hdu 4849 Wow! Such City!（最短路）

Wow! Such City!

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

For the first test case, we have

0       1       2       3       4       5       6       7       8

X 1 2 185180 788997 1483212 4659423 4123738 2178800 219267

Y 3 4 1633196 7845564 2071599 4562697 3523912 317737 1167849

Z 90127 180251 1620338 2064506 625135 5664774 5647950 8282552 4912390

the cost matrix C is

0 180251 1620338

2064506 0 5664774

5647950 8282552 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.

ps：比赛时做了一个小时，结果把题意理解错了，真TTTTTT尴尬！

我的思路：最短路
4000
之后，记录下到每一点的距离对m取余的数共出现多少次，最后出现次数最多的那个数就是“最短距离“

正解思路：最短路之后，起点到每一点的距离对m取余，余数的大小即“最短距离”

代码：

#include<stdio.h>
#include<queue>
#include<string.h>
#include<algorithm>
using namespace std;

#define mem(a,b) memset(a,b,sizeof(a))
#define min(a,b) (a<b?a:b)
#define maxn 1000+5
#define maxv 1000000+5
const int inf=0x3f3f3f3f;
typedef long long  LL;

LL n,m;
LL dis[maxn][maxn];
LL x[maxv],y[maxv],z[maxv];
LL d[maxn];
bool vis[maxn];

void dijkstra()
{
LL i,j,minn,v;
for(i=0; i<n; i++)
{
vis[i]=0;
d[i]=dis[0][i];
}
vis[0]=true;
for(i=0; i<n; i++)
{
minn=inf;
for(j=0; j<n; j++)
if(!vis[j]&&d[j]<minn)
minn=d[j],v=j;
vis[v]=true;
for(j=0; j<n; j++)
{
if(!vis[j]&&d[j]>d[v]+dis[v][j])
d[j]=d[v]+dis[v][j];
}
}
}

void solve()
{
LL maxx=inf;
for(LL i=1;i<n;i++)
maxx=min(maxx,d[i]%m);
printf("%lld\n",maxx);
}

void init()
{
LL sum=n*(n-1)+n-1;
z[0]=(x[0]*90123+y[0])%8475871+1;
z[1]=(x[1]*90123+y[1])%8475871+1;
for(LL i=2; i<sum; i++)
{
x[i]=(12345+(x[i-1]*23456%5837501)+(x[i-2]*34567%5837501)+(x[i-1]*x[i-2]*45678%5837501))%5837501;
y[i]=(56789+(y[i-1]*67890%9860381)+(y[i-2]*78901%9860381)+(y[i-1]*y[i-2]*89012%9860381))%9860381;
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)
dis[i][j]=0;
else
dis[i][j]=z[i*n+j];
}
}

int main()
{
while(~scanf("%lld%lld%lld%lld%lld%lld",&n,&m,&x[0],&x[1],&y[0],&y[1]))
{
init();
dijkstra();
solve();
}
return 0;
}

总结：为什么我理解的意思就和正解不一样，，还是英语太渣！

认真读题！看不懂的单词就查字典！wrong多次后必须要重新读题，判断是否理解错误题意