++) scanf("%d", &B[i]);
            printf("%d\n", calc(1, maxd));
        }
    }
    return 0;
}

忘了多组读入痛失1A...

转载于:https://www.cnblogs.com/DuskOB/p/10669269.html