4152: [AMPPZ2014]The Captain
Time Limit: 20 Sec Memory Limit: 256 MBDescription
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
2 2
1 1
4 5
7 1
6 7
Sample Output
2
HINT
Source
考虑如何减少连边。
对每一维进行分析：只需要连向它最近的一位的就可以了。
1 #include<iostream> 2 #include<algorithm> 3 #include<cstdio> 4 #include<cstring> 5 #include<cmath> 6 #include<cstdlib> 7 #include<vector> 8 using namespace std; 9 typedef long long ll; 10 typedef long double ld; 11 typedef pair<int,int> pr; 12 const double pi=acos(-1); 13 #define rep(i,a,n) for(int i=a;i<=n;i++) 14 #define per(i,n,a) for(int i=n;i>=a;i--) 15 #define Rep(i,u) for(int i=head[u];i;i=Next[i]) 16 #define clr(a) memset(a,0,sizeof(a)) 17 #define pb push_back 18 #define mp make_pair 19 #define fi first 20 #define sc second 21 #define pq priority_queue 22 #define pqb priority_queue <int, vector<int>, less<int> > 23 #define pqs priority_queue <int, vector<int>, greater<int> > 24 #define vec vector 25 ld eps=1e-9; 26 ll pp=1000000007; 27 ll mo(ll a,ll pp){if(a>=0 && a<pp)return a;a%=pp;if(a<0)a+=pp;return a;} 28 ll powmod(ll a,ll b,ll pp){ll ans=1;for(;b;b>>=1,a=mo(a*a,pp))if(b&1)ans=mo(ans*a,pp);return ans;} 29 void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); } 30 //void add(int x,int y,int z){ v[++e]=y; next[e]=head[x]; head[x]=e; cost[e]=z; } 31 int dx[5]={0,-1,1,0,0},dy[5]={0,0,0,-1,1}; 32 ll read(){ ll ans=0; char last=' ',ch=getchar(); 33 while(ch<'0' || ch>'9')last=ch,ch=getchar(); 34 while(ch>='0' && ch<='9')ans=ans*10+ch-'0',ch=getchar(); 35 if(last=='-')ans=-ans; return ans; 36 } 37 const int M=800005,N=200005; 38 int v[M],Next[M],head[N],vis[N],n,e,X[N],Y[N]; 39 ll dis[N],cost[M]; 40 struct nodeQ{ 41 ll v; int i; 42 friend bool operator <(nodeQ a,nodeQ b){ 43 return a.v>b.v; 44 } 45 }; 46 struct node{ 47 int x,y,i; 48 }f[N]; 49 bool cmpx(node a,node b){ 50 return a.x<b.x; 51 } 52 bool cmpy(node a,node b){ 53 return a.y<b.y; 54 } 55 #include<queue> 56 priority_queue<nodeQ> Q; 57 void add(int x,int y){v[++e]=y; Next[e]=head[x]; head[x]=e; cost[e]=min(abs(X[x]-X[y]),abs(Y[x]-Y[y])); } 58 void Dij(){ 59 for (int i=1;i<=n;i++) dis[i]=1e17,vis[i]=0; 60 dis[1]=0; 61 Q.push((nodeQ){dis[1],1}); 62 while (!Q.empty()){ 63 int u=Q.top().i; Q.pop(); if (vis[u]) continue; 64 vis[u]=1; 65 for (int i=head[u];i;i=Next[i]){ 66 if (dis[v[i]]>dis[u]+cost[i]){ 67 dis[v[i]]=dis[u]+cost[i]; 68 Q.push((nodeQ){dis[v[i]],v[i]}); 69 } 70 } 71 } 72 } 73 int main(){ 74 n=read(); 75 for (int i=1;i<=n;i++) f[i].x=X[i]=read(),f[i].y=Y[i]=read(),f[i].i=i; 76 sort(f+1,f+n+1,cmpx); 77 for (int i=1;i<n;i++) add(f[i].i,f[i+1].i),add(f[i+1].i,f[i].i);; 78 sort(f+1,f+n+1,cmpy); 79 for (int i=1;i<n;i++) add(f[i].i,f[i+1].i),add(f[i+1].i,f[i].i); 80 Dij(); 81 printf("%lld",dis[n]); 82 return 0; 83 }