HDU 4849(最短短絡水~)

校試合以外にACM校チームへの加入を申請する方法はありますか? 
Wow! Such City! Time Limit: 15000/8000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others) Total Submission(s): 1453    Accepted Submission(s): 500
Problem Description
Doge, tired of being a popular image on internet, is considering moving to another city for a new way of life.
In his country there are N (2 ≤N≤ 1000) cities labeled 0 . . . N - 1. He is currently in city 0. Meanwhile, for each pair of cities, there exists a road connecting them, costing C
i,
j (a positive integer) for traveling from city i to city j. Please note that C
i,
j may not equal to C
j,
i for any given i ≠ j.
Doge is carefully examining the cities: in fact he will divide cities (his current city 0 is 
NOT included) into M (2 ≤ M ≤ 10
6) categories as follow: If the minimal cost from his current city (labeled 0) to the city i is Di, city i belongs to category numbered 
Di mod M.Doge wants to know the “minimal” category (a category with minimal number) which contains at least one city.
For example, for a country with 4 cities (labeled 0 . . . 3, note that city 0 is not considered), Doge wants to divide them into 3 categories. Suppose category 0 contains no city, category 1 contains city 2 and 3, while category 2 contains city 1, Doge consider category 1 as the minimal one.
Could you please help Doge solve this problem?
Note:
C
i,
j is generated in the following way:
Given integers X
0, X
1, Y
0, Y
1, (1 ≤ X
0, X
1, Y
0, Y
1≤ 1234567), for k ≥ 2 we have
Xk  = (12345 + X
k-1 * 23456 + X
k-2 * 34567 + X
k-1 * X
k-2 * 45678)  mod  5837501
Yk  = (56789 + Y
k-1 * 67890 + Y
k-2 * 78901 + Y
k-1 * Y
k-2 * 89012)  mod  9860381
The for k ≥ 0 we have
Z
k = (X
k * 90123 + Y
k ) mod 8475871 + 1
Finally for 0 ≤ i, j ≤ N - 1 we have
C
i,
j = Z
i*n+j for i ≠ j
C
i,
j = 0   for i = j
 
Input
There are several test cases. Please process till EOF.
For each test case, there is only one line containing 6 integers N,M,X
0,X
1,Y
0,Y
1.See the description for more details.
 
Output
For each test case, output a single line containing a single integer: the number of minimal category.
 
Sample Input

       
       
       
       

        
        
        
        3 10 1 2 3 4
4 20 2 3 4 5
       
       
       
       

 
Sample Output

       
       
       
       

        
        
        
        1
10

For the first test case, we have


	   0	   1	   2	   3	   4	   5	   6	   7	   8
X	   1	   2	 185180	 788997	1483212	4659423	4123738	2178800	 219267
Y	   3	   4	1633196	7845564	2071599	4562697	3523912	317737	1167849
Z	 90127	 180251	1620338	2064506	 625135	5664774	5647950	8282552	4912390

the cost matrix C is
	         	   0	 180251	1620338
	                2064506	   0	5664774
	                5647950	8282552	   0


        
        
        
        
 
         
 
          Hint 
         
So the minimal cost from city 0 to city 1 is 180251, while the distance to city 2 is 1620338. Given M = 10, city 1 and city 2 belong to category 1 and 8 respectively. Since only category 1 and 8 contain at least one city, the minimal one of them, category 1, is the desired answer to Doge’s question. 
        
        
        
        
         
       
       
       
       

 
题意:题目によってxyzの3つの配列を求めて、それから距离の行列を算出して、それから0から最もショートして、すべての対のショートモードmの后ですべての値の中を求めます
最小
裸题

#include <bits/stdc++.h>
using namespace std;
#define maxn 1111

long long n, m;
long long x[maxn*maxn], y[maxn*maxn], z[maxn*maxn];
long long mp[maxn][maxn];
long long d[maxn];
const long long INF = 1e18;
bool vis[maxn];

void dij () {
    for (int i = 0; i < n; i++)
        d[i] = INF;
    memset (vis, 0, sizeof vis);
    d[0] = 0;
    for (int i = 0; i < n; i++) {
        long long Min = INF, pos;
        for (int j = 0; j < n; j++) if (!vis[j] && d[j] < Min) {
            Min = d[j];
            pos = j;
        }
        vis[pos] = 1;
        for (int j = 0; j < n; j++) if (!vis[j]) {
            d[j] = min (d[j], d[pos]+mp[pos][j]);
        }
    }
    for (int i = 0; i < n; i++) d[i] %= m;
    long long ans = d[1];
    //for (int i = 0; i < n; i++) cout << d[i] << endl;
    for (int i = 1; i < n; i++) {
        ans = min (ans, d[i]);
    }
    cout << ans << endl;
}

int main () {
    //freopen ("in.txt", "r", stdin);
    while (cin >> n >> m >> x[0] >> x[1] >> y[0] >> y[1]) {
        for (int i = 0; i < n*n; i++) {
            if (i >= 2)
                x[i] = (12345 + x[i-1] * 23456 + x[i-2] * 34567 + x[i-1] * x[i-2] * 45678)  %  5837501;
            if (i >= 2)
                y[i] = (56789 + y[i-1] * 67890 + y[i-2] * 78901 + y[i-1] * y[i-2] * 89012)  %  9860381;
            z[i] = (x[i]*90123 + y[i]) % 8475871 + 1;
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (i == j)
                    mp[i][j] = 0;
                else
                    mp[i][j] = z[i*n+j];
            }
        }
        dij ();
    }
    return 0;
}