Description
The Pizazz Pizzeria prides itself in delivering pizzas to its customers as fast as possible. Unfortunately, due to cutbacks, they can afford to hire only one driver to do the deliveries. He will wait for 1 or more (up to 10) orders to be processed before he starts any deliveries. Needless to say, he would like to take the shortest route in delivering these goodies and returning to the pizzeria, even if it means passing the same location(s) or the pizzeria more than once on the way. He has commissioned you to write a program to help him.
Input
Input will consist of multiple test cases. The first line will contain a single integer n indicating the number of orders to deliver, where 1 ≤ n ≤ 10. After this will be n + 1 lines each containing n + 1 integers indicating the times to travel between the pizzeria (numbered 0) and the n locations (numbers 1 to n). The jth value on the ith line indicates the time to go directly from location i to location j without visiting any other locations along the way. Note that there may be quicker ways to go from i to j via other locations, due to different speed limits, traffic lights, etc. Also, the time values may not be symmetric, i.e., the time to go directly from location i to j may not be the same as the time to go directly from location j to i. An input value of n = 0 will terminate input.
Output
For each test case, you should output a single number indicating the minimum time to deliver all of the pizzas and return to the pizzeria.
Sample Input
3 0 1 10 10 1 0 1 2 10 1 0 10 10 2 10 0 0
Sample Output
8
Source
inf开太大竟然WA了！！！
【题目大意】类似于TSP问题，只是每个点可以走多次，比经典TSP问题不同的是要先用弗洛伊的预处理一下两两之间的距离。求最短距离。
【解析】可以用全排列做，求出一个最短的距离即可。或者用状态压缩DP.用一个二进制数表示城市是否走过
【状态表示】dp[state][i]表示到达i点状态为state的最短距离
【状态转移方程】dp[state][i] =min{dp[state][i],dp[state'][j]+dis[j][i]} dis[j][i]为j到i的最短距离
【DP边界条件】dp[state][i] =dis[0][i] state是只经过i的状态
1 #pragma comment(linker, "/STACK:1024000000,1024000000") 2 #include<iostream> 3 #include<cstdio> 4 #include<cstring> 5 #include<cmath> 6 #include<math.h> 7 #include<algorithm> 8 #include<queue> 9 #include<set> 10 #include<bitset> 11 #include<map> 12 #include<vector> 13 #include<stdlib.h> 14 using namespace std; 15 #define max(a,b) (a) > (b) ? (a) : (b) 16 #define min(a,b) (a) < (b) ? (a) : (b) 17 #define ll long long 18 #define eps 1e-10 19 #define MOD 1000000007 20 #define N 16 21 #define M 1<<N 22 #define inf 1<<26 23 int n; 24 int mp[N][N]; 25 void flyod(){ 26 for(int k=0;k<=n;k++){ 27 for(int i=0;i<=n;i++){ 28 for(int j=0;j<=n;j++){ 29 if(mp[i][j]>mp[i][k]+mp[k][j]){ 30 mp[i][j]=mp[i][k]+mp[k][j]; 31 } 32 } 33 } 34 } 35 } 36 int dp[M][N]; 37 int main() 38 { 39 while(scanf("%d",&n)==1){ 40 if(n==0) 41 break; 42 43 for(int i=0;i<=n;i++){ 44 for(int j=0;j<=n;j++){ 45 scanf("%d",&mp[i][j]); 46 } 47 } 48 flyod(); 49 50 //int m=1<<n; 51 //memset(dp,inf,sizeof(dp)); 52 for(int S=0;S<(1<<n);S++){//i表示状态 53 for(int i=1;i<=n;i++){ 54 if(S&(1<<(i-1))){ 55 if(S==(1<<(i-1))){ 56 dp[S][i]=mp[0][i]; 57 } 58 else{ 59 dp[S][i]=(int)inf; 60 for(int j=1;j<=n;j++){ 61 if(S&(1<<(j-1)) && j!=i){ 62 dp[S][i]=min(dp[S][i],dp[S^(1<<(i-1))][j]+mp[j][i]); 63 } 64 } 65 } 66 } 67 } 68 } 69 int ans=dp[(1<<n)-1][1]+mp[1][0]; 70 for(int i=2;i<=n;i++){ 71 ans=min(ans,dp[(1<<n)-1][i]+mp[i][0]); 72 } 73 printf("%d ",ans); 74 } 75 return 0; 76 }
无耻地贴上大神的代码
1 #include<iostream> 2 #define INF 100000000 3 using namespace std; 4 int dis[12][12]; 5 int dp[1<<11][12]; 6 int n,ans,_min; 7 int main() 8 { 9 //freopen("in.txt","r",stdin); 10 while(scanf("%d",&n) && n) 11 { 12 for(int i = 0;i <= n;++i) 13 for(int j = 0;j <= n;++j) 14 scanf("%d",&dis[i][j]); 15 for(int k = 0;k <= n;++k) 16 for(int i = 0;i <= n;++i) 17 for(int j = 0;j <=n;++j) 18 if(dis[i][k] + dis[k][j]< dis[i][j]) 19 dis[i][j] = dis[i][k] +dis[k][j]; 20 21 for(int S = 0;S <= (1<<n)-1;++S)//枚举所有状态，用位运算表示 22 for(int i = 1;i <= n;++i) 23 { 24 if(S & (1<<(i-1)))//状态S中已经过城市i 25 { 26 if(S ==(1<<(i-1))) dp[S][i] =dis[0][i];//状态S只经过城市I，最优解自然是从0出发到i的dis,这也是DP的边界 27 else//如果S有经过多个城市 28 { 29 dp[S][i] = INF; 30 for(int j = 1;j <=n;++j) 31 { 32 if(S &(1<<(j-1)) && j != i)//枚举不是城市I的其他城市 33 dp[S][i] =min(dp[S^(1<<(i-1))][j] + dis[j][i],dp[S][i]); 34 //在没经过城市I的状态中，寻找合适的中间点J使得距离更短 35 } 36 } 37 } 38 } 39 ans = dp[(1<<n)-1][1] + dis[1][0]; 40 for(int i = 2;i <= n;++i) 41 if(dp[(1<<n)-1][i] + dis[i][0] < ans) 42 ans = dp[(1<<n)-1][i] +dis[i][0]; 43 printf("%d ",ans); 44 } 45 return 0; 46 }