Design the city
Time Limit: 1000 msMemory Limit: 32768 KB
Cerror is the mayor of city HangZhou. As you may know, the traffic system of this city is so terrible, that there are traffic jams everywhere. Now, Cerror finds out that the main reason of them is the poor design of the roads distribution, and he want to change this situation.
In order to achieve this project, he divide the city up to N regions which can be viewed as separate points. He thinks that the best design is the one that connect all region with shortest road, and he is asking you to check some of his designs.
Now, he gives you an acyclic graph representing his road design, you need to find out the shortest path to connect some group of three regions.
Input
The input contains multiple test cases! In each case, the first line contian a interger N (1 < N < 50000), indicating the number of regions, which are indexed from 0 to N-1. In each of the following N-1 lines, there are three interger Ai, Bi, Li (1 < Li < 100) indicating there's a road with length Li between region Ai and region Bi. Then an interger Q (1 < Q < 70000), the number of group of regions you need to check. Then in each of the following Q lines, there are three interger Xi, Yi, Zi, indicating the indices of the three regions to be checked.
Process to the end of file.
Output
Q lines for each test case. In each line output an interger indicating the minimum length of path to connect the three regions.
Output a blank line between each test cases.
Sample Input
4
0 1 1
0 2 1
0 3 1
2
1 2 3
0 1 2
5
0 1 1
0 2 1
1 3 1
1 4 1
2
0 1 2
1 0 3
Sample Output
3
2
2
2
Author: HE, Zhuobin
題意：
n 個點，n - 1條邊，給出 m 次詢問，詢問連通三點的最短距離
思路：
a, b, c三點距離即（a, b的距離 + a, c的距離 + b, c的距離） / 2，舉個栗子就明白了
tarjin實現：
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int N = 5e4 + 10;
const int M = 7e4 + 10;
int n, m, dis[N], head[N], qu[M], tot, fa[N];
bool vis[N];
struct node
{
int to, w, next;
}edge[N << 1];
struct query
{
int u, v, lca, next;
}q[6 * M];
void init()
{
tot = 0;
memset(vis, 0, sizeof(vis));
memset(head, -1, sizeof(head));
memset(qu, -1, sizeof(qu));
}
void add(int u, int v, int w)
{
edge[tot].to = v;
edge[tot].next = head[u];
edge[tot].w = w;
head[u] = tot++;
}
void add2(int u, int v)
{
q[tot].u = u;
q[tot].v = v;
q[tot].next = qu[u];
qu[u] = tot++;
}
int Find(int x)
{
if(fa[x] != x)
fa[x] = Find(fa[x]);
return fa[x];
}
void tarjin(int u)
{
fa[u] = u;
vis[u] = 1;
for(int i = head[u]; ~i; i = edge[i].next)
{
int v = edge[i].to;
if(!vis[v])
{
dis[v] = dis[u] + edge[i].w;
tarjin(v);
fa[v] = u;
}
}
for(int i = qu[u]; ~i; i = q[i].next)
{
int v = q[i].v;
if(vis[v])
{
q[i].lca = q[i ^ 1].lca = Find(v);
}
}
}
int main()
{
int kcase = 0;
while(~scanf("%d", &n))
{
init();
int u, v, w;
for(int i = 1; i < n; ++i)
{
scanf("%d%d%d", &u, &v, &w);
add(u, v, w);
add(v, u, w);
}
tot = 0;
int a, b, c;
scanf("%d", &m);
for(int i = 1; i <= m; ++i)
{
scanf("%d%d%d", &a, &b, &c);
add2(a, b);
add2(b, a);
add2(a, c);
add2(c, a);
add2(b, c);
add2(c, b);
}
tarjin(0);
if(kcase)
cout<<'\n';
kcase++;
for(int i = 0; i < tot; i += 6)
{
int ans = 0, u, v, lca;
u = q[i].u;
v = q[i].v;
lca = q[i].lca;
ans += dis[u] + dis[v] - 2 * dis[lca];
u = q[i + 2].u;
v = q[i + 2].v;
lca = q[i + 2].lca;
ans += dis[u] + dis[v] - 2 * dis[lca];
u = q[i + 4].u;
v = q[i + 4].v;
lca = q[i + 4].lca;
ans += dis[u] + dis[v] - 2 * dis[lca];
cout << ans / 2 << '\n';
}
}
return 0;
}
樹上倍增實現
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int N = 5e4 + 10;
//dis[i]:點i到根節點的距離
//fa[i][j]:節點i的第2^j個父親，每個點最多2^(logN)個父親，所以第二維開logN
//dep[i]:當前節點的深度
int n, m, dis[N], head[N], fa[N][21], tot, dep[N];
struct Edge
{
int to, next, w;
}edge[N << 1];
void init()
{
tot = 0;
memset(dis, 0, sizeof(dis));
memset(head, -1, sizeof(head));
memset(fa, 0, sizeof(fa));
}
void add(int u, int v, int w)
{
edge[tot].next = head[u];
edge[tot].to = v;
edge[tot].w = w;
head[u] = tot++;
}
void dfs(int u, int father) //當前節點和它的父親
{
dep[u] = dep[father] + 1;
for(int i = 1; (1 << i) <= dep[u]; ++i)
{
if(fa[u][i - 1])
fa[u][i] = fa[fa[u][i - 1]][i - 1];
else break; //如果該點沒有第2^(i - 1)個父親了，也不會有更遠的父親
}
for(int i = head[u]; ~i; i = edge[i].next) //遍歷相鄰的所有點
{
int v = edge[i].to;
if(v != father) // v 不是 u 的父親，就是 u 的兒子
{
dis[v] = dis[u] + edge[i].w; //更新距離
fa[v][0] = u; // v 的第2^0個父親即第一個父親是 u
dfs(v, u);
}
}
}
int lca(int u, int v)
{
if(dep[u] < dep[v]) //默認 u 比 v 深
swap(u, v);
for(int i = 20; i >= 0; --i) //從大到小枚舉使 x 和 y 到達同一層
{
if(dep[fa[u][i]] >= dep[v])
u = fa[u][i];
if(u == v)
return u;
}
for(int i = 20; i >= 0; --i)
{
if(fa[u][i] != fa[v][i])
{
u = fa[u][i];
v = fa[v][i];
}
}
return fa[u][0];
}
int main()
{
int u, v, w, a, b, c, kcase = 0;
while(~scanf("%d", &n))
{
init();
for(int i = 1; i < n; ++i)
{
scanf("%d%d%d", &u, &v, &w);
add(u, v, w);
add(v, u, w);
}
dfs(0, 0); //選取1爲根節點，1的父親是0
if(kcase) cout<<'\n';
kcase++;
scanf("%d", &m);
while(m--)
{
scanf("%d%d%d", &a, &b, &c);
int ans = dis[a] + dis[b] - 2 * dis[lca(a, b)] + dis[a] + dis[c] - 2 * dis[lca(a, c)] + dis[b] + dis[c] - 2 * dis[lca(b, c)];
ans /= 2;
cout<<ans<<'\n';
}
}
return 0;
}