标签：-- ret 表示 clu content space article 现在 end
参考了别人的代码 https://blog.csdn.net/u010372095/article/details/38474721
深感自己的弱小
这是tsp问题，和基本的tsp问题没什么大的区别，唯一的区别就是采用了三进制
原来的二进制的某一位只能表示到达或没到达过，现在加了三进制，就能表示到达过几次了
tsp问题网上的讲解我推荐这一篇 http://www.360doc.com/content/17/0826/11/36546539_682232069.shtml
我个人感觉讲的算是很详细了，剩下的就是代码实现了。
#include<iostream> #include<queue> #include<cstring> #include<vector> #include<cstdio> #include<cmath> #include<map> #include<string> using namespace std; #define ll long long #define se second #define fi first #define oo 0x3fffffff int n,m; int dp[15][60000]; int dis[15][15]; int three[12]; int in[12]; int tothree(int n); int main() { in[0] = 1; for(int i = 1; i <= 10; ++i) in[i] = in[i-1]*3; while(scanf("%d%d",&n,&m) != EOF) { for(int i = 0; i < n; ++i) { for(int j = 0; j < in[n]; ++j) dp[i][j] = -1; for(int j = 0; j < n; ++j) dis[i][j] = -1; } //cout << 1 << endl; for(int i = 0; i < m; ++i) { int a,b,c; scanf("%d%d%d",&a,&b,&c); a --; b --; if(dis[a][b] == -1) dis[a][b] = dis[b][a] = c; else dis[a][b] = dis[b][a] = min(dis[a][b],c); } int minn = -1; for(int l = 1; l < in[n]; ++l) { int k = tothree(l); for(int i = 0; i < n; ++i) { if(three[i]) { if(k == 1) dp[i][l] = 0; if(dp[i][l] == -1) continue; if(k == n) { if(minn == -1) minn = dp[i][l]; else minn = min(minn,dp[i][l]); } for(int j = 0; j < n; ++j) if(i != j && three[j] < 2 && dis[i][j] != -1) { int tog = l + in[j]; if(dp[j][tog] == -1) dp[j][tog] = dp[i][l] + dis[i][j]; else dp[j][tog] = min(dp[j][tog],dp[i][l] + dis[i][j]); } } } } printf("%d\n",minn); } } int tothree(int n) { int k = 0; for(int i = 0; i < 10; ++i) { three[i] = n%3; n/=3; if(three[i]) k++; } return k; }
标签：-- ret 表示 clu content space article 现在 end
原文地址：https://www.cnblogs.com/mltang/p/9740108.html
