Travelling (三进制+状压dp)

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\n");
95         else printf("%d\n", sum);
96     }
97     return 0;
98 }

原文地址：https://www.cnblogs.com/ouyang_wsgwz/p/9823539.html