HDU 2874 - Connections between cities(LCA‘离线算法Tarjan)

题目：http://acm.hdu.edu.cn/showproblem.php?pid=2874

题意：n个点m条边的森林，c个询问，求出两个点的最短路径，若不连接则输出 Not connected。

思路：由于询问c（1~1000000），所以采用LCA的离线算法

由于是森林，则再建立一个点n+1将所以点连接起来，若两个点的最近公共祖先是n+1，则两个点 Not connected。

本题还容易爆内存，要把询问节点的数组设为 short型。

详见http://noalgo.info/476.html

AC.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const int INF = 0x3f3f3f;
const int maxn = 10010;
const int maxq = 1000010;

int far[maxn], rnk[maxn];
int fg(int x)
{
if(far[x] == x) return x;
return far[x] = fg(far[x]);
}
void unite(int a, int b)
{
int x = fg(a);
int y = fg(b);
if(x == y) return;
if(rnk[x] <= rnk[y]) far[x] = y, rnk[y] += rnk[x];
else far[y] = x, rnk[x] += rnk[y];
}

struct Edge {
int to, next, w;
}edge[maxn*4];
int head[maxn], tot;
void addedge(int u, int v, int w)
{
edge[tot].to = v;
edge[tot].w = w;
edge[tot].next = head[u];
head[u] = tot++;
}

short asku[maxq*2], askv[maxq*2];
int askid[maxq*2], asknext[maxq*2], h[maxn];
int tt;

void addask(int u, int v, int id)
{
asku[tt] = u;
askv[tt] = v;
askid[tt] = id;
asknext[tt] = h[u];
h[u] = tt++;
}

int n, m;
int ans[maxq], dis[maxn], anc[maxn];
bool vis[maxn];

void init()
{
tot = 0;
tt = 0;
memset(head, -1, sizeof(head));
memset(h, -1, sizeof(h));
memset(vis, 0, sizeof(vis));
memset(dis, 0, sizeof(dis));
memset(anc, 0, sizeof(anc));
memset(ans, 0, sizeof(ans));

for(int i = 0; i <= n+1; ++i) {
far[i] = i;
rnk[i] = 1;
}
}

void LCA(int u)
{
anc[u] = u;
vis[u] = 1;
for(int i = head[u] ; i != -1 ; i = edge[i].next){
int v = edge[i].to;
if(!vis[v]){
dis[v] = dis[u] + edge[i].w;
LCA(v);
unite(u, v);
anc[fg(u)] = u;
}
}

for(int i = h[u] ; i != -1 ; i = asknext[i]){
int v = askv[i];
if(vis[v]){
ans[askid[i]] = anc[fg(v)];
}
}
}

int main()
{
//freopen("in", "r", stdin);
int k;
while(~scanf("%d %d %d", &n, &m, &k)) {
init();
int u, v, w;
for(int i = 0; i < m; ++i) {
scanf("%d %d %d", &u, &v, &w);
addedge(u, v, w);
addedge(v, u, w);
}

for(int i = 0; i < k; ++i) {
scanf("%d %d", &u, &v);
addask(u, v, i);
addask(v, u, i);
}

for(int i = 1; i <= n; ++i) {
if(far[i] == i) {
addedge(i, n+1, INF);
addedge(n+1, i, INF);
}
}

LCA(n+1);

for(int i = 0 ; i < tt; i+=2){
if(ans[askid[i]] != n + 1)
printf("%d\n",dis[asku[i]]+dis[askv[i]]-2*dis[ans[askid[i]]]);
else
printf("Not connected\n");
}
}
}