2021.1.23 周练-解题报告

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A-Tempter of the Bone

详细题解分析：HDU1010 Tempter of the Bone 深搜DFS-题解

#include 
#define xa x + a[i]
#define yb y + b[i]
using namespace std;
char g[8][8];
int a[] = {0, 0, 1, -1};
int b[] = {-1, 1, 0, 0};
int n, m, t, step, remain, destx, desty;
bool flag;
void dfs(int x, int y, int tc)
{
    if (tc == t && x == destx && y == desty){
        flag = true;
        return;
    }
    int v = t - tc - abs(destx - x) - abs(desty - y); 
    if ((abs(destx - x) + abs(desty - y)) % 2 != (t - tc) % 2)  return;                
    for (int i = 0; i = 0 && xa = 0 && yb > n >> m >> t){
        if (n == 0 && m == 0 && t == 0) break;
        step = remain = 0;
        flag = false;
        int icur, jcur;
        for (int i = 0; i  g[i][j];
                if (g[i][j] == 'S') icur = i, jcur = j;
                else if (g[i][j] == 'D') remain++, destx = i, desty = j;
                else if (g[i][j] == '.') remain++;
            }
        g[icur][jcur] = 'X';
        if (remain >= t)  dfs(icur, jcur, 0);
        if (flag) cout > q){
        memset(g, 0, sizeof(g));
        ans = 0;
        for(int i = 0; i > x >> y;
            g[x][y] = 1;
        }
        dfs(0, 0, 0); 
        cout > g[1];
        flag = false;
        if(g[0][0] == '1' || g[0][0] == '2') dfs(0, 0, 'r');
        else dfs(0, 0, 'd');
        if(flag) cout