lightoj1101 ASecret Mission

题意

给你张图，n个点m条双向边，有正边权。然后是Q个询问<u,v>,表示u到v的路径上的最大值的最小。

分析

求一个MST是显然的，按照最小边权的贪心把图连通。然后就是普通的树链剖分。

/*****************************************
Author      :Crazy_AC(JamesQi)
Time        :2016
File Name   :
*****************************************/
// #pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <sstream>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <climits>
using namespace std;
#define MEM(x,y) memset(x, y,sizeof x)
#define pk push_back
#define lson rt << 1
#define rson rt << 1 | 1
#define bug cout << "BUG HERE\n"
#define ALL(v) (v).begin(), (v).end()
#define lowbit(x) ((x)&(-x))
#define showtime printf("time = %.15f\n",clock() / (double)CLOCKS_PER_SEC)
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> ii;
typedef pair<ii,int> iii;
const double eps = 1e-8;
const double pi = 4 * atan(1);
const int inf = 1 << 30;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int nCase = 0;
int dcmp(double x){//精度正负、0的判断
if (fabs(x) < eps) return 0;
return x < 0?-1:1;
}
template<class T>
inline bool Read(T &n)
{
T x = 0, tmp = 1;
char c = getchar();
while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();
if(c == EOF) return false;
if(c == '-') c = getchar(), tmp = -1;
while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();
n = x*tmp;
return true;
}
template <class T>
inline void write(T n)
{
if(n < 0)
{
putchar('-');
n = -n;
}
int len = 0,data[20];
while(n)
{
data[len++] = n%10;
n /= 10;
}
if(!len) data[len++] = 0;
while(len--) putchar(data[len]+48);
}
const int maxn = 5e4 + 100;
struct Edge {
int u, v, c;
bool operator < (const Edge& rhs) const {
return c < rhs.c;
}
void read() {
Read(u), Read(v), Read(c);
}
}e[maxn*20];
int n, m;
int f[maxn];
int find(int x) {return f[x] == -1?x:f[x] = find(f[x]);}
int fa[maxn], dep[maxn], top[maxn];
int size[maxn], son[maxn];
int SegId[maxn], TreeId[maxn];
int times;
int a[maxn];
int ans;
struct SegmentTree {
struct node {int v, l, r;}p[maxn<<2];
void init() {
memset(p, 0, sizeof p);
}
void build(int rt,int L, int R) {
p[rt] = node{0, L, R};
if (L == R) {p[rt].v = a[TreeId[L]];return ;}
int mid = (L + R) >> 1;
build(lson, L, mid), build(rson, mid + 1, R);
p[rt].v = max(p[lson].v, p[rson].v);
}
int Query(int rt,int L, int R) {
if (L <= p[rt].l && p[rt].r <= R) return p[rt].v;
int Max = -INF;
int mid = (p[rt].l + p[rt].r) >> 1;
if (L <= mid) Max = max(Max, Query(lson, L, R));
if (R > mid) Max = max(Max, Query(rson, L, R));
return Max;
}
}solve;

int head[maxn], pnt[maxn*20], nxt[maxn*20], w[maxn*20], ecnt;
void addedge(int u,int v,int c) {
pnt[ecnt] = v, w[ecnt] = c, nxt[ecnt] = head[u], head[u] = ecnt++;
pnt[ecnt] = u, w[ecnt] = c, nxt[ecnt] = head[v], head[v] = ecnt++;
}
int dfs_1(int u,int pre,int depth) {
fa[u] = pre, dep[u] = depth, size[u] = 1;
int tmp = 0;
for (int i = head[u];~i;i = nxt[i]) {
int v = pnt[i];
if (v == pre) continue;
a[v] = w[i];//边权转移到点上
int now = dfs_1(v, u, depth + 1);
size[u] += now;
if (now > tmp) {
tmp = now;
son[u] = v;
}
}
return size[u];
}
void dfs_2(int u,int header) {
top[u] = header;
if (son[u] != -1) {
SegId[u] = ++times;
TreeId[times] = u;
dfs_2(son[u], header);
}else if (son[u] == -1) {
SegId[u] = ++times;
TreeId[times] = u;
return ;
}
for (int i = head[u];~i;i = nxt[i]) {
int v = pnt[i];
if (v == fa[u] || v == son[u]) continue;
dfs_2(v, v);
}
}
int find(int u,int v) {
int Max = -INF;
int p = top[u], q = top[v];
while(p != q) {
if (dep[p] < dep[q]) {
swap(p, q);
swap(u, v);
}
Max = max(solve.Query(1, SegId[p], SegId[u]), Max);
u = fa[p];
p = top[u];
}
if (dep[u] < dep[v]) swap(u, v);
Max = max(Max, solve.Query(1, SegId[v] + 1,SegId[u]));
return Max;
}
int main(int argc, const char * argv[])
{
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);

int kase;Read(kase);
while(kase--) {
Read(n), Read(m);
for (int i = 0;i < m;++i) {
Read(e[i].u), Read(e[i].v), Read(e[i].c);
}
sort(e, e + m);
memset(f, -1, sizeof f);
int cnt = 0;
for (int i = 0;i < m;++i) {
int t1 = find(e[i].u);
int t2 = find(e[i].v);
if (t1 != t2) {
f[t1] = t2;
e[cnt++] = e[i];
if (cnt == n - 1) break;
}
}
memset(head, -1, sizeof head), ecnt = 0;
for (int i = 0;i < cnt;++i) {
addedge(e[i].u, e[i].v, e[i].c);
}
memset(son, -1, sizeof son), times = 0;
dfs_1(1, -1, 0);
dfs_2(1, 1);
solve.init();
solve.build(1, 1, n);
int Q;cin >> Q;
printf("Case %d:\n", ++nCase);
while(Q--) {
int u, v;
Read(u), Read(v);
printf("%d\n", find(u, v));
}
}
// showtime;
return 0;
}