UVA 11354Bond（MST+LCA）

题意:

给定N≤5×104,M≤105的图,Q≤5×104次询问
(u,v)找到一条路径使得最大边权最小

分析:

找最小,显然先搞一遍MST
求(u,v)树上2点间的最大边,倍增搞一下就可以了,当然树剖也是可以的

代码:

////  Created by TaoSama on 2015-11-09//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int n, m, q;struct RawEdge {    int u, v, c;    bool operator<(const RawEdge& rhs) const {        return c < rhs.c;    }};vector<RawEdge> E;struct Edge {    int v, c, nxt;} edge[N];int head[N], cnt;void add_edge(int u, int v, int c) {    edge[cnt] = (Edge) {v, c, head[u]};    head[u] = cnt++;}int par[N];int find(int x) {    return par[x] = par[x] == x ? x : find(par[x]);}void kruskal() {    for(int i = 1; i <= n; ++i) par[i] = i;    sort(E.begin(), E.end());    cnt = 0; memset(head, -1, sizeof head);    int num = 0;    for(auto &e : E) {        int u = find(e.u), v = find(e.v);        if(u == v) continue;        par[u] = v;//      pr(e.u); pr(e.v); prln(e.c);        add_edge(e.u, e.v, e.c);        add_edge(e.v, e.u, e.c);        if(++num == n - 1) break;    }}int dep[N], p[20][N], maxv[20][N];void dfs(int u, int f, int c, int d) {    dep[u] = d;    p[0][u] = f; maxv[0][u] = c;    for(int i = head[u]; ~i; i = edge[i].nxt) {        int v = edge[i].v;        if(v == f) continue;        dfs(v, u, edge[i].c, d + 1);    }}void build() {    dfs(1, -1, 0, 0);    for(int i = 0; i + 1 < 18; ++i) {        for(int v = 1; v <= n; ++v) {            if(p[i][v] < 0) p[i + 1][v] = -1;            else p[i + 1][v] = p[i][p[i][v]];            maxv[i + 1][v] = max(maxv[i][v], maxv[i][p[i][v]]);        }    }}int lca(int u, int v) {    if(dep[u] > dep[v]) swap(u, v);    for(int i = 0; i < 18; ++i)        if(dep[v] - dep[u] >> i & 1)            v = p[i][v];    if(u == v) return u;    for(int i = 17; ~i; --i)        if(p[i][u] != p[i][v])            u = p[i][u], v = p[i][v];    return p[0][u];}int get(int u, int to) {    int ret = 0;    for(int i = 0; i < 18; ++i)        if(dep[u] - dep[to] >> i & 1)            ret = max(ret, maxv[i][u]), u = p[i][u];    return ret;}int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    int kase = 0;    while(scanf("%d%d", &n, &m) == 2) {        E.clear();        while(m--) {            int u, v, c; scanf("%d%d%d", &u, &v, &c);            E.push_back((RawEdge) {u, v, c});        }        kruskal();        build();        scanf("%d", &q);        if(kase++) puts("");        while(q--) {            int u, v; scanf("%d%d", &u, &v);            int _lca = lca(u, v);//          prln(_lca);            printf("%d\n", max(get(u, _lca), get(v, _lca)));        }    }    return 0;}