HDU - 4849 Wow! Such City! （dijkstra+水）

Wow! Such City!

Time Limit: 15000/8000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 1809    Accepted Submission(s): 610

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  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 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.
   
   
   
    
  
  

 

Source

2014西安全国邀请赛

 

Recommend

liuyiding

英语+数据把我给吓住了，其实就是一个简单的最短路问题，把每条边都计算出来，跑一遍dijkstra就行了

#include <iostream>
#include <iomanip>
#include<stdio.h>
#include<string.h>
#include<stack>
#include<stdlib.h>
#include<queue>
#include<map>
#include<math.h>
#include<algorithm>
#include<vector>
#define mem(a,b) memset(a,b,sizeof(a))
#define inf 0x3f3f3f3f
#define LL long long
#define maxn 1000000+5
#define N 1000+5
using namespace std;
LL n,m,xx0,xx1,yy0,yy1;
LL e[N][N];
LL dis[N];
LL vis[N];
LL x[maxn],y[maxn],z[maxn];
void dijkstra()
{
    LL minn,u=0;
    for(LL i=0;i<n;i++)
    {
        dis[i]=e[0][i];
        vis[i]=0;
    }
    dis[0]=0;
    for(LL i=0;i<n;i++)
    {
         minn=inf;
        for(LL j=0;j<n;j++)
        {
            if(minn>dis[j]&&!vis[j])
            {
                minn=dis[j];
                u=j;
            }
        }
        vis[u]=1;
        for(LL k=0;k<n;k++)
        {
            if(!vis[k]&&dis[k]>dis[u]+e[u][k])
            {
                dis[k]=dis[u]+e[u][k];
            }
        }
    }

}
void solve()
{
    LL minn=inf;
    for(LL i=1;i<n;i++)
    {
        minn=min(minn,dis[i]%m);
    }
    printf("%lld\n",minn);
}
void init()
{
    for(LL i=0;i<n;i++)
    {
        for(LL j=0;j<n;j++)
        {
            e[i][j]=inf;
        }
        e[i][i]=0;
    }
    x[0]=xx0;
    x[1]=xx1;
    y[0]=yy0;
    y[1]=yy1;
    for(LL i=2;i<=n*n;i++)
    {
    x[i]=(12345+x[i-1]*23456+x[i-2]*34567+x[i-1]*x[i-2]*45678)%5837501;
    y[i]=(56789+y[i-1]*67890+y[i-2]*78901+y[i-1]*y[i-2]*89012)%9860381;
    }
    for(LL i=0;i<=n*n;i++)
    {
        z[i]=(x[i]*90123+y[i])%8475871+1;
    }
        for(LL i=0;i<n;i++)
        {
            for(LL j=0;j<n;j++)
            {
                if(i==j)
                    e[i][j]=0;
                    else
                e[i][j]=z[i*n+j];
            }
        }
}
int main()
{

    while(~scanf("%lld%lld%lld%lld%lld%lld",&n,&m,&xx0,&xx1,&yy0,&yy1))
    {
        init();
        dijkstra();
        solve();
    }
    return 0;
}