Connections between cities
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Problem Description
After World War X, a lot of cities have been seriously damaged, and we need to rebuild those cities. However, some materials needed can only be produced in certain places. So we need to transport these materials from city to city. For most of roads had been totally destroyed during the war, there might be no path between two cities, no circle exists as well.
Now, your task comes. After giving you the condition of the roads, we want to know if there exists a path between any two cities. If the answer is yes, output the shortest path between them.
Now, your task comes. After giving you the condition of the roads, we want to know if there exists a path between any two cities. If the answer is yes, output the shortest path between them.
Input
Input consists of multiple problem instances.For each instance, first line contains three integers n, m and c, 2<=n<=10000, 0<=m<10000, 1<=c<=1000000. n represents the number of cities numbered from 1 to n. Following m lines, each line has three integers i, j and k, represent a road between city i and city j, with length k. Last c lines, two integers i, j each line, indicates a query of city i and city j.
Output
For each problem instance, one line for each query. If no path between two cities, output “Not connected”, otherwise output the length of the shortest path between them.
Sample Input
5 3 2
1 3 2
2 4 3
5 2 3
1 4
4 5
1 3 2
2 4 3
5 2 3
1 4
4 5
Sample Output
Not connected
6
6
Hint
Hint
Huge input, scanf recommended.
Source
思路：利用并查集判断是否联通，因为可能不联通，所以可能哟偶多颗树，多次dfs，然后就是模版
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
#define true ture
#define false flase
using namespace std;
#define ll long long
#define inf 0xfffffff
int scan()
{
int res = , ch ;
while( !( ( ch = getchar() ) >= '' && ch <= '' ) )
{
if( ch == EOF ) return << ;
}
res = ch - '' ;
while( ( ch = getchar() ) >= '' && ch <= '' )
res = res * + ( ch - '' ) ;
return res ;
}
#define maxn 100010
#define M 22
int father[maxn];
struct is
{
int v,next,w;
} edge[maxn*];
int deep[maxn],jiedge;
int dis[maxn];
int head[maxn];
int fa[maxn][M];
int findd(int x)
{
return x==father[x]?x:father[x]=findd(father[x]);
}
int hebing(int u,int v)
{
int x=findd(u);
int y=findd(v);
if(x!=y)
father[x]=y;
}
void add(int u,int v,int w)
{
jiedge++;
edge[jiedge].v=v;
edge[jiedge].w=w;
edge[jiedge].next=head[u];
head[u]=jiedge;
}
void dfs(int u)
{
for(int i=head[u]; i; i=edge[i].next)
{
int v=edge[i].v;
int w=edge[i].w;
if(!deep[v])
{
dis[v]=dis[u]+edge[i].w;
deep[v]=deep[u]+;
fa[v][]=u;
dfs(v);
}
}
}
void st(int n)
{
for(int j=; j<M; j++)
for(int i=; i<=n; i++)
fa[i][j]=fa[fa[i][j-]][j-];
}
int LCA(int u , int v)
{
if(deep[u] < deep[v]) swap(u , v) ;
int d = deep[u] - deep[v] ;
int i ;
for(i = ; i < M ; i ++)
{
if( ( << i) & d ) // 注意此处，动手模拟一下，就会明白的
{
u = fa[u][i] ;
}
}
if(u == v) return u ;
for(i = M - ; i >= ; i --)
{
if(fa[u][i] != fa[v][i])
{
u = fa[u][i] ;
v = fa[v][i] ;
}
}
u = fa[u][] ;
return u ;
}
void init(int n)
{
for(int i=;i<=n;i++)
father[i]=i;
memset(head,,sizeof(head));
memset(fa,,sizeof(fa));
memset(deep,,sizeof(deep));
jiedge=;
}
int main()
{
int x,n,t;
while(~scanf("%d%d%d",&n,&x,&t))
{
init(n);
for(int i=; i<x; i++)
{
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
add(u,v,w);
add(v,u,w);
hebing(u,v);
}
for(int i=;i<=n;i++)
if(!deep[i])
{
deep[i]=;
dis[i]=;
dfs(i);
}
st(n);
while(t--)
{
int a,b;
scanf("%d%d",&a,&b);
if(findd(a)!=findd(b))
printf("Not connected\n");
else
printf("%d\n",dis[a]-*dis[LCA(a,b)]+dis[b]);
}
}
return ;
}