1111 Online Map（迪杰斯特拉+DFS）
题目：https://pintia.cn/problem-sets/994805342720868352/problems/994805358663417856
输出要求：
翻译结果：
在最短路径不唯一的情况下，输出最短路径中最快的一条，保证唯一。
如果最快的路径不是唯一的，则输出通过最少交叉口的路径，该交叉口保证是唯一的。
如果最短路径和最快路径相同，请按以下格式将它们打印在一行中：
题目思路很简答，使用两个不同度量（距离、时间）的 迪杰斯特拉+DFS 求解即可。
PS：我英语好菜，我翻译错了，导致做不出来，呜呜呜~~
1 #include<iostream> 2 #include<vector> 3 #include<algorithm> 4 using namespace std; 5 const int maxn = 510; 6 const int inf = 0x3fffffff; 7 int G[maxn][maxn],Time[maxn][maxn],n,m,v1,v2,flag,length,TIME,st,ed; 8 bool visited[maxn]; 9 10 int d[maxn],t[maxn]; 11 vector<int> pre[maxn]; 12 vector<int> path1,path2,tempPath; 13 14 void dijkstra1(int st) { 15 fill(visited,visited+maxn,false); 16 fill(d,d+maxn,inf); 17 d[st] = 0; 18 for(int i = 0; i < n; ++i) { 19 int u = -1,MIN = inf; 20 for(int j = 0; j < n; ++j) { 21 if(!visited[j] && d[j] < MIN) { 22 u = j; 23 MIN = d[j]; 24 } 25 } 26 if(u == -1) return ; 27 visited[u] = true; 28 for(int v = 0; v < n; ++v) { 29 if(!visited[v] && G[u][v] != inf) { 30 if(d[u]+G[u][v] < d[v]) { 31 d[v] = d[u]+G[u][v]; 32 pre[v].clear(); 33 pre[v].push_back(u); 34 } else if(d[u]+G[u][v] == d[v]) 35 pre[v].push_back(u); 36 } 37 } 38 } 39 } 40 41 void dijkstra2(int st) { 42 fill(visited,visited+maxn,false); 43 fill(d,d+maxn,inf); 44 d[st] = 0; 45 for(int i = 0; i < n; ++i) { 46 int u = -1,MIN = inf; 47 for(int j = 0; j < n; ++j) { 48 if(!visited[j] && d[j] < MIN) { 49 u = j; 50 MIN = d[j]; 51 } 52 } 53 if(u == -1) return ; 54 visited[u] = true; 55 for(int v = 0; v < n; ++v) { 56 if(!visited[v] && Time[u][v] != inf) { 57 if(d[u]+Time[u][v] < d[v]) { 58 d[v] = d[u]+Time[u][v]; 59 pre[v].clear(); 60 pre[v].push_back(u); 61 } else if(d[u]+Time[u][v] == d[v]) 62 pre[v].push_back(u); 63 } 64 } 65 } 66 } 67 int minLength, minTime; 68 void DFS1(int v) { 69 if(v == st) { 70 tempPath.push_back(v); 71 int sumLength = 0,sumTime = 0; 72 for(int i = tempPath.size()-1; i > 0; --i) { 73 v1 = tempPath[i],v2 = tempPath[i-1]; 74 sumLength += G[v1][v2]; 75 sumTime += Time[v1][v2]; 76 } 77 if(sumLength < minLength) { 78 minLength = sumLength; 79 minTime = sumTime; 80 path1 = tempPath; 81 } else if(sumLength == minLength && sumTime < minTime) { 82 minTime = sumTime; 83 path1 = tempPath; 84 } 85 tempPath.pop_back(); 86 return ; 87 } 88 tempPath.push_back(v); 89 for(int i = 0; i < pre[v].size(); ++i) 90 DFS1(pre[v][i]); 91 tempPath.pop_back(); 92 } 93 94 void DFS2(int v) { 95 if(v == st) { 96 tempPath.push_back(v); 97 int sumTime = 0; 98 for(int i = tempPath.size()-1; i > 0; --i) { 99 v1 = tempPath[i],v2 = tempPath[i-1]; 100 sumTime += Time[v1][v2]; 101 } 102 if(sumTime < minTime) { 103 minTime = sumTime; 104 path2 = tempPath; 105 } else if(sumTime == minTime && tempPath.size() < path2.size()) 106 path2 = tempPath; 107 tempPath.pop_back(); 108 return ; 109 } 110 tempPath.push_back(v); 111 for(int i = 0; i < pre[v].size(); ++i) 112 DFS2(pre[v][i]); 113 tempPath.pop_back(); 114 } 115 int main() { 116 fill(G[0],G[0]+maxn*maxn,inf); 117 fill(Time[0],Time[0]+maxn*maxn,inf); 118 cin>>n>>m; 119 for(int i = 0; i < m; ++i) { 120 cin>>v1>>v2>>flag>>length>>TIME; 121 G[v1][v2] = length,Time[v1][v2] = TIME; 122 if(flag == 0) G[v2][v1] = G[v1][v2],Time[v2][v1] = Time[v1][v2]; 123 } 124 cin>>st>>ed; 125 dijkstra1(st); 126 minLength = inf, minTime = inf; 127 DFS1(ed); 128 tempPath.clear(); 129 minTime = inf; 130 dijkstra2(st); 131 DFS2(ed); 132 if(path1 == path2) printf("Distance = %d; ",minLength); 133 else { 134 printf("Distance = %d: ",minLength); 135 for(int i = path1.size()-1; i >= 0; --i) { 136 if(i < path1.size()-1) printf(" -> "); 137 printf("%d",path1[i]); 138 } 139 printf("\n"); 140 } 141 printf("Time = %d: ",minTime); 142 for(int i = path2.size()-1; i >= 0; --i) { 143 if(i < path2.size()-1) printf(" -> "); 144 printf("%d",path2[i]); 145 } 146 return 0; 147 }
