hdu 4849 Wow! Such City!(最短路)
2017-03-30 23:35
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Wow! Such City!
Problem DescriptionDoge, 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
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