4152: [AMPPZ2014]The Captain

4152: [AMPPZ2014]The Captain

Time Limit: 20 Sec  Memory Limit: 256 MB
Submit: 1561  Solved: 620
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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

 

code

 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #include<cmath>
 #include<queue>
 #define mp(a,b) make_pair(a,b)
 #define pa pair<long long ,int>
 using namespace std;

 typedef long long LL;
 const int N = ;
 const LL INF = 1e18;

 struct Node {
     int x,y,id;
 }d[N];
 int head[N],L,R,tot,n;
 bool vis[N];
 struct Edge{
     int to,nxt,w;
 }e[];
 long long dis[N];
 priority_queue< pa,vector<pa>,greater<pa> >q;

 inline char nc() {
     static char buf[],*p1 = buf,*p2 = buf;
     return p1==p2&&(p2=(p1=buf)+fread(buf,,,stdin),p1==p2)?EOF:*p1++;
 }
 inline int read() {
     int x = ,f = ;char ch = nc();
     for (; ch<''||ch>''; ch = nc()) if (ch=='-') f = -;
     for (; ch>=''&&ch<=''; ch = nc()) x = x *  + ch - '';
     return x * f;
 }
 bool cmp1(Node a,Node b) {
     return a.x < b.x;
 }
 bool cmp2(Node a,Node b) {
     return a.y < b.y;
 }
 void add_edge(int u,int v,int w) {
     e[++tot].to = v;e[tot].w = w;e[tot].nxt = head[u];head[u] = tot;
     e[++tot].to = u;e[tot].w = w;e[tot].nxt = head[v];head[v] = tot;
 }
 void dij() {
     for (int i=; i<=n; ++i) dis[i] = INF;
     L = ;R = ;
     q.push(mp(,));
     dis[] = ;
     while (!q.empty()) {
         pa x = q.top();q.pop();
         int u = x.second;
         if (vis[u]) continue;vis[u] = true;
         for (int i=head[u]; i; i=e[i].nxt) {
             int v = e[i].to,w = e[i].w;;
             if (dis[v]>dis[u]+w) {
                 dis[v] = dis[u] + w;
                 q.push(mp(dis[v],v));
             }
         }
     }
 }
 int main() {
     freopen("1.txt","r",stdin);
     n = read();
     for (int i=; i<=n; ++i)
         d[i].x = read(),d[i].y = read(),d[i].id = i;
     sort(d+,d+n+,cmp1);
     for (int i=; i<n; ++i)
         add_edge(d[i].id,d[i+].id,d[i+].x-d[i].x);
     sort(d+,d+n+,cmp2);
     for (int i=; i<n; ++i)
         add_edge(d[i].id,d[i+].id,d[i+].y-d[i].y);
     dij();
     printf("%d",dis[n]);
     return ;
 }

spfa被卡了QwQ

 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #include<cmath>
 #include<queue>

 using namespace std;

 typedef long long LL;
 const int N = ;
 const LL INF = 1e18;

 struct Node {
     int x,y,id;
 }d[N];
 int head[N],L,R,tot,n;
 bool vis[N];
 struct Edge{
     int to,nxt,w;
 }e[];
 long long dis[N];
 queue<int>q;

 inline char nc() {
     static char buf[],*p1 = buf,*p2 = buf;
     return p1==p2&&(p2=(p1=buf)+fread(buf,,,stdin),p1==p2)?EOF:*p1++;
 }
 inline int read() {
     int x = ,f = ;char ch = nc();
     for (; ch<''||ch>''; ch = nc()) if (ch=='-') f = -;
     for (; ch>=''&&ch<=''; ch = nc()) x = x *  + ch - '';
     return x * f;
 }
 bool cmp1(Node a,Node b) {
     return a.x < b.x;
 }
 bool cmp2(Node a,Node b) {
     return a.y < b.y;
 }
 void add_edge(int u,int v,int w) {
     e[++tot].to = v;e[tot].w = w;e[tot].nxt = head[u];head[u] = tot;
     e[++tot].to = u;e[tot].w = w;e[tot].nxt = head[v];head[v] = tot;
 }
 void spfa() {
     for (int i=; i<=n; ++i) dis[i] = INF;
     L = ;R = ;
     q.push();
     dis[] = ;
     vis[] = true;
     while (!q.empty()) {
         int u = q.front();q.pop();
         for (int i=head[u]; i; i=e[i].nxt) {
             int v = e[i].to,w = e[i].w;;
             if (dis[v]>dis[u]+w) {
                 dis[v] = dis[u] + w;
                 if (!vis[v])q.push(v),vis[v] = true;
             }
         }
         vis[u] = false;
     }
 }
 int main() {
     n = read();
     for (int i=; i<=n; ++i)
         d[i].x = read(),d[i].y = read(),d[i].id = i;
     sort(d+,d+n+,cmp1);
     for (int i=; i<n; ++i)
         add_edge(d[i].id,d[i+].id,d[i+].x-d[i].x);
     sort(d+,d+n+,cmp2);
     for (int i=; i<n; ++i)
         add_edge(d[i].id,d[i+].id,d[i+].y-d[i].y);
     spfa();
     printf("%d",dis[n]);
     return ;
 }