poj1734 Sightseeing trip
Description There is a travel agency in Adelton town on Zanzibar
island. It has decided to offer its clients, besides many other
attractions, sightseeing the town. To earn as much as possible from
this attraction, the agency has accepted a shrewd decision: it is
necessary to find the shortest route which begins and ends at the same
place. Your task is to write a program which finds such a route.In the town there are N crossing points numbered from 1 to N and M
two-way roads numbered from 1 to M. Two crossing points can be
connected by multiple roads, but no road connects a crossing point
with itself. Each sightseeing route is a sequence of road numbers y_1,
…, y_k, k>2. The road y_i (1<=i<=k-1) connects crossing points x_i
and x_{i+1}, the road y_k connects crossing points x_k and x_1. All
the numbers x_1,…,x_k should be different.The length of the
sightseeing route is the sum of the lengths of all roads on the
sightseeing route, i.e. L(y_1)+L(y_2)+…+L(y_k) where L(y_i) is the
length of the road y_i (1<=i<=k). Your program has to find such a
sightseeing route, the length of which is minimal, or to specify that
it is not possible,because there is no sightseeing route in the town.Input The first line of input contains two positive integers: the
number of crossing points N<=100 and the number of roads M<=10000.
Each of the next M lines describes one road. It contains 3 positive
integers: the number of its first crossing point, the number of the
second one, and the length of the road (a positive integer less than
500).Output There is only one line in output. It contains either a string
‘No solution.’ in case there isn’t any sightseeing route, or it
contains the numbers of all crossing points on the shortest
sightseeing route in the order how to pass them (i.e. the numbers x_1
to x_k from our definition of a sightseeing route), separated by
single spaces. If there are multiple sightseeing routes of the minimal
length, you can output any one of them.
floyd求最小环。输出路径的方法写得比较丑，不过总体是分治策略。
#include0][x][y])
dp[0][x][y]=dp[0][y][x]=z;
}
ans=0x3f3f3f3f;
for (k=1;k<=n;k++)
for (i=1;i<=n;i++)
for (j=1;j<=n;j++)
{
if (i==j) continue;
if (dp[k-1][i][j]+dp[0][j][k]+dp[0][k][i]>0&&dp[k-1][i][j]+dp[0][j][k]+dp[0][k][i]1][i][j]+dp[0][j][k]+dp[0][k][i];
ans_i=i;
ans_j=j;
ans_k=k;
}
if (dp[k-1][i][k]+dp[k-1][k][j]>0&&dp[k-1][i][k]+dp[k-1][k][j]1][i][j])
{
dp[k][i][j]=dp[k-1][i][k]+dp[k-1][k][j];
use[k][i][j]=1;
}
else
{
dp[k][i][j]=dp[k-1][i][j];
use[k][i][j]=0;
}
}
if (ans==0x3f3f3f3f) printf("No solution.\n");
else
{
printf("%d",ans_k);
find(ans_k-1,ans_i,ans_j);
for (i=1;i<=cnt;i++)
if (path[i]!=path[i-1]) printf(" %d",path[i]);
printf("\n");
}
}