【AMPPZ 2014】 The Captain
【题目链接】
【算法】
按x轴排序,将相邻点连边
按y轴排序,将相邻点连边
然后对这个图跑最短路就可以了,笔者用的是dijkstra算法
【代码】
#include<bits/stdc++.h> using namespace std; #define MAXN 200000 struct info { int id,x,y; } a[MAXN+10]; int N,i; int dist[MAXN+10]; vector< pair<int,int> > E[MAXN+10]; template <typename T> inline void read(T &x) { int 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; } template <typename T> inline void write(T x) { if (x < 0) { putchar('-'); x = -x; } if (x > 9) write(x/10); putchar(x%10+'0'); } template <typename T> inline void writeln(T x) { write(x); puts(""); } inline bool cmp1(info a,info b) { return a.x < b.x; } inline bool cmp2(info a,info b) { return a.y < b.y; } inline void add_edge(int x,int y,int d) { E[x].push_back(make_pair(y,d)); E[y].push_back(make_pair(x,d)); } inline void dijkstra() { int i,x,to,cost; static int vis[MAXN+10]; priority_queue< pair<int,int> > q; for (i = 2; i <= N; i++) dist[i] = INT_MAX; q.push(make_pair(0,1)); while (!q.empty()) { x = q.top().second; q.pop(); if (vis[x]) continue; for (i = 0; i < E[x].size(); i++) { to = E[x][i].first; cost = E[x][i].second; if (dist[x] + cost < dist[to]) { dist[to] = dist[x] + cost; q.push(make_pair(-dist[to],to)); } } } } int main() { read(N); for (i = 1; i <= N; i++) { read(a[i].x); read(a[i].y); a[i].id = i; } sort(a+1,a+N+1,cmp1); for (i = 2; i <= N; i++) add_edge(a[i-1].id,a[i].id,a[i].x-a[i-1].x); sort(a+1,a+N+1,cmp2); for (i = 2; i <= N; i++) add_edge(a[i-1].id,a[i].id,a[i].y-a[i-1].y); dijkstra(); writeln(dist[N]); return 0; }