hdu--1595-另类最短路

这题一定要好好读题啊不能走马观花...

Mirko overheard in the car that one of the roads is under repairs, and that it is blocked, but didn't konw exactly which road

---有一条路是坏的但是现在不确定究竟是那一条就是任意的

Mirko wants to know how long will it take for her to get to his city in the worst case, Mirko in finding out what is the longest time in minutes it could take for Marica to come by shortest route by non-blocked roads to his city.

--这2段话应该放一起理解比较好我们要选择的还是一条最短路但是最坏的情况下.. 说白了就是假如在所有路都连通的情况下最短路径是

A->B->C->D->E

但是现在会有一条路是closed 是不连通的而且是那条题目没确定那么我们就要尝试删掉A-B / B-C / C-D / D-E

做法就是这样了

  1 #include <iostream>
  2 #include <cstring>
  3 #include <algorithm>
  4 #include <queue>
  5 using namespace std;
  6 
  7 int n;
  8 const int size = 1010;
  9 const int inf = 0x3f3f3f3f;
 10 queue<int>q;
 11 int dist[size];
 12 bool vis[size];
 13 int pre[size];
 14 struct data
 15 {
 16     int num;
 17     int to[size];
 18     int dist[size];
 19 }node[size];
 20 
 21 void add( int from , int to , int val )
 22 {
 23     node[from].to[ node[from].num ] = to;
 24     node[from].dist[ node[from].num ] = val;
 25     node[from].num ++;
 26 }
 27 
 28 void spfa( int flag )
 29 {
 30     int now , next;
 31     for( int i = 0 ; i<=n ; i++ )
 32     {
 33         dist[i] = (i == 1) ? 0 : inf;
 34     }
 35     memset( vis , false , sizeof(vis) );
 36     q.push(1);
 37     vis[1] = true;
 38     while( !q.empty() )
 39     {
 40         now = q.front();
 41         q.pop();
 42         vis[now] = false;
 43         for( int i = 0 ; i < node[now].num ; i++ )
 44         {
 45             next = node[now].to[i];
 46             if( dist[next] > dist[now] + node[now].dist[i] )
 47             {
 48                 dist[next] = dist[now] + node[now].dist[i];
 49                 if(flag)
 50                     pre[next] = now;
 51                 if( !vis[next] )
 52                 {
 53                     vis[next] = true;
 54                     q.push(next);
 55                 }
 56             }
 57         }
 58     }
 59 }
 60 
 61 int main()
 62 {
 63     cin.sync_with_stdio(false);
 64     int m , from , to , val , ans , index , id;
 65     while( cin >> n >> m )
 66     {
 67         memset( pre , -1 , sizeof(pre) );
 68         for( int i = 0 ; i<=n ; i++ )
 69             node[i].num = 0;
 70         for( int i = 0 ; i<m ; i++ )
 71         {
 72             cin >> from >> to >> val;
 73             add( from , to , val );
 74             add( to , from , val );
 75         }
 76         spfa(1);
 77         ans = dist[n];
 78         for( int i = n ; ~pre[i] ; i = pre[i] )
 79         {
 80             for( int j = 0 ; j<node[i].num ; j++ )
 81             {
 82                 if( node[i].to[j] == pre[i] )
 83                 {
 84                     val = node[i].dist[j];
 85                     index = j;
 86                     node[i].dist[j] = inf;
 87                     break;
 88                 }
 89             }
 90             for( int j = 0 ; j<node[ pre[i] ].num ; j++ )
 91             {
 92                 if( node[ pre[i] ].to[j] == i )
 93                 {
 94                     id = j;
 95                     node[ pre[i] ].dist[j] = inf;
 96                     break;
 97                 }
 98             }
 99             spfa(0);
100             node[i].dist[index] = val;
101             node[ pre[i] ].dist[id] = val;
102             if( ans < dist[n] )
103                 ans = dist[n];
104         }
105         cout << ans << endl;
106     }
107     return 0;
108 }

View Code

突然发现最短路有好多变形题啊蛮有意思的

just follow your heart

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原文地址：https://www.cnblogs.com/radical/p/3961913.html