Bzoj4152: [AMPPZ2014]The Captain

题面

传送门

Sol

分别按\(X\)轴,\(Y\)轴从小到大排序,相邻两个点建边权为\(\Delta x\)\(\Delta y\)的边
然后跑\(Dijkstra\)

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(2e5 + 5);
typedef int Arr[_];
 
IL int Input(){
    RG int x = 0, z = 1; RG char c = getchar();
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    return x * z;
}
 
Arr first, dis, vis;
int n, m, cnt;
struct Point{
    int x, y, id;
} p[_];
struct Data{
    int u, dis;
 
    IL int operator <(RG Data B) const{
        return dis > B.dis;
    }
};
priority_queue <Data> Q;
struct Edge{
    int to, next, w;
} edge[_ << 2];
 
IL void Add(RG int u, RG int v, RG int w){
    edge[cnt] = (Edge){v, first[u], w}, first[u] = cnt++;
    edge[cnt] = (Edge){u, first[v], w}, first[v] = cnt++;
}
 
IL int Cmp1(RG Point A, RG Point B){
    return A.x < B.x;
}
 
IL int Cmp2(RG Point A, RG Point B){
    return A.y < B.y;
}
 
IL void Dijkstra(){
    Fill(dis, 63), dis[1] = 0, Q.push((Data){1, 0});
    while(!Q.empty()){
        RG Data x = Q.top(); Q.pop();
        if(vis[x.u]) continue;
        vis[x.u] = 1;
        for(RG int e = first[x.u]; e != -1; e = edge[e].next){
            RG int v = edge[e].to, w = edge[e].w;
            if(dis[x.u] + w < dis[v]) Q.push((Data){v, dis[v] = dis[x.u] + w});
        }
    }
}
 
int main(RG int argc, RG char *argv[]){
    n = Input();
    for(RG int i = 1; i <= n; ++i)
        first[i] = -1, p[i] = (Point){Input(), Input(), i};
    sort(p + 1, p + n + 1, Cmp1);
    for(RG int i = 1; i < n; ++i)
        Add(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)
        Add(p[i].id, p[i + 1].id, p[i + 1].y - p[i].y);
    Dijkstra();
    printf("%d\n", dis[n]);
    return 0;
}
posted @ 2018-03-28 13:20  Cyhlnj  阅读(98)  评论(0编辑  收藏  举报