Codeforces 827D Best Edge Weight 倍增 + 并查集 || 倍增 + 压倍增标记 （看题解）

Best Edge Weight

我们先找出一棵最小生成树，

对于非树边来说， 答案就是两点路径上的最大值 - 1， 这个直接倍增就能处理。

对于树边来说， 就是非树边的路径经过这条边的最小值 - 1， 这个可以用并查集压缩路径 或者 更压st表一样的方式更新。

感觉就是没想到先扣出来一个最小生成树， 而是往克鲁斯卡尔的过程中想了。

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 2e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}

int n, m;
vector<PII> G[N];
bool can[N];
int ans[N];
int pathMin[N];

struct Edge {
    int u, v, c, id;
    bool operator < (const Edge &rhs) const {
        return c < rhs.c;
    }
} e[N];

int fa[N];
int getRoot(int x) {
    return fa[x] == x ? x : fa[x] = getRoot(fa[x]);
}

int pa[N][20], mx[N][20], depth[N];

void dfs(int u, int fa, int w) {
    depth[u] = depth[fa] + 1;
    pa[u][0] = fa; mx[u][0] = w;
    for(int i = 1; i < 20; i++)
        pa[u][i] = pa[pa[u][i - 1]][i - 1];
    for(int i = 1; i < 20; i++)
        mx[u][i] = max(mx[u][i - 1], mx[pa[u][i - 1]][i - 1]);
    for(auto& e : G[u]) {
        if(e.se == fa ) continue;
        dfs(e.se, u, e.fi);
    }
}

int getMaxWei(int u, int v) {
    if(depth[u] < depth[v]) swap(u, v);
    int dis = depth[u] - depth[v];
    int ans = 0;
    for(int i = 19; i >= 0; i--)
        if(dis >> i & 1) chkmax(ans, mx[u][i]), u = pa[u][i];
    if(u == v) return ans;
    for(int i = 19; i >= 0; i--) {
        if(pa[u][i] != pa[v][i]) {
            chkmax(ans, mx[u][i]);
            chkmax(ans, mx[v][i]);
            u = pa[u][i];
            v = pa[v][i];
        }
    }
    chkmax(ans, mx[u][0]);
    chkmax(ans, mx[v][0]);
    return ans;
}

int getLca(int u, int v) {
    if(depth[u] < depth[v]) swap(u, v);
    int dis = depth[u] - depth[v];
    for(int i = 19; i >= 0; i--)
        if(dis >> i & 1) u = pa[u][i];
    if(u == v) return u;
    for(int i = 19; i >= 0; i--)
        if(pa[u][i] != pa[v][i])
            u = pa[u][i], v = pa[v][i];
    return pa[u][0];
}

void gao(int u, int v, int c) {
    while(1) {
        u = getRoot(u);
        if(depth[u] <= depth[v]) break;
        pathMin[u] = c;
        int nex = getRoot(pa[u][0]);
        fa[u] = nex;
        u = nex;
    }
}

int main() {
    memset(pathMin, 0x3f, sizeof(pathMin));
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++) fa[i] = i;
    for(int i = 1; i <= m; i++) {
        scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].c);
        e[i].id = i;
    }
    sort(e + 1, e + 1 + m);
    for(int i = 1; i <= m; i++) {
        int u = e[i].u, v = e[i].v, c = e[i].c, id = e[i].id;
        int x = getRoot(u);
        int y = getRoot(v);
        if(x != y) {
            can[i] = true;
            fa[y] = x;
            G[u].push_back(mk(c, v));
            G[v].push_back(mk(c, u));
        }
    }

    dfs(1, 0, 0);

    for(int i = 1; i <= n; i++) fa[i] = i;
    for(int i = 1; i <= m; i++) {
        if(can[i]) continue;
        int u = e[i].u, v = e[i].v, c = e[i].c, id = e[i].id;
        int lca = getLca(u, v);
        gao(u, lca, c);
        gao(v, lca, c);
    }

    for(int i = 1; i <= m; i++) {
        int u = e[i].u, v = e[i].v, c = e[i].c, id = e[i].id;
        if(!can[i]) {
            ans[id] = getMaxWei(u, v) - 1;
        } else {
            if(depth[u] < depth[v]) swap(u, v);
            ans[id] = pathMin[u] == inf ? -1 : pathMin[u] - 1;
        }
    }

    for(int i = 1; i <= m; i++)
        printf("%d%c", ans[i], " \n"[i == m]);
    return 0;
}

/*

*/

原文地址：https://www.cnblogs.com/CJLHY/p/10956883.html