HDU 4849 nyist省赛选拔题目

C - Wow! Such City!

 HDU - 4849 

Wow! Such City!

Time Limit: 15000/8000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 1740    Accepted Submission(s): 586

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 ≤ 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 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 X₀, X₁, Y₀, Y₁, (1 ≤ X₀, X₁, Y₀, Y₁≤ 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,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 44 20 2 3 4 5

 

Sample Output

110For the first test case, we have   0   1   2   3   4   5   6   7   8X   1   2 185180 7889971483212465942341237382178800 219267Y   3   4163319678455642071599456269735239123177371167849Z 90127 18025116203382064506 6251355664774564795082825524912390the cost matrix C is            0 1802511620338                2064506   05664774                56479508282552   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

没有想到要用最短路

                                                                                                 最短路dijkstra算法；

#include<cstdio>#include<cstring>#include<cmath>#include<cstdlib>#include<iostream>#include<algorithm>#include<queue>#include<stack>#include<map>#include<vector>using namespace std;typedef long long ll;typedef pair<int,int> P;const int inf=2e9;const int maxn=1000005;ll X[maxn],Y[maxn],Z[maxn],C[1005][1005],d[maxn],used[maxn],N;map<ll,int>mp;void dijk(){    memset(used,0,sizeof(used));    for(int i=0;i<N;i++)        d[i]=inf;        for(int i=1;i<N;i++)        d[i]=C[0][i];    used[0]=1;    for(int i=1;i<=N;i++)    {        int u=-1;        for(ll v=0;v<N;v++)        {            if(used[v]==0&&(u==-1||d[v]<d[u]))                u=v;        }        if(u==-1)  break;        used[u]=1;        for(ll v=0;v<N;v++)        {            d[v]=min(d[v],d[u]+C[u][v]);        }    }}int main(){    ll M,x0,x1,y0,y1;    while(~scanf("%lld%lld%lld%lld%lld%lld",&N,&M,&x0,&x1,&y0,&y1))    {        memset(X,0,sizeof(X));        memset(Y,0,sizeof(Y));        memset(Z,0,sizeof(Z));        memset(C,0,sizeof(C));        memset(d,0,sizeof(d));        mp.clear();        X[0]=x0,X[1]=x1,Y[0]=y0,Y[1]=y1;        int siz=N*N;        int a1=12345,b1=23456,c1=34567,d1=45678,mod1=5837501;        int a2=56789,b2=67890,c2=78901,d2=89012,mod2=9860381;        int a3=90123,mod3=8475871;        for(int i=2; i<siz; i++)            X[i]=(a1+(X[i-1]*b1)%mod1+(X[i-2]*c1)%mod1+((X[i-1]*X[i-2])*d1)%mod1)%mod1;        for(int i=2; i<siz; i++)            Y[i]=(a2+(Y[i-1]*b2)%mod2+(Y[i-2]*c2)%mod2+((Y[i-1]*Y[i-2])*d2)%mod2)%mod2;        for(int i=0; i<siz; i++)            Z[i]=(X[i]*a3+Y[i])%mod3+1;        for(int i=0;i<N;i++)        {            for(int j=0;j<N;j++)            {                C[i][j]=(i==j)?0:(Z[i*N+j]);            }        }        dijk();        for(int j=1; j<N; j++)        {//            printf("%lld \n",Z[j]);            ll l=d[j]%M;            mp[l]++;        }        map<ll,int>::iterator it=mp.begin();        cout<<it->first<<endl;    }    return 0;}