HDU

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, and 

a 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 ... a1N 
a21 a22 ... a2N 
............... 
aN1 aN2 ... aNN 
b1 b2 ... bN 

c d 
e f 
... 
g h 

where 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-->d 
Total cost : ...... 
...... 

From e to f : 
Path: e-->e1-->..........-->ek-->f 
Total cost : ...... 

Note: if there are more minimal paths, output the lexically smallest one. Print a blank line after each test case. 

Sample Input

50 3 22 -1 43 0 5 -1 -122 5 0 9 20-1 -1 9 0 44 -1 20 4 05 17 8 3 11 33 52 4-1 -10

Sample Output

From 1 to 3 :Path: 1-->5-->4-->3Total cost : 21From 3 to 5 :Path: 3-->4-->5Total cost : 16From 2 to 4 :Path: 2-->1-->5-->4Total cost : 17

题意就不在这叙述了，就是一个floyd + 打印路径；

这题的输出有点诡异，交了三发PE。

具体代码如下：

#include <map>#include <set>#include <cmath>#include <queue>#include <string>#include <vector>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define inf 1<<30#define eps 1e-10#define maxn 0x3f3f3f3f#define zero(a) fabs(a)<eps#define Min(a,b) ((a)<(b)?(a):(b))#define Max(a,b) ((a)>(b)?(a):(b))#define pb(a) push_back(a)#define mem(a,b) memset(a,b,sizeof(a))#define LL long long#define lson step<<1#define rson step<<1|1#define MOD 1000000009#define sqr(a) ((a)*(a))using namespace std;int e[1005][1005];int p[1005][1005];int a[1005];void init(int n) {for(int i = 1; i <= n; i++) {for(int j = 1; j <= n; j++) {if(e[i][j] == -1) e[i][j] = maxn;}}}void floyd(int n) {for(int i = 1; i <= n; i++) {for(int j = 1; j <= n; j++) {p[i][j] = j;}}for(int k = 1; k <= n; k++) {for(int i = 1; i <= n; i++) {for(int j = 1; j <= n; j++) {int t = e[i][k] + e[k][j] + a[k];if(e[i][j] > t) {e[i][j] = t;p[i][j] = p[i][k];}else if(e[i][j] == t) {if(p[i][j] > p[i][k])p[i][j] = p[i][k];}}}}}int main() {int n;while(~scanf("%d",&n)&&n) {for(int i = 1; i <= n; i++) {for(int j = 1; j <= n; j++) {scanf("%d",&e[i][j]);}}for(int i = 1; i <= n; i++) scanf("%d",&a[i]);init(n);floyd(n);int fr , to;while(~scanf("%d%d",&fr,&to)) {if(fr == -1 && to == -1)break;int end = fr;printf("From %d to %d :\n",fr,to);printf("Path: ");printf("%d",fr);while(end != to) {printf("-->%d",p[end][to]);end = p[end][to];}printf("\n");printf("Total cost : %d\n",e[fr][to]);printf("\n");}}return 0;}

AcceptedTime31msMemory1952kBLength1712LangG++Submitted2017-12-24 20:50:38SharedRemoteRunId23388578