题目链接
1 #include <bits/stdc++.h>
2 using namespace std;
3 typedef long long ll;
4 inline ll read(){
5 int x=0,f=1;char ch=getchar();
6 while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();}
7 while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
8 return x*f;
9 }
10
11 /***********************************************************/
12
13 int n, m; // n < = 10
14 const int maxn = 6e4+5;
15 int d[12][maxn];
16 //用三进制形式来表示状态
17 int dist[15][15];
18 //表示路径
19 int t[12];
20 int cnt, state[maxn];
21 int temp;
22
23 //判断S的三进制是否没有一个0
24 bool legal(int S){
25 bool ok = true;
26 for(int i = 0;i <= n-1;i++){
27 if(S%3 == 0){
28 ok = false;
29 break;
30 }
31 S /= 3;
32 }
33 return ok;
34 }
35
36 void init(){
37 temp = 1;
38 for(int i = 0;i <= 10;i++){
39 t[i] = temp;
40 temp *= 3;
41 }
42 }
43
44 //i点在集合S中是否出现了至少一次
45 inline bool in(int i, int S){
46 for(int j = 0;j < i;j++)
47 S /= 3;
48 if(S%3) return true;
49 else return false;
50 }
51
52 int dp(int i, int S){
53 if(d[i][S] >= 0) return d[i][S];
54 int &ans = d[i][S];
55 ans = 1e9;
56 int S1 = S - t[i];
57 for(int j = 0;j < n;j++){
58 if(j != i && in(j, S1) && dist[j][i] != -1)
59 ans = min(ans, dp(j, S1) + dist[j][i]);
60 }
61 return ans;
62 }
63
64 int main(){
65 init();
66 while(~scanf("%d %d", &n, &m)){
67 int max_3 = 0;
68 for(int i = 0;i < n;i++)
69 max_3 += t[i];
70 cnt = 0;
71 for(int S = max_3;S < t[n];S++){
72 if(legal(S))
73 state[cnt++] = S;
74 }
75 memset(dist, -1, sizeof(dist));
76 for(int i = 0;i < m;i++){
77 int a, b, c;
78 scanf("%d%d%d", &a, &b, &c);
79 a--;b--;
80 //更新长度
81 if(dist[a][b] == -1 || dist[a][b] > c)
82 dist[a][b] = dist[b][a] = c;
83 }
84 memset(d, -1, sizeof(d));
85 for(int i = 0;i < n;i++)
86 d[i][t[i]] = 0;
87
88 int sum = dp(0, max_3);
89 for(int i = 0;i < n;i++){
90 for(int j = 0;j < cnt;j++){
91 sum = min(sum, dp(i, state[j]));
92 }
93 }
94 if(sum == 1e9) printf("-1
");
95 else printf("%d
", sum);
96 }
97 return 0;
98 }