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Wow! Such City!

Time Limit: 15000/8000 MS (Java/Others)????Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 2366????Accepted Submission(s): 808


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 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 HintSo 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西安全國邀請賽
Recommend liuyiding???|???We have carefully selected several similar problems for you:??6275?6274?6273?6272?6271?

題解：

根據(jù)給出的計算公式，計算任意兩點之間的距離，然后求出起點0到其他n-1個點的最短距離，

輸出最短距離對m取余之后的最小值

#include <iostream> #include <cstdio> #include <cstring> #include <queue> using namespace std; typedef long long ll; ll inf=0x3f3f3f3f; ll x[1010*1010+1010],y[1010*1010+1010],z[1010*1010+1010]; int Map[1010][1010]; int ans[1010];//ans[i]表示起點到i的最短距離 int vis[1010];//vis[i]表示i是否在隊列中 int n,m;void init() { //根據(jù)MAP數(shù)組的計算公式，i的最大值為n*nfor(int i=2; i<(n-1)*n+n; i++){ //這里和題目給定的公式不同，實際計算需要邊計算，邊取模x[i]=(12345+x[i-1]*23456%5837501+x[i-2]*34567%5837501+(x[i-1]*x[i-2]%5837501)*45678%5837501)%5837501;y[i]=(56789+y[i-1]*67890%9860381+y[i-2]*78901%9860381+(y[i-1]*y[i-2]%9860381)*89012%9860381)%9860381;}for(int k=0; k<(n-1)*n+n; k++){z[k]= (x[k]*90123%8475871+y[k])%8475871+1;}for(int i=0; i<n; i++){for(int j=0; j<n; j++){if(i==j){Map[i][j]=0;}else{Map[i][j]= z[i*n+j];}}}}void spfa() {for (int i=0; i<n; i++){ans[i]=inf;vis[i]=0;}queue<int>q;vis[0]=1;ans[0]=0;q.push(0);while (!q.empty()){int s=q.front();q.pop();vis[s]=0;for (int i=0; i<n; i++){if (ans[i]>Map[s][i]+ans[s]){ans[i]=Map[s][i]+ans[s];if (!vis[i]){q.push(i);vis[i]=1;}}}} }int main() {while (~scanf("%d%d%lld%lld%lld%lld",&n,&m,&x[0],&x[1],&y[0],&y[1])){init();spfa();int Min=ans[1]%m;for (int i=2; i<n; i++){if (Min>ans[i]%m)Min=ans[i]%m;}printf ("%d\n",Min);/*for(int i=0;i<n;i++)printf("%d %d\n",ans[i],ans[i]%m);*/}return 0; }

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