hdu 5612 Baby Ming and Matrix games

Baby Ming and Matrix games

题意: 给一个矩形，两个0~9的数字之间隔一个数学运算符(‘+’,’-‘,’*’,’/’),其中’/’表示分数除，再给一个目标的值，问是否存在从一个数字出发，以数字之间的运算符为运算，得到这个目标值；(每个数字只能用一次，其实说白了就是dfs..);可以则输出(Impossible),否则输出(Possible);

思路:坑点就是里面本来全是整数，但是一个除法运算却是分数形式，开始使用了很保险的分数保存，来避免误差的。但是无情WA了很多次。。突然发现写成了(ImPossible)…醉了，然后以为现在肯定要A了，但是还是WA了。。。(有兴趣的帮我找找bug哈！真心感谢)之后重写了一个double类型的，无奈啊(太水了)，下面是A和WA两份代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string.h>
#include<algorithm>
#include<map>
#include<queue>
#include<vector>
#include<cmath>
#include<stdlib.h>
#include<time.h>
using namespace std;
#define rep(i,n) for(int i = 1;i <= n;i++)
#define MS0(a) memset(a,0,sizeof(a))
#define esp 1e-6
int n,m,vis[35][35],f;
double sum;
char s[35][35];
int dir[2][4] = {{0,1,0,-1},{1,0,-1,0}};
double cal(double val,double v,char op)
{
    if(op == '+') return val + v;
    else if(op == '-') return val - v;
    else if(op == '*') return val * v;
    return val/v;
}
void dfs(int i,int j,double val)
{
    if(fabs(val - sum) < esp) f = 0;
    for(int k = 0;k < 4 && f;k++){
        int x = i + 2*dir[0][k] , y = j + 2*dir[1][k];
        char op = s[i + dir[0][k]][j + dir[1][k]];
        if(x < 1 || x > n || y < 1 || y > m || vis[x][y]) continue;
        int v = s[x][y] - '0';
        if(op == '/' && v == 0) continue;
        vis[x][y] = 1;
        dfs(x,y,cal(val,v,op));
        vis[x][y] = 0;
    }
}
int main()
{
    int T,i,j;
    cin>>T;
    while(T--){
        f= 1;MS0(vis);
        scanf("%d%d%lf",&n,&m,&sum);
        rep(i,n) scanf("%s",s[i] + 1);
        for(i = 1;i <= n && f;i += 2)
        for(j = 1;j <= m && f;j += 2){
            vis[i][j] = 1;
            dfs(i,j,s[i][j] - '0');
            vis[i][j] = 0;
        }
        puts(f?"Impossible":"Possible");
    }
    return 0;
}

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string.h>
#include<algorithm>
#include<map>
#include<queue>
#include<vector>
#include<cmath>
#include<stdlib.h>
#include<time.h>
using namespace std;
typedef long long ll;
#define MS0(a) memset(a,0,sizeof(a))
ll cal(ll a,ll b,char op)
{
    if(op == '+') return a + b;
    if(op =='-') return a - b;
    if(op == '*') return a * b;
}
int dir[2][4] = {{0,1,0,-1},{1,0,-1,0}};
char s[100][100];
ll sum,f;
int n,m,vis[50][50];
void dfs(int i,int j,ll ele,ll den)
{
    if(ele%den == 0 && ele / den == sum) f = 0;
    ll e = ele, d = den;
    for(int k = 0;k < 4 && f;k++){
        //system("pause");
        int x = i + 2*dir[0][k],y = j + 2*dir[1][k];
        char op = s[i+dir[0][k]][j+dir[1][k]];
        if(x < 1 || x > n || y < 1|| y > m||vis[x][y]) continue;
        ele = e,den = d;
        int val = (s[x][y]-'0');
        if(op =='/'){
            if(val == 0) continue;
            if(ele % val) den *= val;
            else ele /= val;
        }
        else ele = cal(ele,val,op);
        if(ele % den == 0) ele /= den,den = 1;
        vis[x][y] = 1;
        dfs(x,y,ele,den);
        vis[x][y] = 0;
    }
}
int main()
{
    int T,i,j;
    cin>>T;
    while(T--){
        MS0(vis);
        scanf("%d%d%I64d",&n,&m,&sum);
        for(i = 1;i <= n;i++)
            scanf("%s",s[i] + 1);
        f = 1;
        for(i = 1;i <= n && f;i += 2)
            for(j = 1;j <= m && f;j += 2){
                vis[i][j] = 1;
                dfs(i,j,(ll)(s[i][j] - '0'),1);
                vis[i][j] = 0;
        }
        puts(f?"Impossible":"Possible");
    }
    return 0;
}