标签:read max priority font inline ref ace nbsp ext
bzoj4152[AMPPZ2014]The Captain
题意:
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。n≤200000。
题解:
结论:按某维坐标排序后,只有相邻两个点的距离才可能是这两个点的最小距离。故本题只要对所有点先按横坐标排序,将相邻的点连边,再对所有点按纵坐标排序,将相邻的点连边,之后求一次最短路即可。注意,本题数据大,spfa不能过(加了SLF也不行)。
代码:
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <queue> 5 #define inc(i,j,k) for(int i=j;i<=k;i++) 6 #define maxn 200010 7 #define INF 10000000000000000 8 #define ll long long 9 using namespace std; 10 11 inline int read(){ 12 char ch=getchar(); int f=1,x=0; 13 while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1; ch=getchar();} 14 while(ch>=‘0‘&&ch<=‘9‘)x=x*10+ch-‘0‘,ch=getchar(); 15 return f*x; 16 } 17 int n,x[maxn],y[maxn],id[maxn]; struct e{int t; ll w; int n;}es[maxn*4]; int ess,g[maxn]; 18 void pe(int f,int t,ll w){ 19 es[++ess]=(e){t,w,g[f]}; g[f]=ess; es[++ess]=(e){f,w,g[t]}; g[t]=ess; 20 } 21 bool cmp1(int a,int b){return x[a]<x[b];} bool cmp2(int a,int b){return y[a]<y[b];} 22 struct hn{int u; ll d; bool operator < (const hn&a)const{return d>a.d;}}; 23 ll d[maxn]; bool vis[maxn]; priority_queue<hn>q; 24 ll dijkstra(){ 25 inc(i,1,n)d[i]=INF; d[1]=0; q.push((hn){1,0}); 26 while(!q.empty()){ 27 int x; while(!q.empty()&&vis[x=q.top().u])q.pop(); if(vis[x])break; vis[x]=1; 28 for(int i=g[x];i;i=es[i].n)if(d[es[i].t]>d[x]+es[i].w){ 29 d[es[i].t]=d[x]+es[i].w; q.push((hn){es[i].t,d[es[i].t]}); 30 } 31 } 32 return d[n]; 33 } 34 int main(){ 35 n=read(); inc(i,1,n)x[i]=read(),y[i]=read(); inc(i,1,n)id[i]=i; 36 sort(id+1,id+n+1,cmp1); inc(i,1,n-1)pe(id[i],id[i+1],abs(x[id[i+1]]-x[id[i]])); 37 sort(id+1,id+n+1,cmp2); inc(i,1,n-1)pe(id[i],id[i+1],abs(y[id[i+1]]-y[id[i]])); 38 printf("%lld",dijkstra()); return 0; 39 }
20161108
bzoj4152[AMPPZ2014]The Captain*
标签:read max priority font inline ref ace nbsp ext
原文地址:http://www.cnblogs.com/YuanZiming/p/6055354.html