# [bzoj1097][POI2007]旅游景点atr【最短路】【状压dp】

【题目链接】
https://www.lydsy.com/JudgeOnline/problem.php?id=1097
【题解】
首先预处理出前k+1个点到每个点的最短路。
然后状压dp。记$f\left[i\right]\left[j\right]$表示当前停留在i，需要停留的点的状态为j，转移就枚举下一个停留点是哪里。
时间复杂度:$O\left(K\ast kM+{2}^{K}\ast K\right)$ 小写k为spfa的常数。

/* --------------
user Vanisher
problem bzoj-1097
----------------*/
# include <bits/stdc++.h>
# define    ll      long long
# define    inf     0x3f3f3f3f
# define    M       200010
# define    N       20010
# define    T       20
using namespace std;
int tmp=0, fh=1; char ch=getchar();
while (ch<'0'||ch>'9'){if (ch=='-') fh=-1; ch=getchar();}
while (ch>='0'&&ch<='9'){tmp=tmp*10+ch-'0'; ch=getchar();}
return tmp*fh;
}
struct node{
int data,next,vote;
}e[M*2];
void build(int u, int v, int w){
}
void spfa(int id){
memset(use,0,sizeof(use));
memset(dis[id],inf,sizeof(dis[id]));
use[id]=true; dis[id][id]=0;
int pl=1, pr=1; q[1]=id;
while (pl<=pr){
int x=q[(pl++)%n];
if (dis[id][e[ed].data]>dis[id][x]+e[ed].vote){
dis[id][e[ed].data]=dis[id][x]+e[ed].vote;
if (use[e[ed].data]==0){
use[e[ed].data]=1;
q[(++pr)%n]=e[ed].data;
}
}
use[x]=false;
}
}
int main(){
for (int i=1; i<=m; i++){
build(u,v,w);
}
for (int i=1; i<=k+1; i++)
spfa(i);
for (int i=1; i<=g; i++){
ned[t]=ned[t]+(1<<(s-2));
}
if (k==0){
printf("%d\n",dis[1][n]);
return 0;
}
memset(f,inf,sizeof(f));
for (int i=2; i<=k+1; i++)
if (ned[i]==0) f[i][1<<(i-2)]=dis[1][i];
for (int i=0; i<(1<<k); i++)
for (int j=2; j<=k+1; j++){
if ((i&(1<<(j-2)))==0) continue;
for (int t=2; t<=k+1; t++)
if ((i&(1<<(t-2)))==0){
if ((ned[t]&i)==ned[t])
f[t][i+(1<<(t-2))]=min(f[t][i+(1<<(t-2))],f[j][i]+dis[j][t]);
}
}
int ans=inf;
for (int i=2; i<=k+1; i++)
ans=min(ans,f[i][(1<<k)-1]+dis[i][n]);
printf("%d\n",ans);
return 0;
}
posted @ 2018-04-02 21:45  Vanisher  阅读(68)  评论(0编辑  收藏  举报