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  • [Usaco2017 Dec] A Pie for a Pie

    [题目链接]

            https://www.lydsy.com/JudgeOnline/problem.php?id=5140

    [算法]

             最短路

             时间复杂度 : O(N^2)

    [代码]

             

    #include<bits/stdc++.h>
    using namespace std;
    const int MAXN = 1e5 + 10;
    
    struct info
    {
            int x , y , id;
    } a[MAXN << 1] , b[MAXN << 1];
    
    int n , d;
    int dist[MAXN << 1];
    vector< int > G[MAXN << 1];
    queue< int > q;
    
    template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
    template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
    template <typename T> inline void read(T &x)
    {
        T f = 1; x = 0;
        char c = getchar();
        for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
        for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
        x *= f;
    }
    inline bool cmpA(info a,info b)
    {
            return a.x < b.x;
    }
    inline bool cmpB(info a,info b)
    {
            return a.y < b.y;
    }
    
    int main()
    {
            
            read(n); read(d);
            for (int i = 1; i <= n; i++)
            {
                    read(a[i].x);
                    read(a[i].y);
            }
            for (int i = 1; i <= n; i++)
            {
                    read(b[i].x); 
                    read(b[i].y);        
            }
            memset(dist,255,sizeof(dist));
            for (int i = 1; i <= n; i++)
            {
                    if (a[i].y == 0)
                    {
                            q.push(i);
                            dist[i] = 1;
                    }        
                    a[i].id = i;
            }
            for (int i = 1; i <= n; i++)
            {
                    if (b[i].x == 0)
                    {
                            q.push(i + n);
                            dist[i + n] = 1;        
                    }    
                    b[i].id = i + n;    
            }
            sort(a + 1,a + n + 1,cmpA);
            sort(b + 1,b + n + 1,cmpB);
            for (int i = 1; i <= n; i++)
            {
                    int l = 1 , r = n , pos = -1;
                    while (l <= r)    
                    {
                            int mid = (l + r) >> 1;
                            if (b[mid].y >= a[i].y)
                            {
                                    pos = mid;
                                    r = mid - 1;
                            } else l = mid + 1;
                    }
                    if (pos == -1) continue;
                    for (int j = pos; j <= n; j++)
                    {
                            if (b[j].y > a[i].y + d) break;
                            G[b[j].id].push_back(a[i].id);
                    }
            }
            for (int i = 1; i <= n; i++)
            {
                    int l = 1 , r = n , pos = -1;
                    while (l <= r)
                    {
                            int mid = (l + r) >> 1;
                            if (a[mid].x >= b[i].x)
                            {
                                    pos = mid;
                                    r = mid - 1;
                            } else l = mid + 1;
                    }
                    if (pos == -1) continue;
                    for (int j = pos; j <= n; j++)
                    {
                            if (a[j].x > b[i].x + d) break;
                            G[a[j].id].push_back(b[i].id);
                    }
            }
            while (!q.empty())
            {
                    int cur = q.front();
                    q.pop();
                    for (unsigned i = 0; i < G[cur].size(); i++)
                    {
                            int v = G[cur][i];
                            if (dist[v] != -1) continue;
                            dist[v] = dist[cur] + 1;
                            q.push(v);        
                    }        
            }
            for (int i = 1; i <= n; i++) printf("%d
    ",dist[i]);
            
            return 0;
        
    }
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  • 原文地址:https://www.cnblogs.com/evenbao/p/9795119.html
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