hdu 1385 floyd字典序

Minimum Transport Cost

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11017    Accepted Submission(s):
3058

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, 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

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

 

Source

Asia
1996, Shanghai (Mainland China)

 disguss里给的某些数据根本就是错的吧。。我的错误一开始是变量写错结果对着他们的test改半天ccc。一直死循环

floyd不必多数，字典序得话，肯定有最前面的字母决定总体的大小，所以path[i][j]表示i-->j中间经过的第一个点

还有就是单向图这个我倒是没写错。

#include<bits/stdc++.h>
using namespace std;
#define inf 0x3f3f3f3f
#define ql(a,b) memset(a,b,sizeof(a))
int e[105][105],b[105];
int path[105][105];
void floyd(int n)
{
int i,j,k;
for(k=1;k<=n;++k)
for(i=1;i<=n;++i)
for(j=1;j<=n;++j){
if(e[i][j]>b[k]+e[i][k]+e[k][j]){
e[i][j]=b[k]+e[i][k]+e[k][j];
path[i][j]=path[i][k];
}
else if(e[i][j]==b[k]+e[i][k]+e[k][j]&&path[i][j]>path[i][k]){
path[i][j]=path[i][k];
}
}
}
void print(int u,int v)
{
int k;
if(u==v)
{
printf("%d",v);
return ;
}
k=path[u][v];
printf("%d-->",u);
print(k,v);
}
int main()
{
int n,m,i,j,k,a,c,t=1;
while(cin>>n&&n){
for(i=1;i<=n;++i)
for(j=1;j<=n;++j){
scanf("%d",&e[i][j]);
if(e[i][j]==-1) e[i][j]=inf;
path[i][j]=j;
}
for(i=1;i<=n;++i) cin>>b[i];
floyd(n);
while(cin>>a>>c&&(a+1||c+1)){
printf("From %d to %d :\nPath: ",a,c);
print(a,c);
printf("\nTotal cost : %d\n\n",e[a][c]);
}
}
return 0;
}