标签：possible sum i++ 路考 复杂度 which esc memory void
Sightseeing trip
| Time Limit: 1000MS |
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Memory Limit: 65536K |
| Total Submissions: 6919 |
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Accepted: 2646 |
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Special Judge |
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.
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
分析：经典的最小环问题，当然是用经典的算法：floyd解决啦！
一个最小环肯定是由两个点之间的最短路和次短路连接而成，如果我们求出了最短路再一条一条地删边求次短路，复杂度会很高，而且不止要求一对点，所以换个角度，我们考虑点k在不在最短路上。
假设k不在最短路上，那么处理出d[i][j],再求出经过k的路,a[i][k] + a[k][j],这3个数加起来就构成了一个环，这样就把删边操作改成了每次我们“添加”点时分不经过k的最短路和经过k的路考虑，这样一定会得到一条最短路和次短路的组合，至于如何控制到底经不经过k和怎么求最短路，当然是用到floyd算法。n <= 100,这就是很明显的提示，而且要求任意两点之间的最短路，还要控制到底经不经过k,这就是floyd算法的原理啊，每次我们先把没有更新k的结果算出来，然后再更新k.
至于怎么记录路径，类似于spfa的方法，每次记录到达当前点的前驱，不断向前更新就好了.
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<cmath>
#include<map>
using namespace std;
const int inf = 0x7ffffff;
int n, m, a[110][110], d[110][110],pre[110][110],ans = inf,tot,print[110];
void floyd()
{
for (int k = 1; k <= n; k++)
{
for (int i = 1; i < k; i++)
for (int j = i + 1; j < k; j++)
{
if (d[i][j] + a[i][k] + a[k][j] < ans)
{
ans = d[i][j] + a[i][k] + a[k][j];
tot = 0;
int t = j;
while (t != i)
{
print[++tot] = t;
t = pre[i][t];
}
print[++tot] = i;
print[++tot] = k;
}
}
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
if (d[i][j] > d[i][k] + d[k][j])
{
d[i][j] = d[i][k] + d[k][j];
pre[i][j] = pre[k][j];
}
}
}
int main()
{
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
{
a[i][j] = a[j][i] = d[i][j] = d[j][i] = inf;
pre[i][j] = i; //记录了i到j的最短路中j的前驱
}
for (int i = 1; i <= m; i++)
{
int u, v, w;
scanf("%d%d%d", &u, &v, &w);
a[u][v] = a[v][u] = d[u][v] = d[v][u] = min(w,a[u][v]);
}
floyd();
if (ans == inf)
printf("No solution.\n");
else
{
printf("%d", print[1]);
for (int i = 2; i <= tot; i++)
printf(" %d", print[i]);
printf("\n");
}
return 0;
}
poj1734 Sightseeing trip
标签：possible sum i++ 路考 复杂度 which esc memory void
原文地址：http://www.cnblogs.com/zbtrs/p/7398060.html
