Victor and World
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 262144/131072 K (Java/Others)
Total Submission(s): 132 Accepted Submission(s): 66
Problem Description
After trying hard for many years, Victor has finally received a pilot license. To have a celebration, he intends to buy himself an airplane and fly around the world. There are
n countries on the earth, which are numbered from
1 to
n. They are connected by
m undirected flights, detailedly the
i-th flight connects the
ui-th and the
vi-th country, and it will cost Victor's airplane
wi L fuel if Victor flies through it. And it is possible for him to fly to every country from the first country.
Victor now is at the country whose number is
1, he wants to know the minimal amount of fuel for him to visit every country at least once and finally return to the first country.
Input
Output
Your program should print
T lines : the
i-th of these should contain a single integer, denoting the minimal amount of fuel for Victor to finish the travel.
Sample Input
Sample Output
Source
BestCoder Round #52 (div.2)
Recommend
hujie | We have carefully selected several similar problems for you: 5421 5420 5417 5416 5415
题意,要求从1开始起,经过所有的点,最终回到1,所要的最小值。
状态压缩,16位每位上为1表示已走,0表示没有走过,直接用位运算,表示所有的状态。
dp[i][j] 状态为i,最后一次是j的最小值,则由dp[i][j] + land[k][j] 推出dp[t][k] ,t表示,i把第k位置1的数。有2 ^ 16 * n个状态,n种转移,所以复杂度为o(2^16 * n * n);
第一种方法用spaf来解。
#define N 20
#define M 100005
#define maxn 205
#define MOD 1000000007
int n,land[N][N],T,m,a,b,c,dp[M][N],all,t;
queue q;
bool vis[M][N];
int main()
{
//freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
while(S(T)!=EOF)
{
while(T--){
S2(n,m);
FI(n+1) FJ(n+1) land[i][j] = MOD,vis[i][j] = false;
FI(n+1) land[i][i] = 0;
all = 1<
第二种方法,用Floyed来解,由于n很小,可以很快处理出来。
#define N 20
#define M 100005
#define maxn 205
#define MOD 1000000007
int n,land[N][N],T,m,a,b,c,dp[M][N],all,t;
queue q;
bool vis[M][N];
int main()
{
//freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
while(S(T)!=EOF)
{
while(T--){
S2(n,m);
FI(n+1) FJ(n+1) land[i][j] = MOD,vis[i][j] = false;
FI(n+1) land[i][i] = 0;
all = 1<