ZOJ-3318

Strange Country

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Time Limit: 1 Second      Memory Limit: 32768 KB

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There are n cities in the dream country. Let's use integers from 1 to n to denote the cities. There are some roads between cities. To some strange, all the roads are bidirectional and the roads change from time to time. You have m maps of the country of different time. You are going to the dream country soon and you want to start your journey at city s and finish it at city t. Though you cannot predict the condition when you get there, you think it is useful to study the maps carefully. After studying the maps, you find that all the roads in all maps have the same length and there is an s-t path in each map. You want to choose an s-t path in each map and the paths are relatively short. Further more you don't want too many changes in the paths.

Formally suppose you have chosen an s-t path in each map, namely P₁, P₂,...P_m. Let's define a path's length to be simply the number of edges in it and use LEN to denote the total length of all the paths. Let's define a function as follow: CHANGE(P₁, P₂,...P_m) is the number of indices i (0 < i < m) for which P_i != P_i+1. Let's define the cost function as follow: COST(P₁, P₂,...P_m) = LEN + CHANGE(P₁, P₂,...P_m). You are supposed to find the minimum cost.

Input

There are multiple test cases. The first line of input is an integer T (0 < T < 205) indicating the number of test cases. Then T test cases follow. The first line of each test case is 4 integers n, m, s, t (1 < n, m <= 30, 0 < s, t <= n, s != t). Then there are m map descriptions. The first line of each map description is an integer R, the number of roads in the map (0 < R <= n * (n - 1) / 2). Each of the next R lines contains two integers a, b, the two cities that road connects( 0 < a, b <= n, a != b). You can assume that for each test case there is an s-t path in each map.

Output

For each test case, output in a line the minimum cost defined above.

Sample Input

2
3 3 2 3
2
1 2
3 1
3
1 2
2 3
3 2
2
2 1
2 3
4 2 1 4
3
1 2
2 3
3 4
3
1 2
2 3
3 4

Sample Output

5
6

Hint

Test case 1: three paths are 2-1-3, 2-3, 2-3.

Test case 2: both paths are 1-2-3-4.

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Author: CAO, Peng

Source: The 10th Zhejiang University Programming Contest
Submit

Status

#include <iostream>
#include <stdio.h>
#include<cmath>
#include<algorithm>
#include<string.h>
#include<queue>
#include<set>
#define maxn 40
using namespace std;
int dist[maxn];
int mark[maxn][maxn];
int Edge[maxn][maxn];
int mmap[maxn][maxn][maxn];
int dp[maxn];
int n,m,s,t,flag;
int bfs()
{
    memset(dist,0x3f,sizeof(dist));
    flag = dist[0];
    dist[s] = 0;
    queue<int>que;
    while(!que.empty()) que.pop();
    que.push(s);
    while(!que.empty())
    {
                    int tmp = que.front();
                    que.pop();
                    for(int i=1;i<=n;i++)
                    {
                              if(Edge[tmp][i] == 1 && dist[i] > dist[tmp] + 1)
                              {
                                        dist[i] = dist[tmp] + 1;
                                        que.push(i);
                              }
                    }
    }
    return dist[t];
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
#endif // ONLINE_JUDGE
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d %d %d %d",&n,&m,&s,&t);
        int R,u,v;
        memset(mmap,0,sizeof(mmap));
        memset(mark,0,sizeof(mark));
        memset(dp,0,sizeof(dp));
        for(int i=0; i<m; i++)
        {
            scanf("%d",&R);
            for(int j=0; j<R; j++)
            {
                scanf("%d %d",&u,&v);
                mmap[i][u][v] = 1;
                mmap[i][v][u] = 1;
            }
        }
        for(int i=0; i<m; i++)
        {
            memset(Edge,1,sizeof(Edge));
            for(int j=i; j<m; j++)
            {
                for(int k=1; k<=n; k++)
                {
                    for(int t= 1; t <=n; t++)
                    {
                                                            Edge[k][t] = Edge[k][t] & mmap[j][k][t];
                    }
                }
                mark[i][j] = bfs();
            }
        }
        dp[0] = mark[0][0];
        for(int i=1;i<=m;i++)
        {
                              if(mark[0][i] != flag) dp[i] = mark[0][i] *(i+1);
                              else dp[i] = flag;
                              for(int j=0;j<i;j++)
                              {
                              if(mark[j+1][i] != flag)
                                        dp[i] = min(dp[i] ,dp[j] + mark[j+1][i]*(i-j) + 1);
                              }
        }
        printf("%d\n",dp[m-1]);
    }
    return 0;
}