NOIP模拟 path - 按二进制位分组

题目原文：

企鹅豆豆即将要去考长跑了，但是作为一只企鹅，长跑自然比不过鸵鸟和鸡。为了公平起见，教练告诉豆豆，他可以从 K 个指定地点中选择两个不同的地点分别作为起点和终点来考试.考试地图是一个由 N 个点 M 条边组成的没有重边和自环的连通无向图，一条边的长度为 Ai 。豆豆想知道他的长跑考试最少需要跑多远。

【数据范围】

对于 30% 的数据，K≤4；

对于另外 10% 的数据，K=N；

对于另外 30% 的数据，M=N-1；

对于 100% 的数据，1≤N,M≤100000；T≤5；1≤K≤n；1≤边长≤100000。

题目分析：

就是求图中最近的两个特殊点之间的距离。

• 乱搞版：对每个特殊点进行一遍dijkstra，一遇到其他特殊点就break，更新答案。数据没有刻意去卡，加上部分分的优化勉强能卡掉。
• 高大上的正解: 将k个点编号1...k，枚举编号的二进制的每一位，将这一位为1的点连边S（作为起点），为0的点连边T（作为终点），这样跑\(log(k)\)次dijkstra即可得到答案。总复杂度\(O(nlog(n)log(k))\)，还是尽量加上部分分的优化。

code

乱搞版：

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;

namespace IO{
    inline int readint(){
        int i = 0, f = 1; char ch = getchar();
        for(; (ch < '0' || ch > '9') && ch != '-'; ch = getchar());
        if(ch == '-') f = -1, ch = getchar();
        for(; ch >= '0' && ch <= '9'; ch = getchar()) i = (i << 3) + (i << 1) + (ch - '0');
        return i * f;
    }
    inline ll readll(){
        ll i = 0, f = 1; char ch = getchar();
        for(; (ch < '0' || ch > '9') && ch != '-'; ch = getchar());
        if(ch == '-') f = -1, ch = getchar();
        for(; ch >= '0' && ch <= '9'; ch = getchar()) i = (i << 3) + (i << 1) + (ch - '0');
        return i * f;
    }
    inline void wrint(int x){
        if(x < 0) putchar('-'), x = -x;
        if(x > 9) wrint(x / 10);
        putchar(x % 10 + '0');
    }
    inline void wrll(ll x){
        if(x < 0) putchar('-'), x = -x;
        if(x > 9) wrll(x / 10);
        putchar(x % 10 + '0');
    }
}using namespace IO;

const int N = 1e5 + 5, M = 1e5 + 5, OO = 0x3f3f3f3f;
int n, m, k, ecnt, adj[N], go[M << 1], nxt[M << 1], T;
vector<int> keyPoint;
typedef pair<ll, int> P;
priority_queue<P, vector<P>, greater<P> > que;
bool key[N];
ll len[M << 1], dis[N], minn;

inline void addEdge(int u, int v, ll w){ nxt[++ecnt] = adj[u], adj[u] = ecnt, go[ecnt] = v, len[ecnt] = w;}

inline void solve(){
    ll ret = OO;
    for(int i = 0; i < k; i++){
        int u = keyPoint[i];
        memset(dis, OO, sizeof dis), dis[u] = 0;
        while(!que.empty()) que.pop(); que.push(P(0, u));
        while(!que.empty()){
            P t = que.top(); que.pop();
            if(t.second != u && key[t.second]){
                ret = min(ret, t.first);
                break;
            }
            for(int e = adj[t.second]; e; e = nxt[e]){
                int v = go[e];
                if(dis[v] > t.first + len[e]){
                    dis[v] = t.first + len[e];
                    que.push(P(dis[v], v));
                }
            }
        }
    }
    wrll(ret), putchar('\n');
}

int main(){
    freopen("h.in", "r", stdin);
    T = readint();
    while(T--){
        n = readint(), m = readint();
        memset(adj, 0, sizeof adj), ecnt = 0, keyPoint.clear(), memset(key, 0, sizeof key), minn = OO;
        for(int i = 1; i <= m; i++){
            int u = readint(), v = readint();
            ll w = readll(); minn = min(minn, w);
            addEdge(u, v, w), addEdge(v, u, w);
        }
        k = readint();
        for(int i = 1; i <= k; i++){ int x; keyPoint.push_back(x = readint()), key[x] = true;}
        if(k == n){ wrll(minn), putchar('\n');}
        else solve();
    }
    return 0;
}

正解:

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;

namespace IO{
    inline int readint(){
        int i = 0, f = 1; char ch = getchar();
        for(; (ch < '0' || ch > '9') && ch != '-'; ch = getchar());
        if(ch == '-') f = -1, ch = getchar();
        for(; ch >= '0' && ch <= '9'; ch = getchar()) i = (i << 3) + (i << 1) + (ch - '0');
        return i * f;
    }
    inline ll readll(){
        ll i = 0, f = 1; char ch = getchar();
        for(; (ch < '0' || ch > '9') && ch != '-'; ch = getchar());
        if(ch == '-') f = -1, ch = getchar();
        for(; ch >= '0' && ch <= '9'; ch = getchar()) i = (i << 3) + (i << 1) + (ch - '0');
        return i * f;
    }
    inline void wrint(int x){
        if(x < 0) putchar('-'), x = -x;
        if(x > 9) wrint(x / 10);
        putchar(x % 10 + '0');
    }
    inline void wrll(ll x){
        if(x < 0) putchar('-'), x = -x;
        if(x > 9) wrll(x / 10);
        putchar(x % 10 + '0');
    }
}using namespace IO;

const int N = 1e5 + 5, M = 1e5 + 5, OO = 0x3f3f3f3f;
int n, m, k, ecnt, adj[N], go[M << 1], nxt[M << 1], T;
vector<int> keyPoint;
typedef pair<ll, int> P;
priority_queue<P, vector<P>, greater<P> > que;
ll len[M << 1], dis[N], num[N], minn;

inline void addEdge(int u, int v, ll w){ nxt[++ecnt] = adj[u], adj[u] = ecnt, go[ecnt] = v, len[ecnt] = w;}

inline void solve(){
    ll ret = OO;
    for(int i = 0; (1 << i) <= k; i++){
        memset(dis, OO, sizeof dis);
        while(!que.empty()) que.pop();
        for(int j = 0; j < keyPoint.size(); j++)
            if((1 << i) & num[keyPoint[j]]) dis[keyPoint[j]] = 0, que.push(P(0, keyPoint[j]));
        while(!que.empty()){
            P t = que.top(); que.pop();
            if(num[t.second] && !((1 << i) & num[t.second])){ ret = min(ret, t.first); break; }
            for(int e = adj[t.second], v; e; e = nxt[e]){
                if(dis[v = go[e]] > t.first + len[e]){
                    dis[v] = t.first + len[e];
                    que.push(P(dis[v], v));
                }
            }
        }
    }
    wrll(ret), putchar('\n');
}

int main(){
    freopen("h.in", "r", stdin);
    T = readint();
    while(T--){
        n = readint(), m = readint();
        memset(adj, 0, sizeof adj), ecnt = 0, keyPoint.clear(), memset(num, 0, sizeof num), minn = OO;
        for(int i = 1; i <= m; i++){
            int u = readint(), v = readint();
            ll w = readll(); minn = min(minn, w);
            addEdge(u, v, w), addEdge(v, u, w);
        }
        k = readint();
        for(int i = 1; i <= k; i++){ int x; keyPoint.push_back(x = readint()), num[x] = i;}
        if(k == n) { wrll(minn), putchar('\n');}
        else solve();
    }
    return 0;
}