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 }
突然发现 最短路 有好多变形题啊 蛮有意思的