A1111. Online Map
Input our current position and a destination, an online map can recommend several paths. Now your job is to recommend two paths to your user: one is the shortest, and the other is the fastest. It is guaranteed that a path exists for any request.
Input Specification:
Each input file contains one test case. For each case, the first line gives two positive integers N (2 <= N <= 500), and M, being the total number of streets intersections on a map, and the number of streets, respectively. Then M lines follow, each describes a street in the format:
V1 V2 one-way length time
where V1 and V2 are the indices (from 0 to N-1) of the two ends of the street; one-way is 1 if the street is one-way from V1 to V2, or 0 if not; length is the length of the street; and time is the time taken to pass the street.
Finally a pair of source and destination is given.
Output Specification:
For each case, first print the shortest path from the source to the destination with distance D in the format:
Distance = D: source -> v1 -> ... -> destination
Then in the next line print the fastest path with total time T:
Time = T: source -> w1 -> ... -> destination
In case the shortest path is not unique, output the fastest one among the shortest paths, which is guaranteed to be unique. In case the fastest path is not unique, output the one that passes through the fewest intersections, which is guaranteed to be unique.
In case the shortest and the fastest paths are identical, print them in one line in the format:
Distance = D; Time = T: source -> u1 -> ... -> destination
Sample Input 1:
10 15 0 1 0 1 1 8 0 0 1 1 4 8 1 1 1 3 4 0 3 2 3 9 1 4 1 0 6 0 1 1 7 5 1 2 1 8 5 1 2 1 2 3 0 2 2 2 1 1 1 1 1 3 0 3 1 1 4 0 1 1 9 7 1 3 1 5 1 0 5 2 6 5 1 1 2 3 5
Sample Output 1:
Distance = 6: 3 -> 4 -> 8 -> 5 Time = 3: 3 -> 1 -> 5
Sample Input 2:
7 9 0 4 1 1 1 1 6 1 1 3 2 6 1 1 1 2 5 1 2 2 3 0 0 1 1 3 1 1 1 3 3 2 1 1 2 4 5 0 2 2 6 5 1 1 2 3 5
Sample Output 2:
Distance = 3; Time = 4: 3 -> 2 -> 5
1 #include<cstdio> 2 #include<iostream> 3 #include<algorithm> 4 #include<vector> 5 using namespace std; 6 const int INF = 100000000; 7 int GL[501][501], GT[501][501]; 8 int dst[501], visit[501]; 9 vector<int> pre1[501], pre2[501]; 10 int N, M, sour, destn; 11 void dijkstra(int G[501][501], vector<int> pre[], int s){ 12 fill(dst, dst + 501, INF); 13 fill(visit, visit + 501, 0); 14 dst[s] = 0; 15 for(int i = 0; i < N; i++){ 16 int u = -1, minLen = INF; 17 for(int j = 0; j < N; j++){ 18 if(dst[j] < minLen && visit[j] == 0){ 19 minLen = dst[j]; 20 u = j; 21 } 22 } 23 if(u == -1){ 24 return; 25 } 26 visit[u] = 1; 27 for(int j = 0; j < N; j++){ 28 if(visit[j] == 0 && G[u][j] != INF){ 29 if(G[u][j] + dst[u] < dst[j]){ 30 dst[j] = G[u][j] + dst[u]; 31 pre[j].clear(); 32 pre[j].push_back(u); 33 }else if(G[u][j] + dst[u] == dst[j]){ 34 pre[j].push_back(u); 35 } 36 } 37 } 38 } 39 } 40 int minTime = INF, minLen = INF; 41 vector<int> ans1, ans2, temp1, temp2; 42 void DFS1(int d){ //函数前初始化minTime、minLen 43 temp1.push_back(d); 44 if(d == sour){ 45 int tempL = 0, tempT = 0; 46 for(int i = temp1.size() - 1; i > 0; i--){ 47 tempL += GL[temp1[i]][temp1[i - 1]]; 48 tempT += GT[temp1[i]][temp1[i - 1]]; 49 } 50 if(tempL < minLen){ 51 minLen = tempL; 52 minTime = tempT; 53 ans1 = temp1; 54 }else if(tempL == minLen && tempT < minTime){ 55 minLen = tempL; 56 minTime = tempT; 57 ans1 = temp1; 58 } 59 } 60 61 for(int i = 0; i < pre1[d].size(); i++){ 62 DFS1(pre1[d][i]); 63 } 64 temp1.pop_back(); 65 } 66 void DFS2(int d){ 67 temp2.push_back(d); 68 if(d == sour){ 69 int tempL = 0, tempT = 0; 70 for(int i = temp2.size() - 1; i > 0; i--){ 71 tempL += GL[temp2[i]][temp2[i - 1]]; 72 tempT += GT[temp2[i]][temp2[i - 1]]; 73 } 74 if(tempT < minTime){ 75 ans2 = temp2; 76 minTime = tempT; 77 minLen = tempL; 78 }else if(tempT == minTime && temp2.size() < ans2.size()){ 79 ans2 = temp2; 80 minTime = tempT; 81 minLen = tempL; 82 } 83 } 84 for(int i = 0; i < pre2[d].size(); i++){ 85 DFS2(pre2[d][i]); 86 } 87 temp2.pop_back(); 88 } 89 int main(){ 90 scanf("%d%d", &N, &M); 91 fill(GL[0], GL[0] + 501*501, INF); 92 fill(GT[0], GT[0] + 501*501, INF); 93 for(int i = 0; i < M; i++){ 94 int tag, v1, v2, L, T; 95 scanf("%d%d%d%d%d", &v1, &v2, &tag, &L, &T); 96 if(tag == 1){ 97 GL[v1][v2] = L; 98 GT[v1][v2] = T; 99 }else{ 100 GL[v1][v2] = GL[v2][v1] = L; 101 GT[v1][v2] = GT[v2][v1] = T; 102 } 103 } 104 scanf("%d%d", &sour, &destn); 105 dijkstra(GL, pre1, sour); 106 dijkstra(GT, pre2, sour); 107 minTime = INF; minLen = INF; 108 DFS1(destn); 109 int prtLen = minLen; 110 minTime = INF; minLen = INF; 111 DFS2(destn); 112 int prtTim = minTime; 113 if(ans1 == ans2){ 114 int LL = ans1.size(); 115 printf("Distance = %d; Time = %d: %d", prtLen, prtTim, ans1[LL - 1]); 116 for(int i = LL - 2; i >= 0; i--){ 117 printf(" -> %d", ans1[i]); 118 } 119 }else{ 120 int LL1 = ans1.size(), LL2 = ans2.size(); 121 printf("Distance = %d: %d", prtLen, ans1[LL1 - 1]); 122 for(int i = LL1 - 2; i >= 0; i--){ 123 printf(" -> %d", ans1[i]); 124 } 125 printf("\n"); 126 printf("Time = %d: %d", prtTim, ans2[LL2 - 1]); 127 for(int i = LL2 - 2; i >= 0; i--){ 128 printf(" -> %d", ans2[i]); 129 } 130 } 131 cin >> N; 132 return 0; 133 }
总结：
1、题意：用路程作为边权，求最短路径，如果有多条则输出耗时最短的。 再用时间作为边权求最短路，如果有多条则输出经过节点个数最少的。使用两次迪杰斯特拉和DFS即可。最开始没读清题，以为以时间为边权求最短路，如果有多条则选择路程最短的，结果测试点2过不去。
2、注意one-way是单行道标志，为1表示只有v1到v2的路，为0表示v1到v2和v2到v1都是通的。注意有些是单向路径，有些是双向。另外，由于DFS回溯时得到的路径是倒着的、有些路是单向的，所以在累加边权时也是倒序的，是 GL[temp1[i]][temp1[i - 1]] 而非 GL[temp1[i - 1]][temp1[ i ]]。
