HDU 3001 (triplicate)

Question

Solution

When we see this question, we will think of the traveling salesman problem. However, each vertex can go through a maximum of two times, so we can use a three-digit representation.

Code

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <algorithm>#include <cctype>#include <vector>#define INF 2139062143#define MAX 0x7ffffffffffffff#define del(a,b) memset(a,b,sizeof(a))using namespace std;typedef long long ll;template<typename T>inline void read(T&x){    x=0;T k=1;char c=getchar();    while(!isdigit(c)){if(c==‘-‘)k=-1;c=getchar();}    while(isdigit(c)){x=x*10+c-‘0‘;c=getchar();}x*=k;}const int maxn=15;int dis[maxn][maxn];int dp[60000][maxn];int poww[15];void _init() {    poww[0]=1;    for(int i=1;i<=10;i++) poww[i]=poww[i-1]*3;    return;}int query(int s,int k) {    return s/poww[k]%3;}int add(int x,int k) {    return x+poww[k];}int n,m;bool check(int x) {    for(int i=0;i<n;i++) if(!query(x,i)) return 0;    return 1;}int main(){    _init();    while(~scanf("%d %d",&n,&m)) {        del(dis,127);del(dp,127);        for(int i=1,u,v,w;i<=m;i++) {            read(u),read(v),read(w);--u,--v;            dis[u][v]=min(dis[u][v],w);            dis[v][u]=min(dis[v][u],w);        }                for(int i=0;i<n;i++) dp[add(0,i)][i]=0;                int ans=INF;        for(int s=0;s<poww[n];s++) {            for(int i=0;i<n;i++) {                if(!query(s,i)) continue;                for(int j=0;j<n;j++) {                    if(j==i||query(s,j)==2||dis[i][j]==INF) continue;                    dp[add(s,j)][j]=min(dp[s][i]+dis[i][j],dp[add(s,j)][j]);                }                if(check(s)) ans=min(ans,dp[s][i]);            }        }        if(ans==INF) printf("-1\n");        else printf("%d\n",ans);    }    return 0;}

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HDU 3001 (triplicate)