UVA 11354 Bond (MST + LCA)
求出最小生成树， 然后在最小生成树上求LCA。
| Time Limit: 8000MS | Memory Limit: Unknown | 64bit IO Format: %lld & %llu |
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Description
|
Once again, James Bond is on his way to saving the world. Bond's latest mission requires him to travel between several pairs of cities in a certain country.
The country has N cities (numbered by 1, 2, . . ., N), connected by M bidirectional roads. Bond is going to steal a vehicle, and drive along the roads from city s to city t. The country's police will be patrolling the roads, looking for Bond, however, not all roads get the same degree of attention from the police.
More formally, for each road MI6 has estimated its dangerousness, the higher it is, the more likely Bond is going to be caught while driving on this road. Dangerousness of a path from s to t is defined as the maximum dangerousness of any road on this path.
Now, it's your job to help Bond succeed in saving the world by finding the least dangerous paths for his mission.
Input
There will be at most 5 cases in the input file.
The first line of each case contains two integers N, M (2 ≤ N≤ 50000, 1≤ M ≤ 100000) – number of cities and roads. The next M lines describe the roads. The i-th of these lines contains three integers: xi, yi, di (1 ≤ xi, yi ≤ N, 0 ≤ di ≤ 109) - the numbers of the cities connected by the ith road and its dangerousness.
Description of the roads is followed by a line containing an integer Q (1 ≤ Q ≤ 50000), followed by Q lines, the i-th of which contains two integers si and ti (1 ≤ si, ti ≤ N, si !=ti).
Consecutive input sets are separated by a blank line.
Output
For each case, output Q lines, the i-th of which contains the minimum dangerousness of a path between cities si and ti. Consecutive output blocks are separated by a blank line.
The input file will be such that there will always be at least one valid path.
| Sample Input |
Output for Sample Input |
| 4 5 1 2 10 1 3 20 1 4 100 2 4 30 3 4 10 2 1 4 4 1
2 1 1 2 100 1 1 2 |
20 20
100 |
ProblemSetter: Ivan Krasilnikov
Source
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#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
#define prt(k) cout<<#k" = "<<k<<endl;
typedef long long ll;
const int N = 100005;
const int M = N << 2;
int n, m, head[N], mm;
int len[N];
struct Edge {
int to, next, w;
}e[N << 1];
struct EE {
int u, v, w;
}E[M];
bool cmp(EE a, EE b) { return a.w < b.w; }
void add(int u, int v, int w)
{
e[mm].to = v;
e[mm].w = w;
e[mm].next = head[u];
head[u] = mm++;
}
int sz[N], dep[N];
int f[N][22]; /// f[i][j] 表示 i 的第 2^j 个祖先
int dp[N][22];
void dfs(int u, int fa) /// 点从 1 开始标号
{
f[u][0] = fa;
sz[u] = 1;
for (int i=head[u];~i;i=e[i].next)
{
int v = e[i].to;
if (v != fa)
{
dep[v] = dep[u] + 1;
len[v] = len[u] + e[i].w;
dp[v][0] = e[i].w;
dfs(v, u);
sz[u] += sz[v];
}
}
}
int maxh;
void gao()
{
int j;
for (j=1;(1<<j)<n;j++)
for (int i=1;i<=n;i++)
{
int t = f[i][j-1];
f[i][j] = f[f[i][j-1]][j-1];
dp[i][j] = max(dp[i][j-1], dp[t][j-1]);
}
maxh = j - 1;
}
int swim(int x, int k, int &ma) /// 返回 x 的第 k 个祖先
{
ma = 0;
for (int i=0;i<=maxh;i++)
if (k >> i & 1)
{
ma = max(ma, dp[x][i]);
x = f[x][i];
}
return x;
}
int LCA(int x, int y)
{
if (dep[x] > dep[y]) swap(x, y); ///dep[x] <= dep[y];
int ans = 0;
y = swim(y, dep[y] - dep[x], ans);
if (x == y) return ans;
for (int i=maxh; i>=0; i--) {
if (f[x][i] != f[y][i])
{
ans = max(ans, max(dp[x][i], dp[y][i]));
x = f[x][i], y = f[y][i];
}
}
return max(ans, max(dp[x][0], dp[y][0]) ); //f[x][0];
}
int fa[N];
int find(int x) { return x==fa[x] ? x : fa[x]=find(fa[x]); }
int main()
{
int re; bool blank = false;
while (cin >> n >> m) {
dep[1] = 0; len[1] = 0;
if (blank) putchar(10);
blank = true;
dp[1][0] = 0;
for (int i=0;i<m;i++) {
scanf("%d%d%d", &E[i].u, &E[i].v, &E[i].w);
}
sort(E, E+m, cmp);
for (int i=1;i<=n;i++) fa[i] = i;
mm = 0;
memset(head, -1, sizeof head);
int cnt = 0;
for (int i=0;i<m;i++) {
int a=find(E[i].u), b=find(E[i].v);
if (a - b) {
fa[a] = b;
add(E[i].u, E[i].v, E[i].w);
add(E[i].v, E[i].u, E[i].w);
// printf("[%d %d %d]\n", E[i].u, E[i].v, E[i].w);
if (++cnt == n-1) break;
}
}
dfs(1, 0);
// for (int i=1;i<=n;i++) printf("dep[%d] = %d\n",i,dep[i]);
gao();
/** for (int i=1;i<=n;i++)
for (int j=0;j<=maxh;j++)
{
printf("dp[%d][%d] = %d\n", i, j, dp[i][j]);
} */
int Q;
scanf("%d", &Q);
while (Q--) {
int a, b;
scanf("%d%d",&a,&b);
int c = LCA(a, b);
printf("%d\n", c);
}
// putchar(10);
}
return 0;
}