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  • 5.3.4 Connections between cities

    Connections between cities
    Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 444 Accepted Submission(s): 142

    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.
     
    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
     
    Sample Output
    Not connected
6
    Hint
    Hint Huge input, scanf recommended.

    思路：Tarjan求LCA，虚拟一个0号的根节点，如果两个点的lca是0，那么就不连通，否则，任意两点的距离就是：d(u,v)=dist[v]+dist[u]-2*dist[LCA(u,v)];

      1 #include <iostream>
  2 #include <cmath>
  3 #include <cstdio>
  4 #include <algorithm>
  5 #include <cstring>
  6 #include <string>
  7 #include <cstdlib>
  8 using namespace std;
  9 
 10 const int maxn=10100,maxm=30010,maxq=2000010;
 11 int cnt,n,x,y,m,z,node,v;
 12 bool flag[maxn],vis[maxn];
 13 int h[maxn],h1[maxn],f[maxn],ans[maxn],dis[maxn];
 14 struct qq
 15 {
 16     int to,ne,z,lca;
 17 } e[maxm],q[maxq];
 18 
 19 void addedge(int x,int y,int z)
 20 {
 21     cnt++;
 22     e[cnt].to=y;
 23     e[cnt].ne=h[x];
 24     e[cnt].z=z;
 25     h[x]=cnt;
 26 }
 27 
 28 void addquery(int x,int y,int z)
 29 {
 30     q[cnt].to=y;
 31     q[cnt].ne=h1[x];
 32     q[cnt].z=z;
 33     h1[x]=cnt;
 34     cnt++;
 35     q[cnt].to=x;
 36     q[cnt].ne=h1[y];
 37     q[cnt].z=z;
 38     h1[y]=cnt;
 39     cnt++;
 40 }
 41 
 42 void close()
 43 {
 44 exit(0);
 45 }
 46 
 47 int findfather(int k)
 48 {
 49     if (f[k]==k)
 50         return k;
 51     f[k]=findfather(f[k]);
 52     return f[k];
 53 }
 54 
 55 void tarjan(int k)
 56 {
 57     f[k]=k;
 58     vis[k]=true;
 59     for (int p=h[k];p!=-1;p=e[p].ne)
 60     {
 61         if (not vis[e[p].to])
 62         {
 63             v+=e[p].z;
 64             dis[e[p].to]=dis[k]+e[p].z;
 65             tarjan(e[p].to);
 66             v-=e[p].z;
 67             f[e[p].to]=k;
 68         }
 69     }
 70     for (int p=h1[k];p!=-1;p=q[p].ne)
 71     {
 72         node=q[p].to;
 73         if (vis[node])    //询问的令一个点已经询问过了，表示它们的祖先就是f[另一个点]
 74         {
 75             int lca=findfather(node);
 76                 q[p].z=dis[k]+dis[node]-2*dis[lca];
 77                 q[p^1].z=q[p].z;
 78         }
 79     }
 80 }
 81 
 82 
 83 void init()
 84 {
 85     int query=1;
 86     while(scanf("%d %d %d",&n,&m,&query)!=EOF)
 87     {
 88         cnt=0;
 89         memset(flag,false,sizeof(flag));
 90         memset(h,-1,sizeof(h));
 91         memset(h1,-1,sizeof(h1));
 92         memset(dis,0,sizeof(dis));
 93         for (int i=0;i<=n;i++)
 94             f[i]=i;
 95         for (int i=1;i<=m;i++)
 96         {
 97             scanf("%d %d %d",&x,&y,&z);
 98             addedge(x,y,z);
 99             addedge(y,x,z);
100             int fx=findfather(x);
101             int fy=findfather(y);
102             if (fx!=fy)
103                 f[fx]=fy;
104         }
105         for (int i=1;i<=n;i++)
106             flag[findfather(i)]=true;
107         /*
108         for (int i=1;i<=n;i++)
109             if (flag[i])
110                 addedge(0,i,0);
111                 */
112         cnt=0;
113         for (int i=1;i<=query;i++)
114         {
115             scanf("%d %d",&x,&y);
116             if (f[x]!=f[y])
117                 addquery(x,y,-1);
118             else
119                 addquery(x,y,0);
120         }
121         memset(f,0,sizeof(f));
122         for (int i=1;i<=n;i++)
123             if (flag[i])
124             {
125                 memset(vis,false,sizeof(vis));
126                 tarjan(i);
127             }
128         for (int i=0;i<cnt;i+=2)
129         {
130             if (q[i].z!=-1)
131                 printf("%d\n",q[i].z);
132             else
133                 printf("Not connected\n");
134         }
135     }
136 }
137 
138 int main ()
139 {
140     init();
141     close();
142     return 0;
143 }
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  • 原文地址：https://www.cnblogs.com/cssystem/p/3045991.html
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