hdu 6166 Senior Pan（多源最短路径）（二进制划分集合）

Senior Pan

Problem Description
Senior Pan fails in his discrete math exam again. So he asks Master ZKC to give him graph theory problems everyday.
The task is simple : ZKC will give Pan a directed graph every time, and selects some nodes from that graph, you can calculate the minimum distance of every pair of nodes chosen in these nodes and now ZKC only cares about the minimum among them. That is still too hard for poor Pan, so he asks you for help.

Input
The first line contains one integer T, represents the number of Test Cases.1≤T≤5.Then T Test Cases, for each Test Cases, the first line contains two integers n,m representing the number of nodes and the number of edges.1≤n,m≤100000
Then m lines follow. Each line contains three integers xi,yi representing an edge, and vi representing its length.1≤xi,yi≤n,1≤vi≤100000
Then one line contains one integer K, the number of nodes that Master Dong selects out.1≤K≤n
The following line contains K unique integers ai, the nodes that Master Dong selects out.1≤ai≤n,ai!=aj

Output
For every Test Case, output one integer: the answer

Sample Input
1
5 6
1 2 1
2 3 3
3 1 3
2 5 1
2 4 2
4 3 1
3
1 3 5

Sample Output
Case #1: 2

思路：要求k个点中两点的最小值，肯定不能直接遍历，那么就考虑分类了

枚举二进制的每一位，对这k个点进行分类，若第i位为1则分到s1集合中，否则分到集合s2中，同理反着在再来一次（有向图嘛）

这样就保证了枚举所有的情况（假设点s到点t的距离最短，点s肯定不等于点t，也就是说点s和点t在二进制上至少有一位不同）

代码：

#include<bits/stdc++.h>using namespace std;typedef long long LL;const LL inf=0x3f3f3f3f3f3f3f3f;const int maxn=1e5+10;struct node{    int u,v;    LL w;} p[maxn];struct edge{    int v,next;    LL w;} E[maxn<<1];int first[maxn],num[maxn];LL dis[maxn];bool inq[maxn];int n,m,k,len;LL Min_ans;void init(){    len=0;    memset(first,-1,sizeof(first));    memset(dis,inf,sizeof(dis));    memset(inq,false,sizeof(inq));}void add_edge(int u,int v,LL w){    E[len].v=v,E[len].w=w,E[len].next=first[u],first[u]=len++;}void Build_edge(){    init();    for(int j=0; j<m; ++j)        add_edge(p[j].u,p[j].v,p[j].w);}LL spfa(int st,int ed){    queue<int>q;    q.push(st),inq[st]=true,dis[st]=0;    while(!q.empty())    {        st=q.front();        inq[st]=false,q.pop();        for(int i=first[st]; ~i; i=E[i].next)        {            int v=E[i].v;            LL w=E[i].w;            if(dis[v]>dis[st]+w)            {                dis[v]=dis[st]+w;                if(!inq[v])                {                    inq[v]=true;                    q.push(v);                }            }        }    }    return dis[ed];}int main(){    int t,Case=0;    scanf("%d",&t);    while(++Case<=t)    {        Min_ans=inf;        scanf("%d%d",&n,&m);        for(int i=0; i<m; ++i)            scanf("%d%d%lld",&p[i].u,&p[i].v,&p[i].w);        scanf("%d",&k);        for(int i=1; i<=k; ++i)            scanf("%d",&num[i]);        int res=log(n*1.0)/log(2.0);        int cnt=res==(int)res?res:res+1;        for(int i=0; i<=res; ++i)        {            int tot=1<<i;            Build_edge();            for(int j=1; j<=k; ++j)            {                if(num[j]&tot)                    add_edge(0,num[j],0);                else                    add_edge(num[j],n+1,0);            }            Min_ans=min(Min_ans,spfa(0,n+1));            Build_edge();            for(int j=1; j<=k; ++j)            {                if(num[j]&tot)                    add_edge(num[j],n+1,0);                else                    add_edge(0,num[j],0);            }            Min_ans=min(Min_ans,spfa(0,n+1));        }        printf("Case #%d: %lld\n",Case,Min_ans);    }    return 0;}