题目链接

https://www.lydsy.com/JudgeOnline/problem.php?id=1097

思路

kk很小，考虑状压。

预处理出从11～(k+1)(k+1)出发，到其他点的距离，令fsta,ifsta,i表示走过的点状态为stasta，最后一个到达的点为ii，所需要行走的最短举例，操作起来还是有很多要注意的地方的，具体见代码。

代码

#include <cstdio>
#include <queue>
#include <cstring>const int maxn=20000;
const int maxm=400000;
const int maxk=21;
const int inf=0x3f3f3f3f;int read()
{int x=0,f=1;char ch=getchar();while((ch<'0')||(ch>'9')){if(ch=='-'){f=-f;}ch=getchar();}while((ch>='0')&&(ch<='9')){x=x*10+ch-'0';ch=getchar();}return x*f;
}int n,m,k,pre[maxm+10],now[maxn+10],son[maxm+10],tot,val[maxm+10];
int dist[maxk+2][maxn+10],f[(1<<maxk)+10][maxk+2],vis[maxn+10],t;
int from[maxk+2];
std::queue<int> q;inline int ins(int a,int b,int c)
{pre[++tot]=now[a];now[a]=tot;son[tot]=b;val[tot]=c;return 0;
}inline int spfa(int s)
{int* d=dist[s];d[s]=0;q.push(s);vis[s]=1;while(!q.empty()){int u=q.front(),j=now[u];q.pop();while(j){int v=son[j];if(d[v]>d[u]+val[j]){d[v]=d[u]+val[j];if(!vis[v]){vis[v]=1;q.push(v);}}j=pre[j];}vis[u]=0;}return 0;
}int main()
{n=read();m=read();k=read();while(m--){int a=read(),b=read(),c=read();ins(a,b,c);ins(b,a,c);}t=read();for(int i=1; i<=t; ++i){int a=read(),b=read();from[b]|=1<<(a-1);}for(int i=2; i<=k+1; ++i){from[i]|=1;}memset(dist,63,sizeof dist);memset(f,63,sizeof f);for(int i=1; i<=k+1; ++i){spfa(i);f[1<<(i-1)][i]=0;}for(int sta=1; sta<1<<(k+1); ++sta){int flag=0;for(int i=2; i<=k+1; ++i){if((((1<<(i-1))&sta)==(1<<(i-1)))&&((sta^from[i])!=(sta-from[i]))){flag=1;break;}}if(flag){continue;}for(int i=1; i<=k+1; ++i){if(sta&(1<<(i-1))){int s=sta^(1<<(i-1));flag=0;for(int j=2; j<=k+1; ++j){if((((1<<(j-1))&s)==(1<<(j-1)))&&((s^from[j])!=(s-from[j]))){flag=1;break;}}if(flag){continue;}for(int j=1; j<=k+1; ++j){if(s&(1<<(j-1))){f[sta][i]=std::min(f[sta][i],f[s][j]+dist[j][i]);}}}}}int ans=inf;for(int i=1; i<=k+1; ++i){ans=std::min(ans,f[(1<<(k+1))-1][i]+dist[i][n]);}printf("%d\n",ans);return 0;
}