Hdoj 3001 travelling [3 hexadecimal + traveling salesman]

Question: hdoj 3001 travelling

Question: The standard traveling salesman adds a sentence, and each vertex can take up to two times.

Analysis: The state transition equation is exactly the same, but it only requires a three-digit system, because each point has three states: 0, 1, 2.

Define the State: DP [st] [I]: the minimum cost of the current I point when the State is St

Transition equation: DP [now] [J] = min (DP [now] [J], DP [st] [I] + MP [I] [J ]); now is the status that St can transfer at a time.

Pay attention to initialization.

Analysis:

#include <cstdio>#include <algorithm>#include <cstring>#include <string>#include <iostream>#include <vector>using namespace std;const int inf = 0x3f3f3f3f;const int N = 12;int mp[N][N];int n,m;int bit[N],v[N];int dp[60000][N];void isit(){    bit[0]=1;    for(int i=1;i<N;i++)        bit[i]=bit[i-1]*3;}void solve(int st){    memset(v,0,sizeof(v));    int tmp=0;    while(st)    {        v[tmp++]=(st%3);        st/=3;    }}int count(){    int ans=0;    for(int i=0;i<n;i++)        ans+=(v[i]*bit[i]);    return ans;}int main(){    isit();    while(~scanf("%d%d",&n,&m))    {        memset(mp,inf,sizeof(mp));        for(int i=0;i<m;i++)        {            int x,y,z;            scanf("%d%d%d",&x,&y,&z);            x--,y--;            mp[x][y]=mp[y][x]=min(z,mp[x][y]);        }        memset(dp,inf,sizeof(dp));        for(int i=0;i<n;i++)        {            dp[bit[i]][i]=0;        }        int ans=inf;        for(int st=0;st<bit[n];st++)        {            int ok=1;            solve(st);            for(int i=0;i<n;i++)            {                if(v[i]==0)                    ok=0;                if(dp[st][i]==inf)                    continue;                for(int j=0;j<n;j++)                {                    if(v[j]==2 || i==j)                        continue;                    if(mp[i][j]==inf)                        continue;                    v[j]++;                    int no=count();                    dp[no][j]=min(dp[no][j],dp[st][i]+mp[i][j]);                    v[j]--;                }            }            if(ok)                for(int i=0;i<n;i++)                    ans=min(ans,dp[st][i]);        }        if(ans==inf)            ans=-1;        printf("%d\n",ans);    }    return 0;}

Hdoj 3001 travelling [3 hexadecimal + traveling salesman]