BZOJ 4152: [AMPPZ2014]The Captain( 最短路 )

先按x排序, 然后只有相邻节点的边才有用, 我们连起来, 再按y排序做相同操作...然后就dijkstra

------------------------------------------------------------------------

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<iostream>

#include<queue>

#include<cmath>

  

#define rep(i, n) for(int i = 0; i < n; i++)

#define clr(x, c) memset(x, c, sizeof(x))

  

using namespace std;

 

const int inf = 1000000009, maxn = 200009;

 

struct edge {

int to, w;

edge*next;

} E[maxn << 2], *pt = E, *head[maxn];

 

struct node {

int x, d;

bool operator < (const node&o) const {

return d > o.d;

}

};

 

struct P {

int x, y;

inline void Read() {

scanf("%d%d", &x, &y);

}

} A[maxn];

 

inline void add(int u, int v, int w) {

pt->to = v, pt->w = w;

pt->next = head[u];

head[u] = pt++;

}

#define add_edge(u, v, w) add(u, v, w), add(v, u, w)

 

bool cmpX(const int i, const int j) {

return A[i].x < A[j].x;

}

 

bool cmpY(const int i, const int j) {

return A[i].y < A[j].y;

}

 

int d[maxn], n, X[maxn], Y[maxn];

priority_queue<node> Q;

 

void dijkstra() {

rep(i, n) d[i] = inf;

d[0] = 0, Q.push( (node) {0, 0} );

while(!Q.empty()) {

node t = Q.top(); Q.pop();

if(d[t.x] != t.d) continue;

for(edge*e = head[t.x]; e; e = e->next) if(d[e->to] > d[t.x] + e->w) {

d[e->to] = d[t.x] + e->w;

Q.push( (node) {e->to, d[e->to]} );

}

}

}

 

int main() {

freopen("test.in", "r", stdin);

cin >> n;

rep(i, n) {

   A[i].Read();

   X[i] = Y[i] = i;

}

sort(X, X + n, cmpX);

rep(i, n - 1) {

P*a = A + X[i], *b = A + X[i + 1];

if(b->x - a->x <= abs(a->y - b->y)) add_edge(X[i], X[i + 1], b->x - a->x);

}

sort(Y, Y + n, cmpY);

rep(i, n - 1) {

P*a = A + Y[i], *b = A + Y[i + 1];

if(b->y - a->y <= abs(a->x - b->x)) add_edge(Y[i], Y[i + 1], b->y - a->y);

}

dijkstra();

printf("%d\n", d[n - 1]);

return 0;

}

------------------------------------------------------------------------

4152: [AMPPZ2014]The Captain

Time Limit: 20 Sec  Memory Limit: 256 MB
Submit: 272  Solved: 104
[Submit][Status][Discuss]

Description

给定平面上的n个点，定义(x1,y1)到(x2,y2)的费用为min(|x1-x2|,|y1-y2|)，求从1号点走到n号点的最小费用。

Input

第一行包含一个正整数n(2<=n<=200000)，表示点数。

接下来n行，每行包含两个整数x[i],y[i](0<=x[i],y[i]<=10^9)，依次表示每个点的坐标。

Output

一个整数，即最小费用。

Sample Input

5
2 2
1 1
4 5
7 1
6 7

Sample Output

2

HINT

Source

鸣谢Claris上传