bzoj3590(状压DP)

题目链接

https://www.lydsy.com/JudgeOnline/problem.php?id=3590

题解

cf1155F 一样,不过这题有重边,重边唯一的作用就是形成二元环,然后去重边之后就直接做。。




代码

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/**
*         ┏┓    ┏┓
*         ┏┛┗━━━━━━━┛┗━━━┓
*         ┃       ┃  
*         ┃   ━    ┃
*         ┃ >   < ┃
*         ┃       ┃
*         ┃... ⌒ ...  ┃
*         ┃ ┃
*         ┗━┓ ┏━┛
*          ┃ ┃ Code is far away from bug with the animal protecting          
*          ┃ ┃ 神兽保佑,代码无bug
*          ┃ ┃           
*          ┃ ┃       
*          ┃ ┃
*          ┃ ┃           
*          ┃ ┗━━━┓
*          ┃ ┣┓
*          ┃ ┏┛
*          ┗┓┓┏━━━━━━━━┳┓┏┛
*           ┃┫┫ ┃┫┫
*           ┗┻┛ ┗┻┛
*/

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<stack>
#include<cmath>
#include<set>
#include<bitset>
#include<complex>
#include<assert.h>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge *j=h[x];j;j=j->next)
#define mem(a) memset(a,0,sizeof(a))
#define ll long long
#define eps 1e-8
#define succ(x) (1<<x)
#define mid (x+y>>1)
#define lowbit(x) (x&(-x))
#define sqr(x) ((x)*(x))
#define NM 15
#define nm 5005
using namespace std;
const double pi=acos(-1);
const int inf=1e9+7;
ll read(){
ll x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
return f*x;
}



int n,m,tot,_x,_y,_t,d[NM][NM][nm],g[nm],a[NM][NM];
void M(int&x,int y){if(x>y)x=y;}


int main(){
int _=read();while(_--){
n=read();m=read();tot=succ(n)-1;
inc(i,1,n)inc(j,1,n)inc(k,0,tot)d[i][j][k]=inf;
inc(i,0,tot)g[i]=inf;
inc(i,1,n)inc(j,1,n)if(i!=j)a[i][j]=inf;
while(m--){
_x=read();_y=read();_t=read();
if(a[_x][_y]<inf){
M(d[_x][_x][succ(_x-1)|succ(_y-1)],_t+a[_x][_y]);
M(d[_y][_y][succ(_x-1)|succ(_y-1)],_t+a[_x][_y]);
M(g[succ(_x-1)|succ(_y-1)],_t+a[_x][_y]);
}
M(a[_x][_y],_t);M(a[_y][_x],_t);
}
inc(p,1,n){
d[p][p][succ(p-1)]=0;
inc(j,0,tot)if(j&succ(p-1)){
inc(i,1,n)if((i!=p||j==succ(p-1))&&d[p][i][j]<inf){
if(__builtin_popcount(j)>2&&a[i][p]<inf)
M(g[j],d[p][i][j]+a[i][p]);
inc(k,1,n)if(a[i][k]<inf&&(!(j&succ(k-1))||(p==k&&__builtin_popcount(j)>2))){
M(d[p][k][j|succ(k-1)],d[p][i][j]+a[i][k]);
}
}
}
}
inc(i,0,tot)if(g[i]<inf){
inc(x,1,n)if(i&succ(x-1))inc(y,1,n)if(i&succ(y-1)){
int t=tot^i;
for(int j=t;j;j=t&(j-1)){
M(g[i|j],g[i]+d[x][y][j|succ(x-1)|succ(y-1)]);
}
}
}
if(g[tot]<inf)printf("%d\n",g[tot]);else printf("impossible\n");
}
return 0;
}