[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 + ;

struct info
{
        int x , y , id;
} a[MAXN << ] , b[MAXN << ];

int n , d;
int dist[MAXN << ];
vector< int > G[MAXN << ];
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 = ; x = ;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << ) + (x << ) + c - '';
    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 = ; i <= n; i++)
        {
                read(a[i].x);
                read(a[i].y);
        }
        for (int i = ; i <= n; i++)
        {
                read(b[i].x);
                read(b[i].y);
        }
        memset(dist,,sizeof(dist));
        for (int i = ; i <= n; i++)
        {
                if (a[i].y == )
                {
                        q.push(i);
                        dist[i] = ;
                }
                a[i].id = i;
        }
        for (int i = ; i <= n; i++)
        {
                if (b[i].x == )
                {
                        q.push(i + n);
                        dist[i + n] = ;
                }
                b[i].id = i + n;
        }
        sort(a + ,a + n + ,cmpA);
        sort(b + ,b + n + ,cmpB);
        for (int i = ; i <= n; i++)
        {
                int l =  , r = n , pos = -;
                while (l <= r)
                {
                        int mid = (l + r) >> ;
                        if (b[mid].y >= a[i].y)
                        {
                                pos = mid;
                                r = mid - ;
                        } else l = mid + ;
                }
                if (pos == -) 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 = ; i <= n; i++)
        {
                int l =  , r = n , pos = -;
                while (l <= r)
                {
                        int mid = (l + r) >> ;
                        if (a[mid].x >= b[i].x)
                        {
                                pos = mid;
                                r = mid - ;
                        } else l = mid + ;
                }
                if (pos == -) 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 = ; i < G[cur].size(); i++)
                {
                        int v = G[cur][i];
                        if (dist[v] != -) continue;
                        dist[v] = dist[cur] + ;
                        q.push(v);
                }
        }
        for (int i = ; i <= n; i++) printf("%d\n",dist[i]);

        return ;

}