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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?
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.
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
题目分析：按题意将那个表示距离的矩阵处理出来，然后就是单源最短路了。
代码如下：
# include<iostream>
# include<cstdio>
# include<queue>
# include<cstring>
# include<algorithm>
using namespace std;
const int INF=1<<30;
int n,m;
long long x[1001005],y[1001005],z[1001005];
int a[1005][1005],dis[1005];
void init()
{
for(int i=2;i<n*n;++i)
x[i]=(12345+((x[i-1]%5837501)*23456)%5837501+((x[i-2]%5837501)*34567)%5837501+(((x[i-1]%5837501)*(x[i-2]%5837501))%5837501)*45678)%5837501;
for(int i=2;i<n*n;++i)
y[i]=(56789+((y[i-1]%9860381)*67890)%9860381+((y[i-2]%9860381)*78901)%9860381+(((y[i-1]%9860381)*(y[i-2]%9860381))%9860381)*89012)%9860381;
for(int i=0;i<n*n;++i)
z[i]=(((x[i]%8475871)*90123+y[i])%8475871+1)%8475871;
for(int i=0;i<n;++i){
for(int j=0;j<n;++j)
a[i][j]=(i==j)?0:z[i*n+j];
}
}
void spfa()
{
fill(dis,dis+n,INF);
dis[0]=0;
queue<int>q;
q.push(0);
while(!q.empty())
{
int u=q.front();
q.pop();
for(int i=1;i<n;++i){
if(dis[i]>dis[u]+a[u][i]){
dis[i]=dis[u]+a[u][i];
q.push(i);
}
}
}
}
int main()
{
while(~scanf("%d%d%lld%lld%lld%lld",&n,&m,&x[0],&x[1],&y[0],&y[1]))
{
init();
spfa();
int ans=m;
for(int i=1;i<n;++i)
ans=min(ans,dis[i]%m);
printf("%d\n",ans);
}
return 0;
}
HDU-4849 Wow! Such City! （单源最短路）
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原文地址：http://www.cnblogs.com/20143605--pcx/p/4787111.html
