HDU-4849 Wow! Such City! （单源最短路）

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 ≤ 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 numbered D_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 4

4 20 2 3 4 5

 

Sample Output

1

10

 

题目分析：按题意将那个表示距离的矩阵处理出来，然后就是单源最短路了。

 

 

代码如下：

# include<iostream>
# include<cstdio>
# include<queue>
# include<cstring>
# include<algorithm>
using namespace std;
const int INF=1<<30;
int n,m;
long long x[1001005],y[1001005],z[1001005];
int a[1005][1005],dis[1005];
void init()
{
    for(int i=2;i<n*n;++i)
        x[i]=(12345+((x[i-1]%5837501)*23456)%5837501+((x[i-2]%5837501)*34567)%5837501+(((x[i-1]%5837501)*(x[i-2]%5837501))%5837501)*45678)%5837501;
    for(int i=2;i<n*n;++i)
        y[i]=(56789+((y[i-1]%9860381)*67890)%9860381+((y[i-2]%9860381)*78901)%9860381+(((y[i-1]%9860381)*(y[i-2]%9860381))%9860381)*89012)%9860381;
    for(int i=0;i<n*n;++i)
        z[i]=(((x[i]%8475871)*90123+y[i])%8475871+1)%8475871;
    for(int i=0;i<n;++i){
        for(int j=0;j<n;++j)
            a[i][j]=(i==j)?0:z[i*n+j];
    }
}
void spfa()
{
    fill(dis,dis+n,INF);
    dis[0]=0;
    queue<int>q;
    q.push(0);
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        for(int i=1;i<n;++i){
            if(dis[i]>dis[u]+a[u][i]){
                dis[i]=dis[u]+a[u][i];
                q.push(i);
            }
        }
    }
}
int main()
{
    while(~scanf("%d%d%lld%lld%lld%lld",&n,&m,&x[0],&x[1],&y[0],&y[1]))
    {
        init();
        spfa();
        int ans=m;
        for(int i=1;i<n;++i)
            ans=min(ans,dis[i]%m);
        printf("%d\n",ans);
    }
    return 0;
}