hdu 3001 Travelling TSP变形 三进制状压dp

// hdu 3001 TSP问题的变形// 这次到每个点最多两次，所以可以用三进制的类推// dp[S][u]表示当前在u点访问状态为S时所得到的最小的开销// 采用刷表法，即用当前的状态推出它所能转移的状态// dp[S][u] 可以到达的状态为dp[S+state[v]][v](dist[u][v]!=inf)// dp[S+state[v]][v] = max(dp[S+state[v]][v],dp[S][u]+dist[u][v]);// 其中每个点最多访问2次// 技巧就是先把所有的状态表示出来visit[S][u];// 表示当前在u点访问状态为S时u的访问状态。// 其中我们要求的就是最后每个点都访问过的状态S时的dp[S][v]{v<n};//// 这道题开始完全想套一个裸的TSP模板过的，结果样例都过不了// 哎，发现自己一直很水，继续加油吧。。。// 而且，犯了一个非常严重的错误，导致调了2个小时的bug// 哎。。。加油，不过最终还是ac啦//const int maxn = 15;//int dp[1<<maxn][maxn];//int mp[maxn][maxn];//int n,m;//void init(){//int a,b,c;//memset(mp,0,sizeof(mp));//memset(dp,0x3f,sizeof(dp));//for (int i=0;i<m;i++){//scanf("%d%d%d",&a,&b,&c);//mp[a][b] = mp[b][a] = c;//}//for (int i=0;i<(1<<n);i++)//dp[i][0] = mp[0][i];//}////void print(){//for (int i=0;i<(1<<n);i++){//for (int j=0;j<n;j++)//printf("%d ",dp[i][j]);//puts("");//}//}////void solve(){//for (int S=0;S<(1<<n);S++){//for (int u=1;u<=n;u++)//for (int v=1;v<=n;v++){//if ((S>>v)&1){//dp[S][u] = min(dp[S][u],dp[S^(1<<v)][v]+mp[v][u]);//}//}//}//print();//int ans = 0;//for (int i=0;i<n;i++)//ans = max(ans,dp[(1<<n)-1][i]);//printf("%d\n",ans);//}////int main() {//    freopen("G:\\Code\\1.txt","r",stdin);//while(scanf("%d%d",&n,&m)!=EOF){//init();//solve();//}//return 0;//}#include <algorithm>#include <bitset>#include <cassert>#include <cctype>#include <cfloat>#include <climits>#include <cmath>#include <complex>#include <cstdio>#include <cstdlib>#include <cstring>#include <ctime>#include <deque>#include <functional>#include <iostream>#include <list>#include <map>#include <numeric>#include <queue>#include <set>#include <stack>#include <vector>#define ceil(a,b) (((a)+(b)-1)/(b))#define endl '\n'#define gcd __gcd#define highBit(x) (1ULL<<(63-__builtin_clzll(x)))#define popCount __builtin_popcountlltypedef long long ll;using namespace std;const int MOD = 1000000007;const long double PI = acos(-1.L);template<class T> inline T lcm(const T& a, const T& b) { return a/gcd(a, b)*b; }template<class T> inline T lowBit(const T& x) { return x&-x; }template<class T> inline T maximize(T& a, const T& b) { return a=a<b?b:a; }template<class T> inline T minimize(T& a, const T& b) { return a=a<b?a:b; }const int maxn = 12;const int maxm = 60000;int visit[maxm][maxn];int dp[maxm][maxn];int state[maxn];int mp[maxn][maxn];const int inf = 0x3f3f3f3f;int n,m;void print(){for (int i=0;i<n;i++){for (int j=0;j<n;j++)printf("%d ",mp[i][j]);puts("");}}void init(){memset(mp,inf,sizeof(mp));memset(dp,inf,sizeof(dp));for (int i=0;i<n;i++){dp[state[i]][i] = 0;}int a,b,c;for (int i=0;i<m;i++){scanf("%d%d%d",&a,&b,&c);a--,b--;mp[a][b] = mp[b][a] = min(mp[a][b],c);}//for (int i=0;i<n;i++)//mp[i][i] = 0;//print();}void solve(){int ans = inf;for (int S=0;S<state[n];S++){bool flag = false;for (int u=0;u<n;u++){if (visit[S][u] == 0)flag = true;if (dp[S][u] == inf)continue;for (int v=0;v<n;v++) if (v!=u){if (visit[S][v]>=2)continue;if (mp[u][v] == inf) continue;dp[S+state[v]][v] = min(dp[S+state[v]][v],dp[S][u]+mp[u][v]);}}if (!flag){for (int u=0;u<n;u++)ans = min(ans,dp[S][u]);}}if (ans == inf)ans = -1;printf("%d\n",ans);}void get(){state[0]=1;for (int i=1;i<=10;i++)state[i] = state[i-1] * 3;for (int S=0;S<=state[10];S++){int x = S;for (int i=0;i<=10;i++){visit[S][i] = x%3;x/=3;}}}int main() {//freopen("G:\\Code\\1.txt","r",stdin);get();while(scanf("%d%d",&n,&m)!=EOF){init();solve();}return 0;}