Database Management System 基础01：管理自己的任何事

Jerboas
题目：http://acm.hdu.edu.cn/showproblem.php?pid=2437
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

题意：沙漠中有个小老鼠（姑且认为是老鼠吧），它要找一个窝来在沙漠生存，有一些时临时的窝T，也有永久的窝P，现在它从一个临时的窝出发，去找永久的窝，问：最近的那个窝在哪，距离有多远。

限制：这个老鼠，一次只能走K步，换句话说就是从起点到终点的距离必须是K的倍数。

依旧是DFS题目，当时理解题意，直接做出来，TLE。。。

好吧，后来以为是建立二维数组存储从i点到达j点距离有点太大，于是用结构体存储，提交。。。TLE

终于发现，这道题目是考剪枝额，能剪的地方看只有vis数组上下手了。。

做这么多DFS也做过，vis数组改动的无外乎距离么，这道题也往那方面想，很快就能想出来：

假设从1号点出发，到达2号点有三条路，第一条路长度为4，第二条路长度为11。

假设K为7：那么第一条路和第二条路只需要搜索一次，不需要重复搜索。

为什么呢？因为4%7=4  11%7=4 ，还不懂？

因为最后从起点到终点的距离要是7的倍数，其实加的就是对7的求模，什么时候等于0，并且当时的点为P，就OK

然后当距离相等时，不要忘了，点也要取最近的（因为这个WA一次）

#include <iostream>
#include <string.h>
using namespace std;
char coor[1001];
bool vis[1001][1001];
int N,M,k,finish,minn;
bool ispos;

struct Map
{
    int len;
    int to[1001][2];
}m[1001];

void dfs(int x,int sum)
{
    if(sum>minn)    return;
    if(sum%k==0 && coor[x]==‘P‘ && sum!=0)
    {
        ispos=1;
        // 不光距离要取最小，当距离相同时点也要取最近的
        if(minn>=sum)
        {
            minn=sum;
            if(finish>x)
                finish=x;
        }
        return;
    }
    int i,total;
    for(i=0; i<m[x].len; ++i)
    {
        total=sum+m[x].to[i][1];
        // 对于vis数组的一个剪枝
        if(vis[m[x].to[i][0]][total%k]>total || !vis[m[x].to[i][0]][total%k])
        {
            vis[m[x].to[i][0]][total%k]=total;
            dfs(m[x].to[i][0],total);
        }

    }
}

int main()
{
    int s,i,num,a,b,c,t;
    cin>>t;
    for(num=1;num<=t;++num)
    {
        cin>>N>>M>>s>>k;
        memset(m,0,sizeof(m));
        memset(vis,0,sizeof(vis));

        for(i=1;i<=N;++i)   cin>>coor[i];
        for(i=0;i<M;++i)
        {
            cin>>a>>b>>c;
            m[a].to[m[a].len][0]=b;
            m[a].to[m[a].len][1]=c;
            ++m[a].len;
        }
        ispos=0;
        finish=1002;
        minn=999999;
        dfs(s,0);

        if(!ispos)  cout<<"Case "<<num<<": -1 -1"<<endl;
        else    cout<<"Case "<<num<<": "<<minn<<" "<<finish<<endl;
    }
    return 0;
}

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