[APIO 2015] 雅加达的摩天楼

[题目链接]

        https://www.lydsy.com/JudgeOnline/problem.php?id=4070

[算法]

         考虑将每个"Doge"向其所能到达的楼连边

         直接SPFA求单源最短路可以获得57分

         那么 , 怎样拿到满分呢?

         我们发现这张图的边的数量达到了NM的数量级

         考虑分块 , 将每个点拆成SQRT(N)个点

         将每个Pi <= SQRT(N)的点向(Bi , Pi)连边 , 这样的边不会超过N * SQRT(N)条

         将每个Pi > SQRT(N)的点向其所能到达的所有点连边 , 这样的边不会超过NlogN条

         时间复杂度 : O(N ^ 2) , 实际远不能达到这个上限

[代码]

        

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int inf = 1e9;
const int N = 6000050;

struct edge
{
        int to , w , nxt;
} e[15000005];

int n , m , block , tot , S , T;
int head[N] , dist[N];
bool inq[N];

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 int id(int x , int y)
{
        return y * n + x;
}
inline void addedge(int u , int v , int w)
{
        ++tot;
        e[tot] = (edge){v , w , head[u]};
        head[u] = tot;
}
inline int SPFA()
{
        queue< int > q;
        q.push(S);
        memset(dist , 0x3f , sizeof(dist));
        dist[S] = 0;
        inq[S] = true;
        while (!q.empty())
        {
                int cur = q.front();
                q.pop();
                inq[cur] = false;
                for (int i = head[cur]; i; i = e[i].nxt)
                {
                        int v = e[i].to , w = e[i].w;
                        if (dist[cur] + w < dist[v])
                        {
                                dist[v] = dist[cur] + w;
                                if (!inq[v])
                                {
                                        inq[v] = true;
                                        q.push(v);        
                                }        
                        }        
                }                
        }
        return dist[T] != 0x3f3f3f3f ? dist[T] : -1;
}

int main()
{
        
        read(n); read(m);
        block = min((int)sqrt(n) , 100);
        for (int i = 1; i <= block; ++i)
        {
                for (int j = i; j < n; ++j)
                {
                        addedge(id(j , i) , id(j - i , i) , 1);
                        addedge(id(j - i , i) , id(j , i) , 1);        
                }    
                for (int j = 0; j < n; ++j) addedge(id(j , i) , id(j , 0) , 0);
        }
        for (int k = 1; k <= m; ++k)
        {
                int Bi , Pi;
                read(Bi); read(Pi);
                if (Pi <= block) addedge(id(Bi , 0) , id(Bi , Pi) , 0);    
                else
                {
                        for (int i = Bi + Pi; i < n; i += Pi) addedge(id(Bi , 0) , id(i , 0) , (i - Bi) / Pi);
                        for (int i = Bi - Pi; i >= 0; i -= Pi) addedge(id(Bi , 0) , id(i , 0) , (Bi - i) / Pi);
                }
                if (k == 1) S = id(Bi , 0);
                if (k == 2) T = id(Bi , 0);
        }
        printf("%d\n" , SPFA());
        
        return 0;
    
}

 

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转载自www.cnblogs.com/evenbao/p/10549370.html
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