BZOJ4152: [Amppz2014]the captain

Description The N points on a given plane, the cost of defining (X1,Y1) to (x2,y2) is min (|x1-x2|,|y1-y2|), which is the minimum cost to go from point 1th to point N.

The first line of input contains a positive integer n (2<=n<=200000) that represents the number of points. Next n rows, each row contains two integers x[i],y[i] (0<=x[i],y[i]<=10^9), which in turn represents the coordinates of each point.

Output an integer, which is the minimum cost.

Sample Input5
2 2
1 1
4 5
7 1
6 7Sample Output2
The distance from point test instructions I to J is min (|xi-xj|,|yi-yj|), we can convert it to two sides, a weight of |xi-xj|, a weight of |yi-yj|, which is obviously no problem. Then the problem is transformed into a one-dimensional problem, how to even the least side, we find that by the order of the dimensions of each adjacent points to connect the edge.

#include <cstdio>#include<cctype>#include<queue>#include<cmath>#include<cstring>#include<algorithm>#defineRep (i,s,t) for (int i=s;i<=t;i++)#defineDwn (i,s,t) for (int i=s;i>=t;i--)#defineren for (int i=first[x];i;i=next[i])using namespacestd;Const intBuffersize=1<< -;Charbuffer[buffersize],*head,*Tail;inlineCharGetchar () {if(head==tail) {        intL=fread (Buffer,1, Buffersize,stdin); Tail= (Head=buffer) +l; }    return*head++;} InlineintRead () {intx=0, f=1;CharC=Getchar ();  for(;! IsDigit (c); C=getchar ())if(c=='-') f=-1;  for(; IsDigit (c); C=getchar ()) x=x*Ten+c-'0'; returnx*F;}Const intmaxn=200010;Const intmaxm=800010;Const intinf=1e9;structDijkstra {intN,M,D[MAXN],DONE[MAXN],FIRST[MAXN],NEXT[MAXM]; structEdge {int  from, To,dist;}    EDGES[MAXM]; structHeapnode {intu,d; BOOL operator< (Constheapnode& ths)Const{returnD>THS.D;}    }; voidInitintN) { This->n=n;m=1; }    voidAddedge (intUintVintW) {edges[m]= (Edge) {u,v,w};next[m]=first[u];first[u]=m++; } priority_queue<HeapNode>Q; voidSolveints) {Rep (I,1, N) d[i]=inf; D[s]=0; Q.push ((Heapnode) {s,0});  while(!Q.empty ()) {            intx=q.top (). u;            Q.pop (); if(Done[x])Continue;d one[x]=1; ren {Edge& e=Edges[i]; if(d[e.to]>d[x]+e.dist) {d[e.to]=d[x]+e.dist;                Q.push ((Heapnode) {e.to,d[e.to]}); }}}}}sol;structPoint {intX,y,id;} A[MAXN];BOOLCMP1 (Point A,point b) {returna.x<b.x;}BOOLCMP2 (Point A,point b) {returna.y<b.y;}intMain () {intn=read (); Sol.init (n); Rep (I,1, n) a[a[i].id=i].x=read (), a[i].y=read (); Sort (A+1, a+n+1, CMP1); Rep (I,1, N) {        if(i!=1) Sol. Addedge (a[i].id,a[i-1].id,a[i].x-a[i-1].x); if(i!=n) Sol. Addedge (a[i].id,a[i+1].id,a[i+1].x-a[i].x); } sort (A+1, a+n+1, CMP2); Rep (I,1, N) {        if(i!=1) Sol. Addedge (a[i].id,a[i-1].id,a[i].y-a[i-1].y); if(i!=n) Sol. Addedge (a[i].id,a[i+1].id,a[i+1].y-a[i].y); } sol.solve (1);p rintf ("%d\n", Sol.d[n]); return 0;}

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BZOJ4152: [Amppz2014]the captain