hdu 2437 Jerboas

Jerboas

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 443    Accepted Submission(s): 108

Problem Description

      Jerboas are small desert-living animals, which resemble mice with a long tufted tail and very long hind legs. Jerboas shelter in well-hidden burrows. They create two types of burrow: temporary and permanent. The temporary burrows are plain tubes while the permanent burrows are sealed with a plug of sand to keep heat out and moisture in.



      As far as we know, jerboa burrows in the desert are connected with one-way tunnels. What's more, for some unknown reasons, it's true that start from any burrow, follows the tunnels you can not go back to the starting burrow.
      Summer means last-minute of offers on good times, so of course jerboas could not stay behind. One day, a little jerboa Alice who lived in a temporary burrow S wants to migrate to a permanent one. There are different routes she can take, but Alice is so odd that she only selects those whose total travel distances is a multiple of K. Among all routes that Alice may select, we are interested in the shortest one. Can you help to find it out? Of course different routes may lead to different destinations.

 

Input

      On the first line of input, there is a single positive integer T <= 20 specifying the number of test cases to follow.
      Each test case starts with four integers in the first line: N, M, S, K.
      N is the number of burrows in the desert (burrows are numbered with 1, 2, …, N);
      M is the number of tunnels connecting the burrows;
      S is where Alice lived and K is as described above.
(0 < N <= 1000, 0 <= M <= 20000, 0 < S <= N, 0 < K <= 1000)
      The second line contains N characters each could be ‘T’ or ‘P’. The i-th character specifying the type of the burrow i. ‘T’ means temporary burrow, ‘P’ means permanent burrow. It’s guaranteed that the S-th character is ‘T’.
      Next follow M lines, each line with 3 integers A, B, C. Specifying that there is a tunnel from burrow A to burrow B, and its length is C.
(0 < A, B <= N, A != B, 0 < C < 40000)

 

Output

      For each test case you should output a single line containing "Case X: Y Z" (quotes for clarity) where X is the number of the test case (starting at 1) and Y is the length of the shortest route Alice can select and Z is the destination of the selected route.
      Notice that burrow Z should be a permanent burrow.
      In case there’s more than one solution, Z should be the minimum.
      In case there's no solution, Y and Z should be both equal to -1.

 

Sample Input

2
5 5 1 7
TPPTP
1 2 8
1 4 7
4 3 9
2 3 6
1 5 3
5 5 1 7
TPTTP
1 2 8
1 4 7
4 3 9
2 3 6
1 5 3

 

Sample Output

Case 1: 14 3
Case 2: -1 -1

 

Source

第33届ACM/ICPC亚洲区预选赛_网络选拔赛(成都)

 

Recommend

lcy

 

Statistic | Submit | Back

//1363982 2009-05-13 19:16:28 Time Limit Exceeded 2437 1000MS 4400K 1494 B C++ Wpl 
//1364216 2009-05-13 19:58:46 Accepted 2437 218MS 4412K 1589 B C++ Wpl 
//邻接表+DFS,主要解决从一个点出发,到其它所有终点(所有有'P'标记的都可以作为终点),而且要求这个距离满足一定的条件就是必须是m的整数倍,而且要求是满足所有m倍中最短的那条路径,刚开始用Dijkstra超时,注意…….
#include <iostream>
#include <vector>
#define MAX 1002
using namespace std;
bool ok[MAX],flag;
int n,m,t,kk,start;
int MM,bestend;
typedef struct node
{
    int dis;
    int vector;
    node(){}
    node(int vv,int dd)
    {
        vector=vv;
        dis=dd;
    }
}Point;
vector<Point>G[MAX];
int edges[MAX][MAX];  //还是不太明白edges是用来干什么的
void Init()
{    
    int i,j;
    scanf("%d%d%d%d",&n,&m,&start,&kk);
    char ch;
    for(i=1;i<=n;i++)
    {
        ok[i]=false;
        cin>>ch;
        if(ch=='P')
            ok[i]=true;
        G[i].clear();  //注意清空,切记
    }
    for(i=0;i<=n;i++)
        for(j=0;j<=n;j++)
            edges[i][j]=-1;
    int a,b,s;
    while(m--)
    {
        scanf("%d%d%d",&a,&b,&s);
        G[a].push_back(node(b,s));
    }
}
void DFS(int ssum,int s)
{
    int i;    
    if(ok[s]&&ssum%kk==0)
    {
        if(ssum<MM||MM==-1)
        {
            MM=ssum;
            bestend=s;
            flag=true;
        }
        else if(ssum==MM&&s<bestend)
            bestend=s;
        return ;
    }
    if(G[s].empty())
        return ;
    for(i=0;i<G[s].size();i++)
    {
        int t=ssum+G[s][i].dis;
        int next=G[s][i].vector;
        if(edges[next][t%kk]==-1||edges[next][t%kk]>t)
        {
            edges[next][t%kk]=t;
            DFS(t,next);
        }
    }
}
int main()
{
    scanf("%d",&t);
    int www=1;
    while(t--)
    {
        Init();
        MM=-1;
        flag=false;
        DFS(0,start);
        printf("Case %d: ",www++);
        if(flag)
        {
            printf("%d %d\n",MM,bestend);
        }
        else
        {
            printf("-1 -1\n");
        }
    }
    return 0;
}