HDU 1385 Minimum Transport Cost

Problem Description These are N cities in Spring country. Between each pair of cities there may be one transportation track or none. Now there is some cargo that should be delivered from one city to another. The transportation fee consists of two parts: The cost of the transportation on the path between these cities, anda certain tax which will be charged whenever any cargo passing through one city, except for the source and the destination cities.You must write a program to find the route which has the minimum cost.  

Input First is N, number of cities. N = 0 indicates the end of input.The data of path cost, city tax, source and destination cities are given in the input, which is of the form:a11 a12 ... a1Na21 a22 ... a2N...............aN1 aN2 ... aNNb1 b2 ... bNc de f...g hwhere aij is the transport cost from city i to city j, aij = -1 indicates there is no direct path between city i and city j. bi represents the tax of passing through city i. And the cargo is to be delivered from city c to city d, city e to city f, ..., and g = h = -1. You must output the sequence of cities passed by and the total cost which is of the form:  

Output From c to d :Path: c-->c1-->......-->ck-->dTotal cost : ............From e to f :Path: e-->e1-->..........-->ek-->fTotal cost : ......Note: if there are more minimal paths, output the lexically smallest one. Print a blank line after each test case.  

Sample Input 5 0 3 22 -1 4 3 0 5 -1 -1 22 5 0 9 20 -1 -1 9 0 4 4 -1 20 4 0 5 17 8 3 1 1 3 3 5 2 4 -1 -1 0  

Sample Output From 1 to 3 : Path: 1-->5-->4-->3 Total cost : 21 From 3 to 5 : Path: 3-->4-->5 Total cost : 16 From 2 to 4 : Path: 2-->1-->5-->4 Total cost : 17 分析：典型的floyd（一开始用dijkstra一直RE），这里使用path数组来记录路径，path[a][b]表示a到b的所有路径中的与a最近的一个 结点，path[a][a]则记录当前结点a。另外题目要求按字典序输出结果，只要判断当路径权值相同时，更新结点值小的那个路径即可！ 解决了以上两个问题，这就成了水题： 代码如下： #include<cstdio> #include<algorithm> using namespace std; #define INF 0x3f3f3f3f #define read freopen("zx.in","r",stdin) #define write freopen("zx.out","w",stdout) const int N=1000+10; int nnum,cost[N],path[N][N],map[N][N]; //map[a][b]存放a到b的最短路 void floyd() { for(int i=1;i<=nnum;i++) for(int j=1;j<=nnum;j++) path[i][j]=j;//（路径记录）这里记录路径的方法非常不错 for(int k=1;k<=nnum;k++)//floyd算法 for(int i=1;i<=nnum;i++) for(int j=1;j<=nnum;j++) { int temp=map[i][k]+map[k][j]+cost[k];//（+cost[k]一般没有） if(temp<map[i][j]) { map[i][j]=temp, path[i][j]=path[i][k];//记录当前路径 } else if(temp==map[i][j])//如果最短路有多条 { if(path[i][j]>path[i][k]) path[i][j]=path[i][k];//按路径的字典序输出！ } }//floyd算法 } int main() { read, write; while(scanf("%d",&nnum)!=EOF && nnum) { for(int i=1;i<=nnum;i++) for(int j=1;j<=nnum;j++) { scanf("%d",&map[i][j]); if(map[i][j]==-1) map[i][j]=INF;//按邻接矩阵输入图 } for(int i=1;i<=nnum;i++) scanf("%d",&cost[i]);//可无 floyd();//进行运算 int a,b;//求解a、b两点之间的最短路 while(scanf("%d%d",&a,&b)!=EOF && a!=-1 && b!=-1) { printf("From %d to %d :/n",a,b); printf("Path: %d",a); int xx=a; while(xx!=b) { printf("-->%d",path[xx][b]); xx=path[xx][b]; } printf("/nTotal cost : %d/n/n",map[a][b]); } } return 0; }