Wow! Such City!
Time Limit: 15000/8000 MS (Java/Others) Memory Limit: 102400/102400 K (Java/Others)Total Submission(s): 2366 Accepted Submission(s): 808
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
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.
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
Recommend
题解：
根据给出的计算公式，计算任意两点之间的距离，然后求出起点0到其他n-1个点的最短距离，
输出最短距离对m取余之后的最小值
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
typedef long long ll;
ll inf=0x3f3f3f3f;
ll x[1010*1010+1010],y[1010*1010+1010],z[1010*1010+1010];
int Map[1010][1010];
int ans[1010];//ans[i]表示起点到i的最短距离
int vis[1010];//vis[i]表示i是否在队列中
int n,m;
void init()
{ //根据MAP数组的计算公式，i的最大值为n*n
for(int i=2; i<(n-1)*n+n; i++)
{ //这里和题目给定的公式不同，实际计算需要边计算，边取模
x[i]=(12345+x[i-1]*23456%5837501+x[i-2]*34567%5837501+(x[i-1]*x[i-2]%5837501)*45678%5837501)%5837501;
y[i]=(56789+y[i-1]*67890%9860381+y[i-2]*78901%9860381+(y[i-1]*y[i-2]%9860381)*89012%9860381)%9860381;
}
for(int k=0; k<(n-1)*n+n; k++)
{
z[k]= (x[k]*90123%8475871+y[k])%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]= z[i*n+j];
}
}
}
}
void spfa()
{
for (int i=0; i<n; i++)
{
ans[i]=inf;
vis[i]=0;
}
queue<int>q;
vis[0]=1;
ans[0]=0;
q.push(0);
while (!q.empty())
{
int s=q.front();
q.pop();
vis[s]=0;
for (int i=0; i<n; i++)
{
if (ans[i]>Map[s][i]+ans[s])
{
ans[i]=Map[s][i]+ans[s];
if (!vis[i])
{
q.push(i);
vis[i]=1;
}
}
}
}
}
int main()
{
while (~scanf("%d%d%lld%lld%lld%lld",&n,&m,&x[0],&x[1],&y[0],&y[1]))
{
init();
spfa();
int Min=ans[1]%m;
for (int i=2; i<n; i++)
{
if (Min>ans[i]%m)
Min=ans[i]%m;
}
printf ("%d\n",Min);
/*
for(int i=0;i<n;i++)
printf("%d %d\n",ans[i],ans[i]%m);
*/
}
return 0;
}