HDU - 3078 Network(暴力＋LCA)

题目大意：给出n个点的权值，m条边，2种操作

0 u num，将第u个点的权值改成num

k u v,询问u到v这条路上第k大的权值点

解题思路：该点的话直接该，找第k大的话直接暴力

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

#define N 80010
#define M 160010

struct Edge{
    int to, next, val;
}E[M];

int head[N], val[N], depth[M], ver[M], first[M], dis[M], dp[M][62], pre[N];
int tot, n, m;
bool vis[N];

void AddEdge(int u, int v) {
    E[tot].to = v; E[tot].next = head[u]; E[tot].val = 1; head[u] = tot++;
    u = u ^ v; v = u ^ v; u = u ^ v;
    E[tot].to = v; E[tot].next = head[u]; E[tot].val = 1; head[u] = tot++;
}

void init() {
    for (int i = 1; i <= n; i++)
        scanf("%d", &val[i]);
    int u, v;
    memset(head, -1, sizeof(head));
    tot = 0;

    for (int i = 1; i < n; i++) {
        scanf("%d%d", &u, &v);
        AddEdge(u, v);
    }
}

void dfs(int u, int dep) {
    vis[u] = true; ver[++tot] = u; first[u] = tot; depth[tot] = dep;
    for (int i = head[u]; ~i; i = E[i].next) {
        int v = E[i].to;
        if (!vis[v]) {
            dis[v] = dis[u] + E[i].val;
            pre[v] = u;
            dfs(v, dep+1);
            ver[++tot] = u; depth[tot] = dep;
        }
    }
} 

void RMQ() {
    for (int i = 1; i <= tot; i++)
        dp[i][0] = i;
    int a, b;
    for (int j = 1; (1 << j) <= tot; j++)
        for (int i = 1; i + (1 << j) - 1 <= tot; i++) {
            a = dp[i][j - 1];
            b = dp[i + (1 << (j - 1))][j - 1];
            if (depth[a] < depth[b])
                dp[i][j] = a;
            else
                dp[i][j] = b;
        }
}

int Query(int x, int y) {
    int k = 0;
    while (1 << (k + 1) <= (y - x + 1)) k++;
    int a = dp[x][k];
    int b = dp[y - (1 << k) + 1][k];
    if (depth[a] < depth[b])
        return a;
    return b;
}

int LCA(int a, int b) {
    a = first[a];
    b = first[b];
    if (a > b) {
        a = a ^ b; b = a ^ b; a = a ^ b;
    }
    int c = Query(a, b);
    return ver[c];
}

int num[N], cnt;

void path(int a, int lca) {
    int t = a;
    while (t != lca) {
        num[cnt++] = val[t];
        t = pre[t];
    }
}

bool cmp(const int a, const int b) {
    return a > b;
}

void find(int k, int a, int b) {
    int lca = LCA(a,b);
    if (dis[a] + dis[b] - 2 * dis[lca] + 1 < k) {
        printf("invalid request!\n");
        return ;
    }
    else {
        cnt = 0;
        path(a, lca);
        path(b, lca);
        num[cnt++] = val[lca];
        sort(num, num + cnt, cmp);
        printf("%d\n", num[k - 1]);
    }
}

void solve() {
    memset(vis, 0, sizeof(vis));
    tot = 0;
    dis[1] = 0;
    pre[1] = 1;
    dfs(1,1);
    RMQ();

    int k, a, b;
    while (m--) {
        scanf("%d%d%d", &k, &a, &b);
        if (!k) {
            val[a]=  b;
        }
        else {
            find(k, a, b);
        }
    }
}

int main() {
    while (scanf("%d%d", &n, &m) != EOF) {
        init();
        solve();
    }
    return 0;
}

版权声明：本文为博主原创文章，未经博主允许不得转载。