poj1734Sightseeing trip
Time Limit: 1000MS | Memory Limit: 65536K | |||
Total Submissions: 6811 | Accepted: 2602 | Special Judge |
Description
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
Output
Sample Input
5 7
1 4 1
1 3 300
3 1 10
1 2 16
2 3 100
2 5 15
5 3 20
Sample Output
1 3 5 2
Source
#include <cstring>
#include <cstdio>
#include <iostream>
#include <algorithm> #define inf 0x7ffffff using namespace std; int n, m,a[][],d[][],head[][],res,tot,ans[]; int main()
{
while (~scanf("%d%d", &n, &m))
{
for (int i = ; i <= n; i++)
for (int j = ; j <= n; j++)
{
a[i][j] = d[i][j] = inf;
head[i][j] = i;
}
for (int i = ; i <= m; i++)
{
int x, y, z;
scanf("%d%d%d", &x, &y, &z);
a[x][y] = a[y][x] = d[x][y] = d[y][x] = min(z, a[x][y]);
}
res = inf;
for (int k = ; k <= n; k++)
{
for (int i = ; i < k; i++)
{
for (int j = i + ; j < k; j++)
{
int temp = d[i][j] + a[i][k] + a[k][j];
if (temp < res)
{
res = temp;
tot = ;
int p = j;
while (p != i)
{
ans[++tot] = p;
p = head[i][p];
}
ans[++tot] = i;
ans[++tot] = k;
}
}
}
for (int i = ; i <= n; i++)
for (int j = ; j <= n; j++)
if (d[i][j] > d[i][k] + d[k][j])
{
d[i][j] = d[i][k] + d[k][j];
head[i][j] = head[k][j];
}
}
if (res == inf)
puts("No solution.\n");
else
{
printf("%d", ans[]);
for (int i = ; i <= tot; i++)
printf(" %d", ans[i]);
printf("\n");
}
} return ;
}
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