UVA 10564 Paths through the Hourglass

练习打印路径。。本题居然没有PE，搞死我了。。

后来还是参考了一位名为“加速！！！！！！！！！！”的好人的BLOG，自己改了半天才过了。

思路就是dp[i][j][k]代表从(i,j)出发，到达最底层的权值之和为k的路径条数，细节方面小心点。哎~~

贴代码：

View Code 
  1 #include <iostream>
  2 #include <cstring>
  3 #include <cstdio>
  4 using namespace std;
  5 typedef long long int llg;
  6 llg dp[50][50][501];
  7 int left1[50][50];
  8 int right1[50][50];
  9 int h[50][50];
 10 char answer[50];
 11 int nlines;
 12 bool ok(int x,int y)
 13 {
 14     return y > 0 && y <= h[x][0];
 15 }
 16 llg DP(int i,int j,int k)
 17 {
 18     llg &ans = dp[i][j][k];
 19     if(ans != -1)
 20         return ans;
 21     ans = 0;
 22     int x,y;
 23     x = i + 1;
 24     y = left1[i][j];
 25     if(ok(x,y))
 26     {
 27         dp[i][j][k] += DP(x,y,k - h[i][j]);
 28     }
 29     y = right1[i][j];
 30     if(ok(x,y))
 31     {
 32         dp[i][j][k] += DP(x,y,k - h[i][j]);
 33     }
 34     return ans;
 35 }
 36 void printans(int i,int j,int k)
 37 {
 38     if(i == nlines * 2 - 2)
 39         return;
 40     bool flag = false;
 41     int x,y;
 42     x = i + 1;
 43     y = left1[i][j];
 44     if(ok(x,y) && dp[x][y][k - h[i][j]] > 0)
 45     {
 46         answer[i] = 'L';
 47         flag = true;
 48         printans(x,y,k - h[i][j]);
 49     }
 50     x = i + 1;
 51     y = right1[i][j];
 52     if(!flag && ok(x,y) && dp[x][y][k - h[i][j]] > 0)
 53     {
 54         answer[i] = 'R';
 55         printans(x,y,k - h[i][j]);
 56     }
 57 }
 58 int main()
 59 {
 60     int num,i,j,temp;
 61     llg total;
 62     while(cin>>nlines>>num)
 63     {
 64         if(nlines == 0 && num == 0)
 65             break;
 66         for(i = 0;i < nlines;i++)
 67         {
 68             temp = nlines - i;//本行数字的个数
 69             h[i][0] = temp;
 70             for(j = 1;j <= h[i][0];j++)
 71             {
 72                 left1[i][j] = j - 1;
 73                 right1[i][j] = j;
 74                 cin>>h[i][j];
 75             }
 76         }
 77         left1[nlines - 1][1] = 1;
 78         right1[nlines - 1][1] = 2;
 79         for(i = nlines;i < nlines * 2 - 1;i++)
 80         {
 81             temp = 2 + i - nlines;
 82             h[i][0] = temp;
 83             for(j = 1;j <= h[i][0];j++)
 84             {
 85                 left1[i][j] = j;
 86                 right1[i][j] = j + 1;
 87                 cin>>h[i][j];
 88             }
 89         }
 90         memset(dp,-1,sizeof(dp));
 91         for(i = 1;i <= h[nlines * 2 - 2][0];i++)
 92         {
 93             temp = h[nlines * 2 - 2][i];
 94             dp[nlines * 2 - 2][i][temp] = 1;
 95         }
 96         int start = -1;
 97         total = 0;
 98         for(i = 1;i <= h[0][0];i++)
 99         {
100             total += DP(0,i,num);
101             if(start == -1 && dp[0][i][num] > 0)
102             {
103                 start = i;
104                 printans(0,i,num);
105             }
106         }
107         if(total)
108         {
109             cout<<total<<endl;
110             cout<<start - 1<<" ";
111             for(i = 0;i < 2 * nlines - 2;i++)
112                 cout<<answer[i];
113             cout<<endl;
114         }
115         else
116         {
117             cout<<"0"<<endl;
118             cout<<endl;
119         }
120     }
121     return 0;
122 }

 

posted @ 2012-04-30 23:05  浙西贫农  阅读(231)  评论(0)  编辑  收藏  举报