Ultraviolet A 11354 Bond (MST + LCA)

N <= 50000, m <= 100000 undirected graphs. For Q <= 50000 queries, find the bottleneck path of q-> p each time.

In fact, the bottleneck path array maxcost [u] [v] is used to obtain the prim of the adjacent matrix. However, for n of this question, the adjacent matrix cannot exist... So I copied the algorithm in the big white book... First, kruskal is used to find the MST, and then for the MST tree, find the LCA of p and q each time, and find the bottleneck path in the process of finding the LCA...

 

#include<algorithm>#include<iostream>#include<cstring>#include<cstdlib>#include<fstream>#include<sstream>#include<bitset>#include<vector>#include<string>#include<cstdio>#include<cmath>#include<stack>#include<queue>#include<stack>#include<map>#include<set>#define FF(i, a, b) for(int i=a; i<b; i++)#define FD(i, a, b) for(int i=a; i>=b; i--)#define REP(i, n) for(int i=0; i<n; i++)#define CLR(a, b) memset(a, b, sizeof(a))#define debug puts("**debug**")#define LL long long#define PB push_backusing namespace std;const int maxn = 50005;const int maxm = 100010;const int INF = 1e9;int n, m;int fa[maxn], pa[maxn], cost[maxn], L[maxn], anc[maxn][100], maxcost[maxn][100];struct Edge{    int to, dist;};vector<Edge> edges;vector<int> G[maxn];inline void add(int a, int b, int c){    edges.PB((Edge){b, c});    edges.PB((Edge){a, c});    int nc = edges.size();    G[a].PB(nc-2); G[b].PB(nc-1);}inline void init(){    L[1] = cost[1] = 0;    REP(i, n+1) pa[i] = i;    REP(i, n+1) G[i].clear(); edges.clear();}int findset(int x) {    return x == pa[x] ? x : pa[x] = findset(pa[x]); }struct E{    int u, v, w;    bool operator < (const E rhs) const    {        return w < rhs.w;    }}e[maxm];void MST(){    sort(e, e+m);    int cnt = 0;    REP(i, m)    {        int x = findset(e[i].u), y = findset(e[i].v);        if(x != y)        {            pa[x] = y;            cnt++;            add(e[i].u, e[i].v, e[i].w);        }        if(cnt == n-1) return ;    }}void dfs(int u, int f){    fa[u] = f;    int nc = G[u].size();    REP(i, nc)    {        int v = edges[G[u][i]].to, w = edges[G[u][i]].dist;        if(v != f)        {            cost[v] = w;            L[v] = L[u] + 1;            dfs(v, u);        }    }}void progress(){    FF(i, 1, n+1)    {        anc[i][0] = fa[i], maxcost[i][0] = cost[i];        for(int j=1; (1 << j) < n; j++) anc[i][j] = -1;    }    for(int j=1; (1 << j) < n; j++) FF(i, 1, n+1)    if(anc[i][j-1] != -1)    {        int a = anc[i][j-1];        anc[i][j] = anc[a][j-1];        maxcost[i][j] = max(maxcost[i][j-1], maxcost[a][j-1]);    }}int query(int p, int q){    int lo;    if(L[p] < L[q]) swap(p, q);    for(lo = 1; (1 << lo) <= L[p]; lo++); lo--;    int ans = -INF;    FD(i, lo, 0)        if(L[p] - (1<<i) >= L[q]) ans = max(ans, maxcost[p][i]), p = anc[p][i];    if(p == q) return ans; //LCA -> p    FD(i, lo, 0)    if(anc[p][i] != -1 && anc[p][i] != anc[q][i])    {        ans = max(ans, maxcost[p][i]), p = anc[p][i];        ans = max(ans, maxcost[q][i]), q = anc[q][i];    }    ans = max(ans, cost[p]);    ans = max(ans, cost[q]);    return ans; //LCA -> fa[q] fa[p]}int main(){    int flag = 0;    while(~scanf("%d%d", &n, &m))    {        if(flag++) puts("");        init();        REP(i, m) scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].w);        MST();        dfs(1, -1);        progress();        int Q, p, q;        scanf("%d", &Q);        while(Q--)        {            scanf("%d%d", &p, &q);            printf("%d\n", query(p, q));        }    }    return 0;}