HDU 4849 Wow! Such City! (Dijkstra, 最短路)

Wow! Such City!

Time Limit: 15000/8000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 525    Accepted Submission(s): 188

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 isNOT included) into M (2 ≤ M ≤ 10⁶) categories as follow: If the minimal cost from his current city (labeled 0) to the city i is Di, city i belongs to category numberedD_i 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 X₀, X₁, Y₀, Y₁, (1 ≤ X₀, X₁, Y₀, Y₁≤ 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,X₀,X₁,Y₀,Y₁.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 44 20 2 3 4 5

 

Sample Output

110For the first test case, we have   0   1   2   3   4   5   6   7   8X   1   2 185180 7889971483212465942341237382178800 219267Y   3   4163319678455642071599456269735239123177371167849Z 90127 18025116203382064506 6251355664774564795082825524912390the cost matrix C is            0 1802511620338                2064506   05664774                56479508282552   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.
 

 

Source

2014西安全国邀请赛


题目链接  :http://acm.hdu.edu.cn/showproblem.php?pid=4849


题目大意  :用所给公式生成一个矩阵，求起点到各个点的最短路径，然后将各个最短路径对m取模，求所得模的最小值


题目分析  :用公式生成矩阵，然后迪杰斯特拉算法求出各点到原点的最短路径最后取模比较即可

#include <cstdio>#include <cstring>#include <algorithm>#include <limits.h>using namespace std;int const INF = INT_MAX;int const MAX = 1000 + 2;//这里用int会超范围long long x[MAX * MAX], y[MAX * MAX], z[MAX * MAX];long long c[MAX][MAX]; //生成的矩阵int dist[MAX], vis[MAX];int n, m;int ans;int Dijkstra(int v0) //迪杰斯特拉模版{    for(int i = 0; i < n; i++)    {        dist[i] = c[v0][i];        vis[i] = 0;     }    vis[v0] = 1;    dist[v0] = 0;    for(int i = 0; i < n - 1; i++)    {        int Min = INF, u = v0;        for(int j = 0; j < n; j++)        {            if(!vis[j] && dist[j] < Min)            {                u = j;                Min = dist[j];            }        }        vis[u] = 1;        for(int k = 0; k < n; k++)            if(!vis[k] && c[u][k] < INF && dist[u] + c[u][k] < dist[k])                dist[k] = dist[u] + c[u][k];    }    ans = INF;  //求所得模的最小值    for(int i = 1; i < n; i++)        ans = min(ans, dist[i] % m);}int main(){    int size;    while(scanf("%d%d%lld%lld%lld%lld", &n, &m, &x[0], &x[1], &y[0], &y[1]) != EOF)    {           size = n * n;        for(int k = 2; k < size; k++)        {            x[k] = (12345 + x[k-1] * 23456 + x[k-2] * 34567 + x[k-1] * x[k-2] * 45678) % 5837501;            y[k] = (56789 + y[k-1] * 67890 + y[k-2] * 78901 + y[k-1] * y[k-2] * 89012) % 9860381;          }        for(int k = 0; k < size; k++)            z[k] = (x[k] * 90123 + y[k]) % 8475871 + 1;        //生成矩阵        for(int i = 0; i < n; i++)            for(int j = 0; j < n; j++)                c[i][j] = (i == j ? 0 : z[i * n + j]);        Dijkstra(0);        printf("%d\n", ans);    }}