kuangbin four thematic topics Silver Cow Party POJ - 3268

 

Topic links: https://vjudge.net/problem/POJ-3268

Meaning of the questions: start line at point X, cattle other points to point X to the party, and then back to their point from point X, path is one-way, all cattle must take the shortest path,

Find the shortest road to go all the cattle in the longest.

(Direct look at the code it, is a change it, and other board dijkstra almost)

---

  1 #include <iostream>
  2 #include <cstring>
  3 #include <algorithm>
  4 #include <cstdio>
  5 #include <string>
  6 using namespace std;
  7  
  8 typedef long long LL;
  9 #define inf (1LL << 30) - 1
 10 #define rep(i,j,k) for(int i = (j); i <= (k); i++)
 11 #define rep__(i,j,k) for(int i = (j); i < (k); i++)
 12 #define per(i,j,k) for(int i = (j); i >= (k); i--)
 13 #define per__(i,j,k) for(int i = (j); i > (k); i--)
 14 
 15 const int N = 1010;
 16 int G[N][N];
 17 bool vis[N];
 18 int dis1[N];
 19 int dis2[N];
 20 int u,v,w;
 21 int n,m,X;
 22 
 23 void init(){
 24 
 25     rep(i,1,n) vis[i] = false;
 26     rep(i,1,n) rep(j,1,n){
 27         if(i == j) G[i][j] = 0;
 28         else G[i][j] = inf;
 29     }
 30 }
 31 
 32 void input(){
 33 
 34     rep(i,1,m){
 35         cin >> u >> v >> w;
 36         if(G[u][v] > w) G[u][v] = w;
 37     }
 38 }
 39 
 40 void dijkstra(){
 41 
 42 //////////////////////////////////////////////////////////
 43     //（1）第一部分
 44     REP (I, . 1 , n-) DIS1 [I] G = [I] [X]; // DIS [] is stored to the shortest distance X city point 
45      VIS [X] = to true ;
 46 is  
47      REP (I , 2 , n-) {
 48  
49          int X = - . 1 ;
 50          int W = INF;
 51 is  
52 is          REP (J, . 1 , n-) {
 53 is              IF ! (VIS [J] && W> DIS1 [J]) W = DIS1 [X = J];
 54 is          }
 55          IF (X == - . 1 ) Continue ;
 56 is  
57 is          VIS [X] =to true ;
 58          REP (k, . 1 , n-) {
 59              // if the distance from point k to point X k is greater than the distance from the point x to the point x to point X plus distance DIS then update [] array 
60              IF (VIS! [K] && DIS1 [K]> DIS1 [X] + G [K] [X]) {
 61 is                  DIS1 [K] = DIS1 [X] + G [K] [X];
 62 is              }
 63 is          }
 64      }
 65  
66   //    REP (I,. 1, n-) COUT << DIS1 [I] << endl; 
67  
68  /////////////////////////// /////////////////////////////////
 69      // (2) the second part is exactly the same dijkstra board 
70      REP ( I, . 1 , n-) VIS [I] =false;
 71     rep(i,1,n) dis2[i] = G[X][i];
 72     vis[X] = true;
 73 
 74     rep(i,2,n){
 75         int x = -1;
 76         int w = inf;
 77 
 78         rep(j,1,n){
 79             if(!vis[j] && w > dis2[j]) w = dis2[x = j];
 80         }
 81 
 82         if(x == -1) continue;
 83 
 84         VIS [X] = to true ; 
 85          REP (K, . 1 , n-) {
 86              IF (VIS [K] && DIS2 [K]> DIS2 [X] +! {G [X] [K])
 87                  DIS2 [K] DIS2 = [X] + G [X] [K];
 88              }
 89          }
 90      }
 91 is      int ANS = 0 ;
 92  
93      // the line back and forth together, select the longest round, is the answer 
94      REP ( I, . 1 , n-) IF ! (= X-I) = max ANS (ANS, DIS1 [I] + DIS2 [I]);
 95  
96      COUT ANS << << endl;
 97 }
 98 
 99 int main(){
100  
101     ios::sync_with_stdio(false);
102     cin.tie(0);
103 
104     cin >> n >> m >> X;
105 
106     init();
107     input();
108     dijkstra();
109 
110     getchar();getchar();
111     return 0;
112 }