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「AMPPZ2014」The Captain

作者:互联网

传送门:
这是一道bzoj权限题
Luogu团队题链接

解题思路

直接连边的话边数肯定会爆炸,考虑减少边数。
我们画出坐标系,发现一个东西:
对于两个点 \(A,B\),\(|x_A-y_A|\) 可以经由由他们俩之间的若干点取到,\(y\) 同理。
所以我们只需要先把所有点分别按照 \(x\) 和 \(y\),相邻两点之间连边即可。

细节注意事项

参考代码

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cctype>
#include <cmath>
#include <ctime>
#include <queue>
#define rg register
using namespace std;
template < typename T > inline void read(T& s) {
    s = 0; int f = 0; char c = getchar();
    while (!isdigit(c)) f |= (c == '-'), c = getchar();
    while (isdigit(c)) s = s * 10 + (c ^ 48), c = getchar();
    s = f ? -s : s;
}

typedef long long LL;
const int _ = 200010;
const int __ = 800010;

int tot, head[_], nxt[__], ver[__], w[__];
inline void Add_edge(int u, int v, int d)
{ nxt[++tot] = head[u], head[u] = tot, ver[tot] = v, w[tot] = d; }
inline void link(int u, int v, int d) { Add_edge(u, v, d), Add_edge(v, u, d); }

int n, vis[_]; LL dis[_];
struct node { int x, y, id; }p[_];

inline bool cmp1(const node& a, const node& b) { return a.x < b.x; }

inline bool cmp2(const node& a, const node& b) { return a.y < b.y; }

inline void Dijkstra() {
    priority_queue < pair < LL, int > > Q;
    memset(dis, 0x3f, sizeof dis);
    dis[1] = 0, Q.push(make_pair(0, 1));
    while (!Q.empty()) {
        pair < LL, int > x = Q.top(); Q.pop();
        int u = x.second;
        if (vis[u]) continue; vis[u] = 1;
        for (rg int i = head[u]; i; i = nxt[i]) {
            int v = ver[i];
            if (dis[v] > dis[u] + w[i])
                dis[v] = dis[u] + w[i], Q.push(make_pair(-dis[v], v));
        }
    }
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("in.in", "r", stdin);
#endif
    read(n);
    for (rg int i = 1; i <= n; ++i) read(p[i].x), read(p[i].y), p[i].id = i;
    sort(p + 1, p + n + 1, cmp1);
    for (rg int i = 1; i < n; ++i) link(p[i].id, p[i + 1].id, p[i + 1].x - p[i].x);
    sort(p + 1, p + n + 1, cmp2);
    for (rg int i = 1; i < n; ++i) link(p[i].id, p[i + 1].id, p[i + 1].y - p[i].y);
    Dijkstra();
    printf("%lld\n", dis[n]);
    return 0;
}

完结撒花 \(qwq\)

标签:Captain,const,int,tot,AMPPZ2014,inline,include,dis
来源: https://www.cnblogs.com/zsbzsb/p/11745719.html