HDU

Wow! Such City!

Time Limit: 15000/8000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 1809    Accepted Submission(s): 610

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西安全国邀请赛

 

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liuyiding

英语+数据把我给吓住了，其实就是一个简单的最短路问题，把每条边都计算出来，跑一遍dijkstra就行了

#include <iostream>#include <iomanip>#include<stdio.h>#include<string.h>#include<stack>#include<stdlib.h>#include<queue>#include<map>#include<math.h>#include<algorithm>#include<vector>#define mem(a,b) memset(a,b,sizeof(a))#define inf 0x3f3f3f3f#define LL long long#define maxn 1000000+5#define N 1000+5using namespace std;LL n,m,xx0,xx1,yy0,yy1;LL e[N][N];LL dis[N];LL vis[N];LL x[maxn],y[maxn],z[maxn];void dijkstra(){    LL minn,u=0;    for(LL i=0;i<n;i++)    {        dis[i]=e[0][i];        vis[i]=0;    }    dis[0]=0;    for(LL i=0;i<n;i++)    {         minn=inf;        for(LL j=0;j<n;j++)        {            if(minn>dis[j]&&!vis[j])            {                minn=dis[j];                u=j;            }        }        vis[u]=1;        for(LL k=0;k<n;k++)        {            if(!vis[k]&&dis[k]>dis[u]+e[u][k])            {                dis[k]=dis[u]+e[u][k];            }        }    }}void solve(){    LL minn=inf;    for(LL i=1;i<n;i++)    {        minn=min(minn,dis[i]%m);    }    printf("%lld\n",minn);}void init(){    for(LL i=0;i<n;i++)    {        for(LL j=0;j<n;j++)        {            e[i][j]=inf;        }        e[i][i]=0;    }    x[0]=xx0;    x[1]=xx1;    y[0]=yy0;    y[1]=yy1;    for(LL i=2;i<=n*n;i++)    {    x[i]=(12345+x[i-1]*23456+x[i-2]*34567+x[i-1]*x[i-2]*45678)%5837501;    y[i]=(56789+y[i-1]*67890+y[i-2]*78901+y[i-1]*y[i-2]*89012)%9860381;    }    for(LL i=0;i<=n*n;i++)    {        z[i]=(x[i]*90123+y[i])%8475871+1;    }        for(LL i=0;i<n;i++)        {            for(LL j=0;j<n;j++)            {                if(i==j)                    e[i][j]=0;                    else                e[i][j]=z[i*n+j];            }        }}int main(){    while(~scanf("%lld%lld%lld%lld%lld%lld",&n,&m,&xx0,&xx1,&yy0,&yy1))    {        init();        dijkstra();        solve();    }    return 0;}