4152: [AMPPZ2014]The Captain
4152: [AMPPZ2014]The Captain
Time Limit: 20 Sec Memory Limit: 256 MBSubmit: 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
2 2
1 1
4 5
7 1
6 7
Sample Output
2
code
1 #include<cstdio> 2 #include<algorithm> 3 #include<cstring> 4 #include<cmath> 5 #include<queue> 6 #define mp(a,b) make_pair(a,b) 7 #define pa pair<long long ,int> 8 using namespace std; 9 10 typedef long long LL; 11 const int N = 200100; 12 const LL INF = 1e18; 13 14 struct Node { 15 int x,y,id; 16 }d[N]; 17 int head[N],L,R,tot,n; 18 bool vis[N]; 19 struct Edge{ 20 int to,nxt,w; 21 }e[1000100]; 22 long long dis[N]; 23 priority_queue< pa,vector<pa>,greater<pa> >q; 24 25 inline char nc() { 26 static char buf[100000],*p1 = buf,*p2 = buf; 27 return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++; 28 } 29 inline int read() { 30 int x = 0,f = 1;char ch = nc(); 31 for (; ch<'0'||ch>'9'; ch = nc()) if (ch=='-') f = -1; 32 for (; ch>='0'&&ch<='9'; ch = nc()) x = x * 10 + ch - '0'; 33 return x * f; 34 } 35 bool cmp1(Node a,Node b) { 36 return a.x < b.x; 37 } 38 bool cmp2(Node a,Node b) { 39 return a.y < b.y; 40 } 41 void add_edge(int u,int v,int w) { 42 e[++tot].to = v;e[tot].w = w;e[tot].nxt = head[u];head[u] = tot; 43 e[++tot].to = u;e[tot].w = w;e[tot].nxt = head[v];head[v] = tot; 44 } 45 void dij() { 46 for (int i=1; i<=n; ++i) dis[i] = INF; 47 L = 1;R = 0; 48 q.push(mp(0,1)); 49 dis[1] = 0; 50 while (!q.empty()) { 51 pa x = q.top();q.pop(); 52 int u = x.second; 53 if (vis[u]) continue;vis[u] = true; 54 for (int i=head[u]; i; i=e[i].nxt) { 55 int v = e[i].to,w = e[i].w;; 56 if (dis[v]>dis[u]+w) { 57 dis[v] = dis[u] + w; 58 q.push(mp(dis[v],v)); 59 } 60 } 61 } 62 } 63 int main() { 64 freopen("1.txt","r",stdin); 65 n = read(); 66 for (int i=1; i<=n; ++i) 67 d[i].x = read(),d[i].y = read(),d[i].id = i; 68 sort(d+1,d+n+1,cmp1); 69 for (int i=1; i<n; ++i) 70 add_edge(d[i].id,d[i+1].id,d[i+1].x-d[i].x); 71 sort(d+1,d+n+1,cmp2); 72 for (int i=1; i<n; ++i) 73 add_edge(d[i].id,d[i+1].id,d[i+1].y-d[i].y); 74 dij(); 75 printf("%d",dis[n]); 76 return 0; 77 }
spfa被卡了QwQ
1 #include<cstdio> 2 #include<algorithm> 3 #include<cstring> 4 #include<cmath> 5 #include<queue> 6 7 using namespace std; 8 9 typedef long long LL; 10 const int N = 200100; 11 const LL INF = 1e18; 12 13 struct Node { 14 int x,y,id; 15 }d[N]; 16 int head[N],L,R,tot,n; 17 bool vis[N]; 18 struct Edge{ 19 int to,nxt,w; 20 }e[1000100]; 21 long long dis[N]; 22 queue<int>q; 23 24 inline char nc() { 25 static char buf[100000],*p1 = buf,*p2 = buf; 26 return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++; 27 } 28 inline int read() { 29 int x = 0,f = 1;char ch = nc(); 30 for (; ch<'0'||ch>'9'; ch = nc()) if (ch=='-') f = -1; 31 for (; ch>='0'&&ch<='9'; ch = nc()) x = x * 10 + ch - '0'; 32 return x * f; 33 } 34 bool cmp1(Node a,Node b) { 35 return a.x < b.x; 36 } 37 bool cmp2(Node a,Node b) { 38 return a.y < b.y; 39 } 40 void add_edge(int u,int v,int w) { 41 e[++tot].to = v;e[tot].w = w;e[tot].nxt = head[u];head[u] = tot; 42 e[++tot].to = u;e[tot].w = w;e[tot].nxt = head[v];head[v] = tot; 43 } 44 void spfa() { 45 for (int i=1; i<=n; ++i) dis[i] = INF; 46 L = 1;R = 0; 47 q.push(1); 48 dis[1] = 0; 49 vis[1] = true; 50 while (!q.empty()) { 51 int u = q.front();q.pop(); 52 for (int i=head[u]; i; i=e[i].nxt) { 53 int v = e[i].to,w = e[i].w;; 54 if (dis[v]>dis[u]+w) { 55 dis[v] = dis[u] + w; 56 if (!vis[v])q.push(v),vis[v] = true; 57 } 58 } 59 vis[u] = false; 60 } 61 } 62 int main() { 63 n = read(); 64 for (int i=1; i<=n; ++i) 65 d[i].x = read(),d[i].y = read(),d[i].id = i; 66 sort(d+1,d+n+1,cmp1); 67 for (int i=1; i<n; ++i) 68 add_edge(d[i].id,d[i+1].id,d[i+1].x-d[i].x); 69 sort(d+1,d+n+1,cmp2); 70 for (int i=1; i<n; ++i) 71 add_edge(d[i].id,d[i+1].id,d[i+1].y-d[i].y); 72 spfa(); 73 printf("%d",dis[n]); 74 return 0; 75 }
