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 Ci,j (a positive integer) for traveling from city i to city
j. Please note that Ci,j may not equal to Cj,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 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?
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 Ci,j (a positive integer) for traveling from city i to city
j. Please note that Ci,j may not equal to Cj,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 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:
Ci,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 + Xk-1 * 23456 + Xk-2 * 34567 + Xk-1 * Xk-2 * 45678) mod 5837501
Yk = (56789 + Yk-1 * 67890 + Yk-2 * 78901 + Yk-1 * Yk-2 * 89012) mod 9860381
The for k ≥ 0 we have
Zk = (Xk * 90123 + Yk ) mod 8475871 + 1
Finally for 0 ≤ i, j ≤ N - 1 we have
Ci,j = Zi*n+j for i ≠ j
Ci,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.
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;
}