01背包(类) UVA 10564 Paths through the Hourglass

 

题目传送门

  1 /*
  2     01背包(类)：dp[i][j][k] 表示从(i, j)出发的和为k的方案数，那么cnt = sum (dp[1][i][s])
  3             状态转移方程：dp[i][j][k] = dp[i+1][j][k-c] + dp[i+1][j+1][k-c];(下半部分)    上半部分类似
  4             因为要输出字典序最小的，打印路径时先考虑L   
  5 */
  6 /************************************************
  7  * Author        :Running_Time
  8  * Created Time  :2015-8-9 15:45:11
  9  * File Name     :UVA_10564.cpp
 10  ************************************************/
 11 
 12 #include <cstdio>
 13 #include <algorithm>
 14 #include <iostream>
 15 #include <sstream>
 16 #include <cstring>
 17 #include <cmath>
 18 #include <string>
 19 #include <vector>
 20 #include <queue>
 21 #include <deque>
 22 #include <stack>
 23 #include <list>
 24 #include <map>
 25 #include <set>
 26 #include <bitset>
 27 #include <cstdlib>
 28 #include <ctime>
 29 using namespace std;
 30 
 31 #define lson l, mid, rt << 1
 32 #define rson mid + 1, r, rt << 1 | 1
 33 typedef long long ll;
 34 const int MAXN = 22;
 35 const int MAXS = 5e2 + 10;
 36 const int INF = 0x3f3f3f3f;
 37 const int MOD = 1e9 + 7;
 38 int a[MAXN*2][MAXN];
 39 ll dp[MAXN*2][MAXN][MAXS];
 40 int n, s;
 41 
 42 void print(int x, int y, int sum)    {
 43     if (x >= 2 * n - 1) return ;
 44     int v = a[x][y];
 45     if (x < n)  {
 46         if (y > 1 && dp[x+1][y-1][sum-v])   {
 47             printf ("L");  print (x+1, y-1, sum - v);
 48         }
 49         else    {
 50             printf ("R");  print (x+1, y, sum - v);
 51         }        
 52     }
 53     else    {
 54         if (dp[x+1][y][sum-v])    {
 55             printf ("L");  print (x+1, y, sum - v);
 56         }
 57         else    {
 58             printf ("R");  print (x+1, y+1, sum - v);
 59         }
 60     }
 61 }
 62 
 63 int main(void)    {     //UVA 10564    Paths through the Hourglass
 64     while (scanf ("%d%d", &n, &s) == 2) {
 65         if (!n && !s)   break;
 66         for (int i=1; i<=n; ++i)    {
 67             for (int j=1; j<=n-i+1; ++j)    scanf ("%d", &a[i][j]);
 68         }
 69         for (int i=n+1; i<=2*n-1; ++i)  {
 70             for (int j=1; j<=i-n+1; ++j)    scanf ("%d", &a[i][j]);
 71         }
 72 
 73         memset (dp, 0, sizeof (dp));
 74         for (int i=1; i<=n; ++i)    {
 75             int c = a[2*n-1][i];
 76             dp[2*n-1][i][c] = 1;
 77         }
 78         for (int i=2*n-2; i>=n; --i)  {
 79             for (int j=1; j<=i-n+1; ++j)    {
 80                 int c = a[i][j];
 81                 for (int k=c; k<=s; ++k)  {
 82                     dp[i][j][k] = dp[i+1][j][k-c] + dp[i+1][j+1][k-c];
 83                 }
 84             }
 85         }
 86         ll cnt = 0;
 87         for (int i=n-1; i>=1; --i)    {
 88             for (int j=1; j<=n-i+1; ++j)    {
 89                 int c = a[i][j];
 90                 for (int k=c; k<=s; ++k)    {
 91                     if (j > 1)  dp[i][j][k] += dp[i+1][j-1][k-c];
 92                     if (j < n - i + 1)  dp[i][j][k] += dp[i+1][j][k-c];
 93                 }
 94                 if (i == 1) cnt += dp[i][j][s];
 95             }
 96         }
 97         printf ("%lld\n", cnt);
 98         for (int i=1; i<=n; ++i)    {
 99             if (dp[1][i][s])    {
100                 printf ("%d ", i-1);
101                 print (1, i, s);    break;
102             }
103         }
104         puts ("");
105     }
106 
107     return 0;
108 }

 

posted @ 2015-08-09 16:48  Running_Time  阅读(378)  评论(0)  编辑  收藏  举报