You are given undirected weighted graph. Find the length of the shortest cycle which starts from the vertex 1 and passes throught all the edges at least once. Graph may contain multiply edges between a pair of vertices and loops (edges from the vertex to itself).
Input
The first line of the input contains two integers n and m (1 ≤ n ≤ 15, 0 ≤ m ≤ 2000), n is the amount of vertices, and m is the amount of edges. Following m lines contain edges as a triples x, y, w (1 ≤ x, y ≤ n, 1 ≤ w ≤ 10000), x, y are edge endpoints, and w is the edge length.
Output
Output minimal cycle length or -1 if it doesn't exists.
Example
3 3
1 2 1
2 3 1
3 1 1
3
3 2
1 2 3
2 3 4
14
题意是给一个无向图，要求从1出发回到1，但是必须经过图中每一条边，只上一次
这从不从1出发有什么用呢？这就是个欧拉回路。
然后判一下每个点的度数是不是奇数，如果是奇数就需要再找个奇数度数的点匹配。
最后，匹配直接搜索就行。
1 #include<cstdio> 2 #include<iostream> 3 #include<cstring> 4 #include<cstdlib> 5 #include<algorithm> 6 #include<cmath> 7 #include<queue> 8 #include<deque> 9 #include<set> 10 #include<map> 11 #include<ctime> 12 #define LL long long 13 #define inf 0x7ffffff 14 #define pa pair<int,int> 15 #define mkp(a,b) make_pair(a,b) 16 #define pi 3.1415926535897932384626433832795028841971 17 using namespace std; 18 inline LL read() 19 { 20 LL x=0,f=1;char ch=getchar(); 21 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 22 while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} 23 return x*f; 24 } 25 int n,m; 26 int v[110]; 27 int fa[110]; 28 int p[110],cnt; 29 bool mrk[110]; 30 int dis[110][110]; 31 LL ans,ans2; 32 struct edg{int x,y,z;}e[10010]; 33 inline int getfa(int x){return fa[x]==x?x:fa[x]=getfa(fa[x]);} 34 inline void dfs(int now,LL tot) 35 { 36 if (now==cnt/2+1){ans2=min(ans2,tot);return;} 37 int x,y; 38 for (x=1;x<=cnt;x++)if (!mrk[x])break; 39 mrk[x]=1; 40 for (y=x+1;y<=cnt;y++)if (!mrk[y]&&dis[p[x]][p[y]]<=1000000) 41 { 42 mrk[y]=1; 43 dfs(now+1,tot+dis[p[x]][p[y]]); 44 mrk[y]=0; 45 } 46 mrk[x]=0; 47 } 48 int main() 49 { 50 n=read();m=read(); 51 for (int i=1;i<=n;i++)fa[i]=i; 52 for (int i=1;i<n;i++) 53 for (int j=i+1;j<=n;j++) 54 dis[i][j]=dis[j][i]=1<<28; 55 for (int i=1;i<=m;i++) 56 { 57 e[i].x=read(); 58 e[i].y=read(); 59 e[i].z=read();v[e[i].x]++;v[e[i].y]++; 60 if (e[i].z<dis[e[i].x][e[i].y])dis[e[i].x][e[i].y]=dis[e[i].y][e[i].x]=e[i].z; 61 ans+=e[i].z; 62 int xx=getfa(e[i].x),yy=getfa(e[i].y); 63 if (xx>yy)swap(xx,yy); 64 fa[xx]=yy; 65 } 66 for (int i=1;i<=m;i++) 67 { 68 if (getfa(e[i].x)==getfa(1)&&getfa(e[i].y)==getfa(1))continue; 69 puts("-1"); 70 return 0; 71 } 72 for (int k=1;k<=n;k++) 73 for (int i=1;i<=n;i++) 74 for (int j=1;j<=n;j++) 75 if (dis[i][j]>dis[i][k]+dis[k][j]) 76 dis[i][j]=dis[i][k]+dis[k][j]; 77 for (int i=1;i<=n;i++)if (v[i]&1)p[++cnt]=i; 78 if (cnt&1){puts("-1");return 0;} 79 ans2=1ll<<50; 80 dfs(1,ans); 81 printf("%lld ",ans2); 82 }
You are given undirected weighted graph. Find the length of the shortest cycle which starts from the vertex 1 and passes throught all the edges at least once. Graph may contain multiply edges between a pair of vertices and loops (edges from the vertex to itself).
Input
The first line of the input contains two integers n and m (1 ≤ n ≤ 15, 0 ≤ m ≤ 2000), n is the amount of vertices, and m is the amount of edges. Following m lines contain edges as a triples x, y, w (1 ≤ x, y ≤ n, 1 ≤ w ≤ 10000), x, y are edge endpoints, and w is the edge length.
Output
Output minimal cycle length or -1 if it doesn't exists.
Example
3 3
1 2 1
2 3 1
3 1 1
3
3 2
1 2 3
2 3 4
14