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 ≤ 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.
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。。虽然不是纯最短路（不过也差不多了。。）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
#include
#include
#include
#include
#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
dis[i]) { minx=dis[i]; k=i; } } v[k]=1; for(int i=0; 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