1018. Public Bike Management (30)
There is a public bike service in Hangzhou City which provides great convenience to the tourists from all over the world. One may rent a bike at any station and return it to any other stations in the city.
The Public Bike Management Center (PBMC) keeps monitoring the real-time capacity of all the stations. A station is said to be in perfect condition if it is exactly half-full. If a station is full or empty, PBMC will collect or send bikes to adjust the condition of that station to perfect. And more, all the stations on the way will be adjusted as well.
When a problem station is reported, PBMC will always choose the shortest path to reach that station. If there are more than one shortest path, the one that requires the least number of bikes sent from PBMC will be chosen.
Figure 1
Figure 1 illustrates an example. The stations are represented by vertices and the roads correspond to the edges. The number on an edge is the time taken to reach one end station from another. The number written inside a vertex S is the current number of bikes stored at S. Given that the maximum capacity of each station is 10. To solve the problem at S3, we have 2 different shortest paths:
1. PBMC -> S1 -> S3. In this case, 4 bikes must be sent from PBMC, because we can collect 1 bike from S1 and then take 5 bikes to S3, so that both stations will be in perfect conditions.
2. PBMC -> S2 -> S3. This path requires the same time as path 1, but only 3 bikes sent from PBMC and hence is the one that will be chosen.
Input Specification:
Each input file contains one test case. For each case, the first line contains 4 numbers: Cmax (<= 100), always an even number, is the maximum capacity of each station; N (<= 500), the total number of stations; Sp, the index of the problem station (the stations are numbered from 1 to N, and PBMC is represented by the vertex 0); and M, the number of roads. The second line contains N non-negative numbers Ci (i=1,...N) where each Ci is the current number of bikes at Si respectively. Then M lines follow, each contains 3 numbers: Si, Sj, and Tij which describe the time Tij taken to move betwen stations Si and Sj. All the numbers in a line are separated by a space.
Output Specification:
For each test case, print your results in one line. First output the number of bikes that PBMC must send. Then after one space, output the path in the format: 0->S1->...->Sp. Finally after another space, output the number of bikes that we must take back to PBMC after the condition of Sp is adjusted to perfect.
Note that if such a path is not unique, output the one that requires minimum number of bikes that we must take back to PBMC. The judge's data guarantee that such a path is unique.
Sample Input:10 3 3 5 6 7 0 0 1 1 0 2 1 0 3 3 1 3 1 2 3 1Sample Output:
3 0->2->3 0
又是属于比较烦的题目，直接dijkstra。但在比较路径大小的时候需要考虑：1）先比较到达时间，若时间小则路径小；2）若时间一样，则比较send的自行车数量，数量小的路径小；3）若send
的自行车数量也一样，则比较back的自行车数量，back的数量小则路径小。另外也要注意send与back的更新，后面station多出来的自行车是无法补给前面station的。
代码
1 #include <stdio.h> 2 #include <string.h> 3 4 #define MAXV 501 5 6 int map[MAXV][MAXV]; 7 int reminder[MAXV]; 8 int send[MAXV]; 9 int back[MAXV]; 10 int times[MAXV]; 11 int flag[MAXV]; 12 int path[MAXV][MAXV]; 13 14 int findMinTimesPoint(int); 15 int main() 16 { 17 int Cmax,N,Sp,M,s,e,i; 18 while(scanf("%d%d%d%d",&Cmax,&N,&Sp,&M) != EOF){ 19 for(i=1;i<=N;++i) 20 scanf("%d",&reminder[i]); 21 memset(map,0,sizeof(map)); 22 for(i=0;i<M;++i){ 23 scanf("%d%d",&s,&e); 24 scanf("%d",&map[s][e]); 25 map[e][s] = map[s][e]; 26 } 27 memset(send,0,sizeof(send)); 28 memset(back,0,sizeof(back)); 29 memset(flag,0,sizeof(flag)); 30 memset(path,0,sizeof(path)); 31 s = 0; 32 send[s] = 0; 33 back[s] = 0; 34 flag[s] = 1; 35 path[0][0] = 1; 36 path[0][1] = 0; 37 for(i=1;i<=N;++i){ 38 times[i] = -1; 39 if(map[s][i]){ 40 times[i] = map[s][i]; 41 send[i] = Cmax / 2 - reminder[i]; 42 if(send[i] < 0) 43 send[i] = 0; 44 back[i] = reminder[i] - Cmax / 2; 45 if(back[i] < 0) 46 back[i] = 0; 47 path[i][0] = 2; 48 path[i][1] = 0; 49 path[i][2] = i; 50 } 51 } 52 times[s] = 0; 53 while(!flag[Sp]){ 54 s = findMinTimesPoint(N); 55 flag[s] = 1; 56 for(i=1;i<=N;++i){ 57 if(!flag[i] && map[s][i]){ 58 if(times[i] == -1 || (times[i] > times[s] + map[s][i])){ 59 times[i] = times[s] + map[s][i]; 60 path[i][0] = path[s][0] + 1; 61 int j; 62 for(j=1;j<=path[s][0];++j) 63 path[i][j] = path[s][j]; 64 path[i][j] = i; 65 int t = Cmax / 2 - reminder[i]; 66 if(t == 0){ 67 send[i] = send[s]; 68 back[i] = back[s]; 69 } 70 else if(t > 0){ 71 if(t-back[s] >= 0){ 72 back[i] = 0; 73 send[i] = send[s] + t - back[s]; 74 } 75 else{ 76 send[i] = send[s]; 77 back[i] = back[s] - t; 78 } 79 } 80 else{ 81 send[i] = send[s]; 82 back[i] = back[s] - t; 83 } 84 } 85 else if(times[i] == times[s] + map[s][i]){ 86 int t = Cmax / 2 - reminder[i]; 87 int tSend,tBack; 88 if(t==0){ 89 tSend = send[s]; 90 tBack = back[s]; 91 } 92 else if(t > 0){ 93 if(t-back[s] >= 0){ 94 tBack = 0; 95 tSend = send[s] + t - back[s]; 96 } 97 else{ 98 tSend = send[s]; 99 tBack = back[s] - t; 100 } 101 } 102 else{ 103 tSend = send[s]; 104 tBack = back[s] - t; 105 } 106 if(tSend < send[i] || (tSend == send[i] && tBack < back[i])){ 107 send[i] = tSend; 108 back[i] = tBack; 109 path[i][0] = path[s][0] + 1; 110 int j; 111 for(j=1;j<=path[s][0];++j) 112 path[i][j] = path[s][j]; 113 path[i][j] = i; 114 } 115 } 116 } 117 } 118 } 119 printf("%d %d",send[Sp],path[Sp][1]); 120 for(i = 2;i<=path[Sp][0];++i){ 121 printf("->%d",path[Sp][i]); 122 } 123 printf(" %d\n",back[Sp]); 124 } 125 return 0; 126 } 127 128 int findMinTimesPoint(int n) 129 { 130 int i = 1; 131 int minPointInd; 132 while(i<=n && (times[i] == -1 || flag[i])) 133 ++i; 134 if(i > n) 135 return -1; 136 minPointInd = i; 137 for(;i<=n;++i){ 138 if(!flag[i] && times[i] != -1 && (times[i] < times[minPointInd] || 139 (times[i] == times[minPointInd] && send[i]<send[minPointInd]) || 140 (send[i]==send[minPointInd] && back[i]<back[minPointInd]))) 141 minPointInd = i; 142 } 143 return minPointInd; 144 }
