| Time Limit: 1000MS | Memory Limit: 65536K | |||
| Total Submissions: 7483 | Accepted: 2827 | 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
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
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
#include
#include
#include
#include
using namespace std;
const int maxn=300+10;
int ans=0x3f3f3f3f;
int a[maxn][maxn],d[maxn][maxn],pos[maxn][maxn];
vector<int>path;
void getpath(int x,int y){
if(pos[x][y]==0) return ;
getpath(x,pos[x][y]);
path.push_back(pos[x][y]);
getpath(pos[x][y],y);
}
int main(){
int n,m;
scanf("%d%d",&n,&m);
int u,v,w;
memset(d,0x3f3f3f3f,sizeof(d));
memset(a,0x3f3f3f3f,sizeof(a));
while(m--){
scanf("%d%d%d",&u,&v,&w);
a[u][v]=a[v][u]=d[u][v]=d[v][u]=min(a[u][v],w);
}
for (int k=1;k<=n;k++){
for (int i=1;i
for (int j=i+1;j
if((long long)d[i][j]+a[k][i]+a[j][k]<ans){
ans=d[i][j]+a[k][i]+a[j][k];
path.clear();
path.push_back(i);
getpath(i,j);
path.push_back(j);
path.push_back(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];
pos[i][j]=k;
}
}
}
}
if(ans==0x3f3f3f3f) printf("No solution.\n");
else {
for (int i=0;i
printf("%d ",path[i]);
}
printf("\n");
}
return 0;
}
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