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Test instructions
Given the 01 matrix, ask the minimum number of digits represented by all 01 binary strings from (N,M) through the road.

Analysis:
First, the prefix 0 is to be removed, but also found that only constantly down and to the right to achieve the shortest path. So first DFS finds all the 0 positions that can go, minus the prefix 0. Then BFS to the lower right corner, 0 of the place to go 0, do not walk 1, Without any 0 of the cases to go 1. If there are more than 0 go, see who is behind 0. If there is no 0, all 1 will go, and see who may be behind 0. It's Jiangzi.
In addition, the problem uses the nature of the diagonal, when the X+y==tot, all the points to the end of the condition of the distance are equal. In fact, this is also a backslash, equivalent to a layer of anti-slash search.
In addition, judging tot==0 and tot==n+m when the order is different, an AC one WA. Self-study to think about is a special situation,
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At this time both are suitable, so should first Judge Tot==n+m.

#include <cstdio>#include <cstring>#include <cmath>#include <iostream>#include <vector>#include <map>#include <algorithm>#define Read Freopen ("Q.in", "R", stdin)#define LL Long LongConst intMAXN =1003;using namespace STD;intN,m;CharMP[MAXN][MAXN];intdx[]={0,0,-1,1};intdy[]={1,-1,0,0};intVIS[MAXN][MAXN];intTotvoidDfsintXintY) {if(Vis[x][y])return;if(mp[x][y]==' 1 ')return; vis[x][y]=1;//vis represents all of the 0 locations visited    if(X+y>tot) Tot=x+y;//tot represents the coordinates of the currently traversed road and    if(x+1<=n) DFS (x+1, y);if(y+1&LT;=M) DFS (x,y+1);if(X-1>=1) DFS (X-1, y);if(Y1>=1) DFS (x,y-1);}voidBFS () {intI,j; for(i=tot;i<n+m;i++) {intflag=1;//To determine if there are 0 can walk if there are more than 0 can go?          for(J=max (1, i-m); J<=min (n,i-1); J + +)if(Vis[j][i-j]) {inty=i-j+1<=m?mp[j][i-j+1]-' 0 ':1;intx=j+1<=n?mp[j+1][i-j]-' 0 ':1;if(x==0|| y==0) flag=0;//cout<<endl<< "x=" <<x<< "y=" <<y<<endl;} for(J=max (1, i-m); J<=min (n,i-1); J + +)if(Vis[j][i-j]) {intx=i-j+1<=m?mp[j][i-j+1]-' 0 ':1;inty=j+1<=n?mp[j+1][i-j]-' 0 ':1;if(X==flag) vis[j][i-j+1]=1;if(Y==flag) vis[j+1][i-j]=1;//If there are more than 0, two-pronged}printf("%d", flag); }}intMain () {//Read;    intCAsscanf("%d", &cas); while(cas--) {intI,j;memset(Vis,0,sizeof(VIS));scanf("%d%d", &n,&m); for(i=1; i<=n;i++)scanf('%s ', mp[i]+1); tot=0; Dfs1,1);if(tot==n+m) {printf("0\n");Continue; }if(tot==0)        {printf("1"); vis[1][1]=1; tot=2; } BFS ();printf("\ n"); }}

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