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  • 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
    ", Case, ans);
        }
        
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/CzxingcHen/p/9843478.html
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