HDU 6166 Senior Pan

首先考虑到把集合拆成两部分然后代进去跑多源多汇最短路，这样子求出来的最小值就是答案。

然后考虑怎么拆分集合能够计算到所有的答案，其实对于规模为$n$的点拆分$log$次就够了，因为两个点的编号的二进制表示一定有一位是不同的，每一次选取所有该位为$1$的点作源，所有该位为$0$的点作汇，然后跑一跑最短路就好了。

事实上随机化拆分也是可以过的233。

时间复杂度$O(nlog^2n)$。

Code：

#include <cstdio>
#include <cstring>
#include <queue>
#include <iostream>
using namespace std;
typedef long long ll;
typedef pair <ll, int> pin;

const int N = 1e5 + 5;
const ll inf = 0x3f3f3f3f3f3f3f3f;

int testCase, n, m, K, tot = 0, head[N], a[N];
ll dis[N];
bool vis[N], mark[N];

struct Edge {
    int to, nxt;
    ll val;
} e[N << 2];

inline void add(int from, int to, ll val) {
    e[++tot].to = to;
    e[tot].val = val;
    e[tot].nxt = head[from];
    head[from] = tot;
} 

template <typename T>
inline void read(T &X) {
    X = 0; char ch = 0; T op = 1;
    for(; ch > '9' || ch < '0'; ch = getchar())
        if(ch == '-') op = -1;
    for(; ch >= '0' && ch <= '9'; ch = getchar())
        X = (X << 3) + (X << 1) + ch - 48;
    X *= op;
}

template <typename T>
inline void chkMin(T &x, T y) {
    if(y < x) x = y;
}

priority_queue <pin> Q;
void dij() {
    memset(vis, 0, sizeof(vis));    
    for(; !Q.empty(); ) {
        int x = Q.top().second; Q.pop();
        if(vis[x]) continue;
        vis[x] = 1;
        for(int i = head[x]; i; i = e[i].nxt) {
            int y = e[i].to;
            if(dis[y] > dis[x] + e[i].val) {
                dis[y] = dis[x] + e[i].val;
                Q.push(pin(-dis[y], y));
            }
        }
    }
}

int main() {
    read(testCase);
    for(int Case = 1; Case <= testCase; ++Case) {
        read(n), read(m);
        
        tot = 0; memset(head, 0, sizeof(head));
        for(int i = 1; i <= m; i++) {
            int x, y; ll v;
            read(x), read(y), read(v);
            add(x, y, v);
        }
        
        read(K);
        for(int i = 1; i <= K; i++)    read(a[i]);
        
        ll ans = inf;
        for(int i = 0; i <= 18; i++) {        
            memset(mark, 0, sizeof(mark));
            memset(dis, 0x3f, sizeof(dis));
            for(int j = 1; j <= K; j++) {
                if((a[j] >> i) & 1) Q.push(pin(dis[a[j]] = 0LL, a[j]));
                else mark[j] = 1;
            }
            
            dij();
            
            for(int j = 1; j <= K; j++)
                if(mark[j]) chkMin(ans, dis[a[j]]);
            
            memset(mark, 0, sizeof(mark));
            memset(dis, 0x3f, sizeof(dis));
            for(int j = 1; j <= K; j++) {
                if((a[j] >> i) & 1) mark[j] = 1;
                else Q.push(pin(dis[a[j]] = 0LL, a[j]));
            }
                
            dij();
            
            for(int j = 1; j <= K; j++)
                if(mark[j]) chkMin(ans, dis[a[j]]);
        }
        
        printf("Case #%d: %lld\n", Case, ans);
    }
    
    return 0;
}