[BZOJ4152][AMPPZ2014]The Captain（堆优化dijkstra）

题目描述

传送门

题目大意：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。

题解

分别按照xy排序，然后相邻点连边，跑最短路就行了
写了一发堆优化dijkstra，竟然把大小记反了！

代码

#include<algorithm>#include<iostream>#include<cstring>#include<cstdio>#include<cmath>#include<queue>using namespace std;#define LL long long#define N 1000005int n;struct data{LL x,y;int id;}p[N];int tot,point[N],nxt[N],v[N];LL c[N];LL dis[N];bool vis[N];priority_queue < pair<LL,int> > q;int cmpx(data a,data b){return a.x<b.x||(a.x==b.x&&a.y<b.y);}int cmpy(data a,data b){return a.y<b.y||(a.y==b.y&&a.x<b.x);}LL Abs(LL x){return (x>0)?x:-x;}LL calc(data a,data b){return min(Abs(a.x-b.x),Abs(a.y-b.y));}void add(int x,int y,LL z){//  printf("%d %d %lld\n",x,y,z);    ++tot; nxt[tot]=point[x]; point[x]=tot; v[tot]=y; c[tot]=-z;    ++tot; nxt[tot]=point[y]; point[y]=tot; v[tot]=x; c[tot]=-z;}void dijkstra(int s,int t){    memset(dis,128,sizeof(dis));    dis[s]=0;q.push(make_pair(0,s));    while (!q.empty())    {        pair<LL,int> now=q.top();q.pop();        int x=now.second;        if (vis[x]) continue;vis[x]=1;        for (int i=point[x];i;i=nxt[i])            if (dis[v[i]]<dis[x]+c[i])            {                dis[v[i]]=dis[x]+c[i];                q.push(make_pair(dis[v[i]],v[i]));            }    }}int main(){    scanf("%d",&n);    for (int i=1;i<=n;++i) scanf("%lld%lld",&p[i].x,&p[i].y),p[i].id=i;    sort(p+1,p+n+1,cmpx);    for (int i=1;i<n;++i) add(p[i].id,p[i+1].id,calc(p[i],p[i+1]));    sort(p+1,p+n+1,cmpy);    for (int i=1;i<n;++i) add(p[i].id,p[i+1].id,calc(p[i],p[i+1]));    dijkstra(1,n);    printf("%lld\n",-dis[n]);}