HDU 2821 Pusher

dfs.枚舉每個點作爲起點，然後四個方向移動就行。先把字符串數組轉化爲整數，方便判斷，每次dfs前後，注意修改與恢復。

關於pile能否出界，網上很多人說什麼：不能推出邊界；還有人說什麼：如果只有一個，就可以推出界，多了就不行，等等。但題目中明確要求：

And if a whole pile is pushed outside the grid, it will be considered as cleared.

就是說可以推出界，不要人云亦云，更不要自己還沒弄明白就抄了份代碼發題解。。。

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<string>
#include<cmath>
#include<map>
///LOOP
#define REP(i, n) for(int i = 0; i < n; i++)
#define FF(i, a, b) for(int i = a; i < b; i++)
#define FFF(i, a, b) for(int i = a; i <= b; i++)
#define FD(i, a, b) for(int i = a - 1; i >= b; i--)
#define FDD(i, a, b) for(int i = a; i >= b; i--)
///INPUT
#define RI(n) scanf("%d", &n)
#define RII(n, m) scanf("%d%d", &n, &m)
#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)
#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)
#define RFI(n) scanf("%lf", &n)
#define RFII(n, m) scanf("%lf%lf", &n, &m)
#define RFIII(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define RFIV(n, m, k, p) scanf("%lf%lf%lf%lf", &n, &m, &k, &p)
#define RS(s) scanf("%s", s)
///OUTPUT
#define PN printf("\n")
#define PI(n) printf("%d\n", n)
#define PIS(n) printf("%d ", n)
#define PS(s) printf("%s\n", s)
#define PSS(s) printf("%s ", n)
#define PC(cas) printf("Game %d:\n", cas)
///OTHER
#define PB(x) push_back(x)
#define CLR(a, b) memset(a, b, sizeof(a))
#define CPY(a, b) memcpy(a, b, sizeof(b))
#define display(A, n, m) {REP(i, n){REP(j, m)PIS(A[i][j]);PN;}}

using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int MOD = 1000000;
const int INFI = 1e9 * 2;
const LL LINFI = 1e17;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int N = 55;
const int M = 11111;
const int move[8][2] = {0, 1, 0, -1, 1, 0, -1, 0, 1, 1, 1, -1, -1, 1, -1, -1};

int p[N][N], ans[M];
int n, m, num;
char s[N][N], a[5] = "RLDU";

bool isclear()
{
    REP(i, n)REP(j, m)if(p[i][j])return 0;
    return 1;
}

bool check(int x, int y)
{
    return x >= 0 && y >= 0 && x < n && y < m;
}

bool dfs(int x, int y, int k)
{
    if(isclear())return 1;
    int tx, ty;
    REP(i, 4)FF(j, 1, N)
    {
        tx = x + j * move[i][0];
        ty = y + j * move[i][1];
        if(!check(tx, ty))break;
        if(p[tx][ty])
        {
            if(j == 1)break;
            int x1 = tx + move[i][0], y1 = ty + move[i][1];
            int t1 = p[tx][ty], t2 = p[x1][y1];
            p[x1][y1] += t1 - 1;
            p[tx][ty] = 0;
            ans[k] = i;num = k + 1;
            if(dfs(tx, ty, k + 1))return 1;
            p[x1][y1] = t2;
            p[tx][ty] = t1;
            break;
        }
    }
    return 0;
}

int main()
{
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);

    while(RII(m, n) != EOF)
    {
        CLR(p, 0);
        CLR(ans, -1);
        REP(i, n)
        {
            RS(s[i]);
            REP(j, m)
            {
                if(s[i][j] == '.')p[i][j] = 0;
                else p[i][j] = s[i][j] - 'a' + 1;
            }
        }
        REP(i, n)REP(j, m)if(!p[i][j])
        {
            if(dfs(i, j, 0))
            {
                PI(i);PI(j);
                REP(k, num)putchar(a[ans[k]]);puts("");
                goto end;
            }
        }
        end: ;
    }
    return 0;
}