HDU5335 Work out Hierarchy traversal

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HDu5335

Test instructions

1000X1000 's map asks what is the smallest binary sequence that goes through (1000,1000) through four directions.

Problem Solving Ideas:

First, you should find a value of 0 and closest to the lower-right corner of the BFS (x+y value is the largest, with more than one saved).

This will ensure that the next sequence is the shortest.

The next best strategy for each step is only two possible: right or down, that is, the next layer of new nodes are on the same slash

If you encounter 0, then you must go to save, and if the current layer has 0 and 1, then the node with 1 does not need to continue to expand

At the end of the search, the minimum value of the current layer is output each time, and the smallest binary sequence can be obtained.

Code:

#include <iostream> #include <cstdio> #include <cstring> #include <queue> #include <vector > #define MAXN 1005using namespace std;struct node{int x, y;} head,t;int map[maxn][maxn];int Vis[maxn][maxn];int dir[ 4][2]= {1,0,0,1,-1,0,0,-1};int n,m;int ok () {if (t.x<1| | t.y<1| | t.x>n| | t.y>m| |    VIS[T.X][T.Y]) return 0; return 1;}    void BFs () {vector<node>w;    int maxx=0;    memset (vis,0,sizeof (VIS));    queue<node>q;   Q.push (node) {0,1});        Starting from outside the graph while (!q.empty ()) {Head=q.front ();        Q.pop ();                if (Head.x+head.y>maxx) {if (head.x==n&&head.y==m)//has a path that is all 0 {                printf ("0\n");            Return            } maxx=head.x+head.y;            W.clear ();        W.push_back (head);        } if (Head.x+head.y==maxx) W.push_back (head);            for (int i=0; i<4; i++) {t.x=head.x+dir[i][0]; T.y=head.y+dIR[I][1]; if (!ok () | | |            Map[t.x][t.y]==1) continue;            Vis[t.x][t.y]=1;        Q.push (t);    }} vector<node>q2[2];    int now=0,next=1;    int flag;    if (map[1][1]!=1) {for (int i=0; i<w.size (); i++) Q2[0].push_back (W[i]);    } else//does not have a leading 0 q2[0].push_back ((node) {0,1});        while (1)//hierarchy traversal {flag=0;            for (int i=0;i<q2[now].size (); i++) {head=q2[now][i];                for (int i=0;i<2;i++) {t.x=head.x+dir[i][0];                T.Y=HEAD.Y+DIR[I][1]; if (!ok () | | |                VIS[T.X][T.Y]) continue;                Vis[t.x][t.y]=1;                    if (t.x==n&&t.y==m) {printf ("%d\n", Map[t.x][t.y]);                Return                  } if (map[t.x][t.y]==0) {if (!flag)      Q2[next].clear ();                    flag=1;                Q2[next].push_back (t);            } else if (!flag&&map[t.x][t.y]==1) q2[next].push_back (t);        }} q2[now].clear ();        if (flag)//The layer has 0 printf ("0");        else printf ("1");    Swap (Now,next);    }}int Main () {//Freopen ("In.txt", "R", stdin);    int T;    Char STR[MAXN];    int Len;    scanf ("%d", &t);        while (t--) {scanf ("%d%d", &n,&m);            for (int i=1; i<=n; i++) {scanf ("%s", str+1);            Len=strlen (str+1);        for (int j=1; j<=len; j + +) map[i][j]=str[j]-' 0 ';    } BFS (); } return 0;}

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HDU5335 Work out Hierarchy traversal