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countries on the earth, which are numbered from
to
. They are connected by
undirected flights, detailedly the
-th flight connects the 
-th and the 
-th country, and it will cost Victor‘s airplane 
L fuel if Victor flies through it. And it is possible for him to fly to every country from the first country.
, 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
, denoting the number of test cases.
and
in the first line, denoting the number of the countries and the number of the flights.
lines, each line contains three integers 
, 
and 
, describing a flight. 




. 




. 








. 






. 







.Output
lines : the
-th of these should contain a single integer, denoting the minimal amount of fuel for Victor to finish the travel.Sample Input
1
3 2
1 2 2
1 3 3
Sample Output
10
/*/ 一开始题目没读仔细,以为是一个最小树,秒WA一发; 后来想半天发现这个有环,就不是最小树了,搜了一下是Floyd+dp状压。 写完之后一直发现输出的是INF=0x3f3f3f3f ,找了半天最后想到二进制标记状态这里,maps标记用0 0开始会舒服好多。改过来就对了。题目很有意思。。 AC代码: /*/
#include"algorithm"
#include"iostream"
#include"cstring"
#include"cstdlib"
#include"string"
#include"cstdio"
#include"vector"
#include"cmath"
#include"queue"
using namespace std;
typedef long long LL;
#define memset(x,y) memset(x,y,sizeof(x))
#define memcpy(x,y) memcpy(x,y,sizeof(x))
#define MX 401
#define INF 0x3f3f3f3f
int maps[20][20];
int dp[200000][20],vis[20];
void init() {
memset(maps,0x3f);
memset(vis,0x3f);
memset(dp ,0x3f);
}
int main() {
int T;
scanf("%d",&T);
while(T--) {
init();
int n,m,u,v,w;
scanf("%d%d",&n,&m);
for(int i=0; i<m; i++) {
scanf("%d%d%d",&u,&v,&w);
if(maps[--u][--v] > w) //状态压缩从0开始会好写一些
maps[v][u]=maps[u][v]= w;
}
for(int i=0; i<n; i++)maps[i][i]=0; //标记自己到自己距离为0 //后面会要加到这个数字。。
for(int k=0; k<n; k++) {
for(int i=0; i<n; i++) {
for(int j=0; j<n; j++) {
maps[i][j]=min(maps[i][j],maps[i][k]+maps[k][j]); //Floyd 把去某一点的最小路程计算出来
}
}
}
dp[1][0]=0;
vis[0]=0;
m=1<<n;
for(int i=1; i<m; i++) {
for(int j=0; j<n; j++) {
if(dp[i][j]==INF)continue;
for(int k=0; k<n; k++) {
if(i&(1<<k)||maps[j][k]==INF)continue; //二进制 1 表示该点走过,0表示没走过,第二维表示现在所在的点。压缩状态
if( dp[i|(1<<k)][k]>dp[i][j]+maps[j][k]) {
dp[i|(1<<k)][k]=dp[i][j]+maps[j][k];
vis[k]=min(vis[k],dp[i|(1<<k)][k]); //比较,去到下一点需要的最小路
}
}
}
}
int minn=1e9+100;;
for(int i=0; i<n; i++) {
minn=min(minn,dp[m-1][i]+vis[i]);
}
printf("%d\n",minn);
}
return 0;
}
ACM: HDU 5418 Victor and World - Floyd算法+dp状态压缩
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原文地址:http://www.cnblogs.com/HDMaxfun/p/5759221.html