poj1734Sightseeing trip【floyd+最小环+路径记录】
2015-11-24 14:01
417 查看
Language: Default Sightseeing trip
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 |
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#define inf 99999999
using namespace std;
const int maxn=110;
int path[maxn];
int map[maxn][maxn];
int dist[maxn][maxn];
int road[maxn][maxn];
int num;
void record(int s,int t){
if(road[s][t]){
record(s,road[s][t]);
record(road[s][t],t);
}
else path[num++]=t;
}
int floyd(int n){
memset(path,0,sizeof(path));
memset(road,0,sizeof(road));
int ans=inf;
for(int k=1;k<=n;++k){
for(int i=1;i<k;++i){
for(int j=i+1;j<k;++j){
if(ans>dist[i][j]+map[i][k]+map[k][j]){
ans=dist[i][j]+map[i][k]+map[k][j];
num=0;
path[num++]=i;
record(i,j);
path[num++]=k;
}
}
}
for(int i=1;i<=n;++i){
for(int j=1;j<=n;++j){
if(dist[i][j]>dist[i][k]+dist[k][j]){
dist[i][j]=dist[i][k]+dist[k][j];
road[i][j]=k;
}
}
}
}
return ans;
}
int main()
{
int i,j,n,m;
while(scanf("%d%d",&n,&m)!=EOF){
for(i=1;i<=n;++i){
for(j=1;j<=n;++j){
map[i][j]=dist[i][j]=inf;
}
}
int a,b,c;
for(i=0;i<m;++i){
scanf("%d%d%d",&a,&b,&c);
if(map[a][b]>c){
map[a][b]=map[b][a]=c;
dist[a][b]=dist[b][a]=c;
}
}
int ans=floyd(n);
if(ans<inf){
for(i=0;i<num;++i){
if(i){
printf(" %d",path[i]);
}
else {
printf("%d",path[i]);
}
}
printf("\n");
}
else {
printf("No solution.\n");
}
}
return 0;
}
相关文章推荐
- LVS持久连接
- nginx 配置
- 阿里巴巴离职DBA职业生涯总结:突然35岁~
- 1月31号CCIE捷报
- T264_param_t, T264_t结构体成员解析
- HP Cello 1999 trace解析
- 用ActionBar+Fragment实现Tab分类列表
- way and meritorious coach Carlos
- java性能优化之java web项目性能优化(一)
- Excel Sheet Column Title (easy)
- 【CSDN常见问题解答】如何避免表单重复提交
- 使用Matlab相机标定库(Camera Calibration Toolbox)问题小记
- xcodebuild 命令 后面加一个 GCC_OPTIMIZATION_LEVEL=0,这样 就可以调试 了 静态库了
- 测试一下代码片
- 泛型程序设计
- Java泛型<T>,<E>,<k,v>含义
- tx
- hdoj Easier Done Than Said?
- hdoj3974Assign the task【胡搞+邻接表】
- java /n /r /t