ZOJ 2110 DFS

狗要出门，且正好在T秒

就是DFS + 剪枝， 联系一下剪枝技巧

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int step[][2] = {0,1,0,-1,1,0,-1,0};
const int maxn = 10 + 7;

char Map[maxn][maxn];
int m,n,t;
int aimx,aimy;
int bex,bey;

bool DFS(int x,int y,int tt)
{
    if(x == 0 || x == m+1 || y == 0 || y == n+1) return false;
    if(x == aimx && y == aimy && tt == t) return true;
    int Dis = ((t - tt) - abs(x-aimx) - abs(y -aimy));
    if(Dis % 2 || Dis < 0)  return false;
    for(int i = 0; i <4; ++i)
    {
        if(Map[x+step[i][0]][y+step[i][1]] != ‘X‘)
        {
            Map[x+step[i][0]][y+step[i][1]] = ‘X‘;
            //cout << " X : " << x << "  Y : " << y << endl;
            if(DFS(x+step[i][0], y+step[i][1], tt+1)) return true;
            Map[x+step[i][0]][y+step[i][1]] = ‘.‘;
        }
    }
    return false;
}

int main()
{
    while(cin >> m >> n >> t && (n + m + t))
    {
        for(int i = 0; i < maxn; ++i)
            for(int j = 0; j < maxn; ++j) Map[i][j] = ‘X‘;
        int cnt = 0;
        for(int i = 1; i <= m; ++i)
        {
            scanf("%s",Map[i] + 1);
            for(int j = 1; j <= n; ++j) if(Map[i][j] == ‘S‘) bex = i, bey = j;
            else if(Map[i][j] == ‘D‘) aimx = i, aimy = j;
            else if(Map[i][j] == ‘X‘) cnt++;
        }
        if(m*n - cnt < t) {
            printf("NO\n");
            continue;
        }
        Map[bex][bey] = ‘X‘;
        if(DFS(bex,bey,0)) {
            printf("YES\n");
        } else printf("NO\n");
    }
    return 0;
}