hdu3001 Travelling
Travelling
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2520 Accepted Submission(s): 735
Problem Description
After coding so many days,Mr Acmer wants to have a good rest.So travelling is the best choice!He has decided to visit n cities(he insists on seeing all the cities!And he does not mind which city being his start station because superman can bring him to any city at first but only once.), and of course there are m roads here,following a fee as usual.But Mr Acmer gets bored so easily that he doesn't want to visit a city more than twice!And he is so mean that he wants to minimize the total fee!He is lazy you see.So he turns to you for help.
Input
There are several test cases,the first line is two intergers n(1<=n<=10) and m,which means he needs to visit n cities and there are m roads he can choose,then m lines follow,each line will include three intergers a,b and c(1<=a,b<=n),means there is a road between a and b and the cost is of course c.Input to the End Of File.
Output
Output the minimum fee that he should pay,or -1 if he can't find such a route.
Sample Input
2 1
1 2 100
3 2
1 2 40
2 3 50
3 3
1 2 3
1 3 4
2 3 10
Sample Output
100
90
7
用三进制来进行壮态压缩，这样就可以大大减少状态数，在预处理一下，就可以了，其次，每一个点都要走一次！
#include <iostream> #include <stdio.h> using namespace std; int threenum[11]={1,3,9,27,81,243,729,2187,6561,19683,59049}; #define inf 0x4f4f4f4f int dp[600000][15],n,m; int threeevery[600000][15],dis[15][15]; int fmin(int x,int y){if(x<y)return x;return y;} void makethree()//先预处理，算出所有的值 { int i,temp,j; for(i=0;i<threenum[10];i++)//3的１１次 { temp=i; for(j=0;j<10;j++) { threeevery[i][j]=temp%3; temp=temp/3; } } } void init() { int i,j,s,e,val; for(i=0;i<threenum[n];i++) { for(j=0;j<n;j++) { dp[i][j]=inf; } } for(i=0;i<n;i++) for(j=0;j<n;j++) { dis[i][j]=inf; } for(i=0;i<n;i++) { dp[threenum[i]][i]=0;//到自已的点初始化为０ } for(i=0;i<m;i++) { scanf("%d%d%d",&s,&e,&val); s--;e--;//全部从0开 #include <iostream>
#include <stdio.h>
using namespace std;
int threenum[11]={1,3,9,27,81,243,729,2187,6561,19683,59049};
#define inf 0x4f4f4f4f
int dp[600000][15],n,m;
int threeevery[600000][15],dis[15][15];
int fmin(int x,int y){if(x<y)return x;return y;}
void makethree()//先预处理，算出所有的值
{
int i,temp,j;
for(i=0;i<threenum[10];i++)//3的１１次
{
temp=i;
for(j=0;j<10;j++)
{
threeevery[i][j]=temp%3;
temp=temp/3;
}
}
}
void init()
{
int i,j,s,e,val;
for(i=0;i<threenum[n];i++)
{
for(j=0;j<n;j++)
{
dp[i][j]=inf;
}
}
for(i=0;i<n;i++)
for(j=0;j<n;j++)
{
dis[i][j]=inf;
}
for(i=0;i<n;i++)
{
dp[threenum[i]][i]=0;//到自已的点初始化为０
}
for(i=0;i<m;i++)
{
scanf("%d%d%d",&s,&e,&val);
s--;e--;//全部从0开[html] view plaincopyprint? if(val<dis[s][e]) dis[s][e]=dis[e][s]=val;//排除重边取最小值 } } int makedp() { int i,j,k,minx; bool flag; minx=inf; for(i=0;i<threenum[n];i++)//对于每一位都要计算 { flag=true;//标记是否每一个点都已经经过 for(j=0;j<n;j++) { if(threeevery[i][j]==0)//是这个点没有走 flag=false; if(dp[i][j]==inf)//没有走过 { continue; } for(k=0;k<n;k++)//加入k这个点 { if((j==k)||threeevery[i][k]>=2||(dis[j][k]==inf))//自已到自已的地方，经过超过两次，不相连通去掉 continue; dp[i+threenum[k]][k]=fmin(dp[i+threenum[k]][k],dp[i][j]+dis[j][k]);//从j到k } } if(flag)//满足条件，对最小值进行更新 { for(j=0;j<n;j++) { minx=fmin(dp[i][j],minx); } } } if(minx==inf) { minx=-1; } printf("%d\n",minx); return 1; } int main() { makethree(); while(scanf("%d%d",&n,&m)!=EOF) { init(); makedp(); } return 0; } if(val<dis[s][e])
dis[s][e]=dis[e][s]=val;//排除重边取最小值
}
}
int makedp()
{
int i,j,k,minx;
bool flag;
minx=inf;
for(i=0;i<threenum[n];i++)//对于每一位都要计算
{
flag=true;//标记是否每一个点都已经经过
for(j=0;j<n;j++)
{
if(threeevery[i][j]==0)//是这个点没有走
flag=false;
if(dp[i][j]==inf)//没有走过
{
continue;
}
for(k=0;k<n;k++)//加入k这个点
{
if((j==k)||threeevery[i][k]>=2||(dis[j][k]==inf))//自已到自已的地方，经过超过两次，不相连通去掉
continue;
dp[i+threenum[k]][k]=fmin(dp[i+threenum[k]][k],dp[i][j]+dis[j][k]);//从j到k
}
}
if(flag)//满足条件，对最小值进行更新
{
for(j=0;j<n;j++)
{
minx=fmin(dp[i][j],minx);
}
}
}
if(minx==inf)
{
minx=-1;
}
printf("%d\n",minx);
return 1;
}
int main()
{
makethree();
while(scanf("%d%d",&n,&m)!=EOF)
{
init();
makedp();
}
return 0;
}
补充：软件开发 , C++ ,
