HDNOIP201404 Shortest Path

HDNOIP201404 Shortest Path

Difficulty level: A; programming language: Unlimited; run time limit: 1000ms; run space limit: 51200KB; code length limit: 2000000B

Question Description

A, B, and C are 3 non-equal 1-bit positive numbers, with them and the number 0 can fill an n row n column of the grid array, each lattice has 4 kinds of digital one. A grid filled with 0 indicates an obstacle and cannot belong to any path. Can you find a path that starts at 1 rows, 1 columns, and reaches n rows n columns with the least cost? Note: Each lattice can only go to the adjacent upper, lower, left, and right non-0 and not out of bounds. The so-called path cost refers to the sum of the numbers in all the squares through which the path passes. You can program the minimum path cost to reach N rows n columns from the position of 1 rows and 1 columns, and output-1 if you cannot reach it.

Input

The first line enters the number n. The next n rows are a number string of length n, and the N strings form the square of a number symbol. In addition to ' 0 ', the matrix can contain up to 3 numbers of characters.

Output

There is only a minimum cost or-1 this integer.

Input example

"Input Sample 1" 4 1231 2003 1002 1113 "Input Sample 2" 4 3150 1153 3311 0530

Output example

"Output Example 1" 10 "Output Example 2" -1

Other Notes

60% of data, n<10,80% data, n<100,100% data, n<1000

It is a good question indeed.

1000*1000 the shortest possible short-circuit may be a bit difficult, measured card when 1000s+. So how do you do it?

In addition to ' 0 ', the matrix can contain up to 3 numbers of characters.

This reminds us that we can do some articles on it. Consider why the heap is used to optimize Dijkstra because some edges are very long and some edges are very short, and for all the same points in the edge, the distance to them is incremented.

The algorithm is out of the way, with 3 monotone queues instead of Heap, pay attention to how to take the team head and how to join the team tail each time.

#include <cstdio>#include<cctype>#include<queue>#include<cstring>#include<algorithm>#defineRep (i,s,t) for (int i=s;i<=t;i++)#defineren for (int i=first[x];i!=-1;i=next[i])using namespaceStd;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=1010;CharA[MAXN][MAXN];intN,D[MAXN][MAXN],VIS[MAXN][MAXN],IDX[MAXN],TP;structPoint {intx, y; BOOL operator< (Constpoint& ths) {returnd[x][y]<d[ths.x][ths.y];}; Queue<Point> q[3];intGetfront () {intc=-1; if(q[0].size ()) c=0; if(q[1].size () && (c<0|| q[1].front () <q[c].front ()) c=1; if(q[2].size () && (c<0|| q[2].front () <q[c].front ()) c=2; returnC;}intmx[]={1,-1,0,0},my[]={0,0,1,-1};intsolve () {if(a[1][1]=='0'|| a[n][n]=='0')return-1; q[idx[a[1][1]]].push (point) {1,1});d [1][1]=a[1][1]-'0';  while(q[0].size () +q[1].size () +q[2].size ()) {intT=getfront ();intX=q[t].front (). x,y=Q[t].front (). Y;q[t].pop (); if(x==n&&y==n)returnD[x][y]; if(Vis[x][y])Continue; vis[x][y]=1; Rep (dir,0,3) {            intnx=x+mx[dir],ny=y+My[dir]; if(nx>=1&&nx<=n&&ny>=1&&ny<=n&&a[nx][ny]!='0'&&d[x][y]+a[nx][ny]-'0'<D[nx][ny]) {D[nx][ny]=d[x][y]+a[nx][ny]-'0';                                           Q[idx[a[nx][ny]]].push (point) {nx,ny}); }                  }                                             }    return-1;}intMain () {n=read (); Rep (I,1, N) scanf ("%s", a[i]+1); memset (IDX,-1,sizeof(IDX)); Rep (I,1, N) Rep (J,1, N) {        if(idx[a[i][j]]<0&&a[i][j]!='0') idx[a[i][j]]=tp++; D[I][J]=1<< -; } printf ("%d\n", Solve ()); return 0;}

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HDNOIP201404 Shortest Path