题意:旅游公司要开发一条新的路线 , 要求这是一个总路程尽可能短的环 , 并且不能只含两个城市 , 除开起点外 ,
不能重复走之前走过的城市 , 输出这条路线?
Floyd算法求最小环
代码:
//用floyd算法 , 求有向图的最小环
#include
#include
#include
#include
using namespace std;
#define INF 0xfffffff
#define maxn 110
int grap[maxn][maxn] , n , m;
int dist[maxn][maxn];
int past[maxn][maxn];
int mincircle;
int path[maxn] , k1 ;
void init()
{
int i ,
j;
for(i = 1; i
<= n; i++)
for(j = 1; j
<= n; j++)
grap[i][j] =
INF;
}
void floyd()
{
mincircle =
INF;
int i ,
j;
for(i = 1; i
<= n; i++)
for(j = 1; j
<= n; j++)
{
dist[i][j] =
grap[i][j];
past[i][j] =
i;
}
for(int k =
1; k <= n; k++) //每个点都成为一次中间点 ,
和bellman-ford不一样
{
for(i = 1; i
<= n; i++) //判断是不是最小环
for(j = 1; j
<= n; j++)
{
if(i ==
j) continue;
if(dist[i][j] != INF &&
grap[j][k]!=INF&&grap[k][i]!=INF && mincircle >
dist[i][j]+grap[j][k]+grap[k][i])
{
mincircle =
dist[i][j]+grap[j][k]+grap[k][i];
k1 =
0;
path[k1++] =
i;
path[k1++] =
k;
path[k1++] =
j;
while(past[i][path[k1-1]] != i)
{
path[k1] =
past[i][path[k1-1]];
k1 +=
1;
}
}
}
for(i = 1; i
<= n; i++) //Floyd算法
for(j = 1; j
<= n; j++)
{
if(i == k ||
j == k) continue;
if(dist[i][k] != INF && dist[k][j] != INF &&
dist[i][k]+dist[k][j]
{
dist[i][j] =
dist[i][k]+dist[k][j];
past[i][j] =
past[k][j];
}
}
}
}
int main()
{
while(scanf("%d %d" , &n , &m) != EOF)
{
init();
int i , x ,
y , z;
for(i = 1; i
<= m; i++)
{
scanf("%d %d
%d" , &x , &y , &z);
if(grap[x][y] > z)
grap[x][y] =
grap[y][x] = z;
}
floyd();
if(mincircle
== INF || mincircle < 0)
cout<<"No solution."<<endl;
else
{
printf("%d"
, path[0]);
for(i = 1; i
< k1; i++)
printf(" %d"
, path[i]);
cout<<endl;
}
}
return
0;
}
Floyd算法求最小环
代码:
//用floyd算法 , 求有向图的最小环
#include
#include
#include
#include
using namespace std;
#define INF 0xfffffff
#define maxn 110
int grap[maxn][maxn] , n , m;
int dist[maxn][maxn];
int past[maxn][maxn];
int mincircle;
int path[maxn] , k1 ;
void init()
{
}
void floyd()
{
}
int main()
{
}