HDU 3631 Shortest Path

Shortest Path

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3636    Accepted Submission(s): 857

Problem Description

When YY was a boy and LMY was a girl, they trained for NOI (National Olympiad in Informatics) in GD team. One day, GD team’s coach, Prof. GUO asked them to solve the following shortest-path problem.
There is a weighted directed multigraph G. And there are following two operations for the weighted directed multigraph:
(1) Mark a vertex in the graph.
(2) Find the shortest-path between two vertices only through marked vertices.
For it was the first time that LMY faced such a problem, she was very nervous. At this moment, YY decided to help LMY to analyze the shortest-path problem. With the help of YY, LMY solved the problem at once, admiring YY very much. Since then, when LMY meets problems, she always calls YY to analyze the problems for her. Of course, YY is very glad to help LMY. Finally, it is known to us all, YY and LMY become programming lovers.
Could you also solve the shortest-path problem?

 

Input

The input consists of multiple test cases. For each test case, the first line contains three integers N, M and Q, where N is the number of vertices in the given graph, N≤300; M is the number of arcs, M≤100000; and Q is the number of operations, Q ≤100000. All vertices are number as 0, 1, 2, … , N - 1, respectively. Initially all vertices are unmarked. Each of the next M lines describes an arc by three integers (x, y, c): initial vertex (x), terminal vertex (y), and the weight of the arc (c). (c > 0) Then each of the next Q lines describes an operation, where operation “0 x” represents that vertex x is marked, and operation “1 x y” finds the length of shortest-path between x and y only through marked vertices. There is a blank line between two consecutive test cases.
End of input is indicated by a line containing N = M = Q = 0.

 

Output

Start each test case with "Case #:" on a single line, where # is the case number starting from 1.
For operation “0 x”, if vertex x has been marked, output “ERROR! At point x”.
For operation “1 x y”, if vertex x or vertex y isn’t marked, output “ERROR! At path x to y”; if y isn’t reachable from x through marked vertices, output “No such path”; otherwise output the length of the shortest-path. The format is showed as sample output.
There is a blank line between two consecutive test cases.

 

Sample Input

5 10 10

1 2 6335

0 4 5725

3 3 6963

4 0 8146

1 2 9962

1 0 1943

2 1 2392

4 2 154

2 2 7422

1 3 9896

0 1

0 3

0 2

0 4

0 4

0 1

1 3 3

1 1 1

0 3

0 4

0 0 0

 

Sample Output

Case 1:

ERROR! At point 4

ERROR! At point 1

0

0

ERROR! At point 3

ERROR! At point 4

 

Source

2010 Asia Regional Tianjin Site —— Online Contest

 

Recommend

lcy   |   We have carefully selected several similar problems for you:  3621 3629 3622 3625 3626 

 

思路：Floyd，Floyd算法本来就是使用每个点对每一条路经进行

更新，如果按照常规写法去写，每查询一次运行一次Floyd算法，

那么超时好像是注定的的，因为Floyd使用邻接表实现时时间复杂

度是N^3，所以每对一个新加入的点进行路径更新，这样就快多了

 

超时代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
using namespace std;
const int MAXN = 100000000;
int n,m,q;
int a,b,c;
int map[310][310];
int pre[310][310];
int change;
int cast;
int mark[310];
int Floyd(int n,int map[310][310],int pre[310][310],int x,int y)
{
    int dist[310][310];
    for(int i = 1;i <= n;i ++)
       for(int j = 1;j <= n;j ++)
           dist[i][j] = map[i][j];
    for(int k = 1;k <= n;k ++)
    {
        if(mark[k] == 0)
            continue ;
        for(int i = 1;i <= n;i ++)
        {
            if(mark[i] == 0)
                continue ;
            for(int j = 1;j <= n;j ++)
            {
                if(mark[j] == 0)
                    continue ;
                if(dist[i][k] != MAXN && dist[k][j] != MAXN &&
                   dist[i][k] + dist[k][j] < dist[i][j])
                {
                    dist[i][j] = dist[i][k] + dist[k][j];
                    pre[i][j] = pre[k][j];
                }
            }
        }
    }
    return dist[x][y];
}
int main()
{
    cast = 0;
    while(scanf("%d%d%d",&n,&m,&q))
    {
        cast ++;
        if(n == 0 && m == 0 && q == 0)
             break ;
        memset(mark,0,sizeof(mark));
        for(int i = 1;i <= n;i ++)
           for(int j = 1;j <= n;j ++)
           {
               pre[i][j] = i;
               if(i == j)
                  map[i][j] = 0;
               else
                  map[i][j] = MAXN;
           }
        for(int i = 1;i <= m;i ++)
        {
            scanf("%d%d%d",&a,&b,&c);
            a ++;b ++;
            if(a == b)
                continue ;
            map[a][b] = map[b][a] = c;
        }
        printf("Case %d:
",cast);
        while(q --)
        {
            scanf("%d",&change);
            if(change == 0)
            {
                scanf("%d",&a);
                if(mark[a] == 1)
                   printf("ERROR! At point %d
",a);
                else
                   mark[a] = 1;
            }
            else
            {
                scanf("%d%d",&a,&b);
                if(mark[a] == 0 || mark[b] == 0)
                    printf("ERROR! At path %d to %d
",a,b);
                else
                {
                    int sum;
                    sum = Floyd(n,map,pre,a,b);
                    if(sum == MAXN)
                        printf("No such path
");
                    else
                        printf("%d
",sum);
                }
            }
        }
    }
    return 0;
}

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
using namespace std;
const int MAXN = 0x3f3f3f3f;
int n,m,q;
int a,b,c;
int map[310][310];
int change;
int cast;
int mark[310];
void Floyd(int k)
{
    for(int i = 1;i <= n;i ++)
       for(int j = 1;j <= n;j ++)
       {
           if(map[i][k] + map[k][j] < map[i][j])
              {
                  map[i][j] = map[i][k] + map[k][j];
              }
       }
}
int main()
{
    cast = 1;
    while(scanf("%d%d%d",&n,&m,&q),n,m,q)
    {
        if(cast > 1)
            printf("
");
        printf("Case %d:
",cast);
        cast ++;
        for(int i = 1;i <= n;i ++)
            for(int j = 1;j <= n;j ++)
            {
                if(i == j)
                   map[i][j] = 0;
                else
                   map[i][j] = MAXN;
            }
        memset(mark,0,sizeof(mark));
        for(int i = 1;i <= m;i ++)
        {
            scanf("%d%d%d",&a,&b,&c);
            a ++;b ++;
            if(a == b)
               continue ;
            else
            {
                if(map[a][b] > c)
                     map[a][b] = c;
            }
        }
        while(q --)
        {
            scanf("%d",&change);
            if(change == 0)
            {
                scanf("%d",&a);
                a ++;
                if(mark[a] == 1)
                {
                    a --;
                    printf("ERROR! At point %d
",a);
                }
                else
                {
                    mark[a] = 1;
                    Floyd(a);
                }
            }
            else
            {
                scanf("%d%d",&a,&b);
                a ++;b ++;
                if(mark[a] == 0 || mark[b] == 0)
                {
                    a --;b --;
                    printf("ERROR! At path %d to %d
",a,b);
                }
                else
                {
                    if(map[a][b] == MAXN)
                        printf("No such path
");
                    else
                        printf("%d
",map[a][b]);
                }
            }
        }
    }
    return 0;
}