题解-雅加达的摩天楼 (APIO2015)

分块思想，机智的建图。

1.n = min(sqrt(N), 100) ……设一个玄学限制，跑得会快很多。

2.SPFA 不要加 LLL 也不要加 SLF 优化！千万不要加！不然九十八！

我写了发 SPFA ，时间略卡。可能 Dijkstra 在这道题会更优秀一些吧……

#include <stdio.h>
#include <deque>
#include <math.h>

using namespace std;

const int _V = 30005*105;
const int _E = 30005*500;
const int INF = 1e9;

deque<int> Q;
int v[_E], w[_E], nxt[_E], fst[_V], dis[_V];
int Tot, N, n, M;
bool mk[_V];

void Ins(int t1, int t2, int t3)
{
    v[++Tot] = t2, w[Tot] = t3;
    nxt[Tot] = fst[t1];
    fst[t1] = Tot;
    return;
}

void SPFA(int beg)
{
    int t = (N-1)*(n+1)+n+1, i;
    for (i = 1; i <= t; ++i)
        dis[i] = INF, mk[i] = false;
    while (!Q.empty()) Q.pop_back();
    dis[beg] = 0, mk[beg] = true, Q.push_back(beg);
    while (!Q.empty()) {
        int p = Q.front();
        Q.pop_front(), mk[p] = false;
        for (i = fst[p]; i; i = nxt[i]) {
            if (dis[v[i]] > w[i]+dis[p]) {
                dis[v[i]] = w[i]+dis[p];
                if (!mk[v[i]]) {
                    mk[v[i]] = true;
                    Q.push_back(v[i]);
                }
                }
            }
        }
    }
    return;
}

int main()
{
    int i, j, k, beg, end;
    scanf("%d%d", &N, &M);
    n = min((int)sqrt(N+0.5), 100);
    for (i = 0; i < N; ++i)
        for (j = 1; j <= n; ++j) {
            Ins(i*(n+1)+j, i*(n+1)+n+1, 0);
            if (i+j < N) {
                Ins(i*(n+1)+j, (i+j)*(n+1)+j, 1);
                Ins((i+j)*(n+1)+j, i*(n+1)+j, 1);
            }
        }
    for (i = 1; i <= M; ++i) {
        int pos, p;
        scanf("%d%d", &pos, &p);
        if (i == 1) beg = pos;
        else if (i == 2) end = pos;
        if (p <= n) {
            Ins(pos*(n+1)+n+1, pos*(n+1)+p, 0);
        } else {
            for (j = pos+p; j < N; j += p)
                Ins(pos*(n+1)+n+1, j*(n+1)+n+1, (j-pos)/p);
            for (j = pos-p; j >= 0; j -= p)
                Ins(pos*(n+1)+n+1, j*(n+1)+n+1, (pos-j)/p);
        }
    }
    SPFA(beg*(n+1)+n+1);
    printf("%d
", dis[end*(n+1)+n+1] >= INF ? -1 : dis[end*(n+1)+n+1]);
    return 0;
}