HDU 4849-Wow! Such City!（最短路）

Wow! Such City!

Time Limit: 15000/8000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)

Total Submission(s): 824    Accepted Submission(s): 310

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

头一次遇到区域赛的最短路的题目。。

尽管不是纯最短路（只是也差点儿相同了。。）权值由递推公式（题目中已给出）生成，然后跑一遍dijkstra，起点为1，求dis[i]%m的最小值。权值注意超出int范围要用lld

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;
ll dis[1010],v[1010];
ll map[1010][1010];
int x0,x1,y0,y1,n,m;
void dijkstra()
{
    int minx,k=0;
    for(int i=0; i<=n; i++)
    {
        dis[i]=map[0][i];;
        v[i]=0;
    }
    dis[0]=0;
    for(int j=0; j<n; j++)
    {
        minx=inf;
        for(int i=0; i<n; i++)
        {
            if(v[i]==0&&minx>dis[i])
            {
                minx=dis[i];
                k=i;
            }
        }
        v[k]=1;
        for(int i=0; i<n; i++)
        {
            if(v[i]==0&&dis[i]>dis[k]+map[k][i])
            {
                dis[i]=dis[k]+map[k][i];
            }
        }
    }
    return ;
}
ll xx[1002000],yy[1002000],zz[1002000];
int main()
{
    while(scanf("%d%d%d%d%d%d",&n,&m,&x0,&x1,&y0,&y1)!=EOF)
    {
        for(int i=0; i<n; i++)
        {
            for(int j=0; j<n; j++)
            {
                map[i][j]=inf;
            }
            map[i][i]=0;
        }
        xx[0]=x0;xx[1]=x1;
        yy[0]=y0;yy[1]=y1;
        for(int i=2; i<=n*(n-1)+n; i++)
        {
            xx[i]=(12345+xx[i-1]*23456+xx[i-2]*34567+xx[i-1]*xx[i-2]*45678)%5837501;
            yy[i]=(56789+yy[i-1]*67890+yy[i-2]*78901+yy[i-1]*yy[i-2]*89012)%9860381;
        }
        for(int i=0; i<=n*(n-1)+n; i++)
        {
            zz[i]=(xx[i]*90123+yy[i])%8475871+1;
        }
        for(int i=0; i<n; i++)
        {
            for(int j=0; j<n; j++)
            {
                if(i==j)
                    map[i][j]=0;
                else map[i][j]=zz[i*n+j];
            }
        }
        dijkstra();
        ll minn=inf;
        for(int i=1; i<n; i++)
        {
            minn=min(minn,dis[i]%m);
        }
        printf("%lld\n",minn);
    }
    return 0;
}