挑战第二章 poj 3009 水题DFS

题目给出剪枝条件10步就剪。。直接暴力

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <cmath>
#include <vector>
#include <bitset>
#include <bitset>
#include <ctime>
using namespace std;

#define LL long long
#define SZ(v) ((int)(v).size())
#define FOR(i,st,ed) for((i)=(st);(i)!=(ed);++(i))
#define REP(i,ed) FOR(i,0,ed)
#define FORE(i,a,b) for(int i=(a);i<=(b);++i)
#define REPE(i,n) FORE(i,0,n)
#define FORSZ(i,a,v) FOR(i,a,SZ(v))
#define REPSZ(i,v) REP(i,SZ(v))

void nextInt(int &x)
{
scanf("%d", &x);
}

void nextInt(int &x, int &y)
{
scanf("%d%d", &x, &y);
}

int g[30][30];
int nowx, nowy;
int n, m;
int xbegin,xend,ybegin,yend;

void init()
{
memset(g,0,sizeof(g));
int i,j;
swap(n,m);
REP (i,n)
{
REP(j, m)
{
nextInt(g[i][j]);
if (g[i][j] == 2)
{
xbegin = i;
ybegin = j;
}
if (g[i][j] == 3)
{
xend = i;
yend = j;
}
}
}
}

int ans = 0;
const int dx[]={0,0,-1,1};
const int dy[]={1,-1,0,0};

void dfs(int nx, int ny, int dir,int bu)
{

if (bu > 10) return;
if (nx < 0 || ny <0 || nx >= n || ny >=m) return;//出界，不合法
//cout<<nx<<" "<<ny<<" "<<dir<<" "<<bu<<endl;

if (nx == xend && ny == yend)
{
ans = min(ans, bu);
}

int i;

if (dir != -1)
{
int wx = nx + dx[dir];
int wy = ny + dy[dir];
if (g[wx][wy] == 0)
{
dfs(wx,wy,dir,bu);
}
else
{
g[wx][wy] = 0;
dfs(nx,ny,-1,bu);
g[wx][wy] = 1;
}
return;
}

REP(i,4)
{
int wx = nx + dx[i];
int wy = ny + dy[i];
if (g[wx][wy]==1) continue;
dfs(wx,wy, i, bu + 1);
}
}

void doit()
{
ans = 0x7fffffff;
//cout << xbegin<<" "<<ybegin<<" "<<xend<<" "<<yend<<endl;
g[xbegin][ybegin] = g[xend][yend] = 0;
dfs(xbegin,ybegin,-1,0);
if (ans == 0x7fffffff) ans = -1;
printf("%d\n", ans);
}

int main()
{

while (~scanf("%d %d",&n, &m))
{
double begin = clock();
if (!n && !m) break;
init();
doit();
//cout << (clock()-begin)/1000 << endl;
}

}

/*
6 6
1 0 0 2 1 0
1 1 0 0 0 0
0 0 0 0 0 3
0 0 0 0 0 0
1 0 0 0 0 1
0 1 1 1 1 1

*/