01背包(类) UVA 10564 Paths through the Hourglass

题目传送门

 /*
     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])
             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
             因为要输出字典序最小的，打印路径时先考虑L
 */
 /************************************************
  * Author        :Running_Time
  * Created Time  :2015-8-9 15:45:11
  * File Name     :UVA_10564.cpp
  ************************************************/

 #include <cstdio>
 #include <algorithm>
 #include <iostream>
 #include <sstream>
 #include <cstring>
 #include <cmath>
 #include <string>
 #include <vector>
 #include <queue>
 #include <deque>
 #include <stack>
 #include <list>
 #include <map>
 #include <set>
 #include <bitset>
 #include <cstdlib>
 #include <ctime>
 using namespace std;

 #define lson l, mid, rt << 1
 #define rson mid + 1, r, rt << 1 | 1
 typedef long long ll;
 const int MAXN = ;
 const int MAXS = 5e2 + ;
 const int INF = 0x3f3f3f3f;
 const int MOD = 1e9 + ;
 int a[MAXN*][MAXN];
 ll dp[MAXN*][MAXN][MAXS];
 int n, s;

 void print(int x, int y, int sum)    {
     if (x >=  * n - ) return ;
     int v = a[x][y];
     if (x < n)  {
         if (y >  && dp[x+][y-][sum-v])   {
             printf ("L");  print (x+, y-, sum - v);
         }
         else    {
             printf ("R");  print (x+, y, sum - v);
         }
     }
     else    {
         if (dp[x+][y][sum-v])    {
             printf ("L");  print (x+, y, sum - v);
         }
         else    {
             printf ("R");  print (x+, y+, sum - v);
         }
     }
 }

 int main(void)    {     //UVA 10564    Paths through the Hourglass
     while (scanf ("%d%d", &n, &s) == ) {
         if (!n && !s)   break;
         for (int i=; i<=n; ++i)    {
             for (int j=; j<=n-i+; ++j)    scanf ("%d", &a[i][j]);
         }
         for (int i=n+; i<=*n-; ++i)  {
             for (int j=; j<=i-n+; ++j)    scanf ("%d", &a[i][j]);
         }

         memset (dp, , sizeof (dp));
         for (int i=; i<=n; ++i)    {
             int c = a[*n-][i];
             dp[*n-][i][c] = ;
         }
         for (int i=*n-; i>=n; --i)  {
             for (int j=; j<=i-n+; ++j)    {
                 int c = a[i][j];
                 for (int k=c; k<=s; ++k)  {
                     dp[i][j][k] = dp[i+][j][k-c] + dp[i+][j+][k-c];
                 }
             }
         }
         ll cnt = ;
         for (int i=n-; i>=; --i)    {
             for (int j=; j<=n-i+; ++j)    {
                 int c = a[i][j];
                 for (int k=c; k<=s; ++k)    {
                     if (j > )  dp[i][j][k] += dp[i+][j-][k-c];
                     if (j < n - i + )  dp[i][j][k] += dp[i+][j][k-c];
                 }
                 if (i == ) cnt += dp[i][j][s];
             }
         }
         printf ("%lld\n", cnt);
         for (int i=; i<=n; ++i)    {
             if (dp[][i][s])    {
                 printf ("%d ", i-);
                 print (, i, s);    break;
             }
         }
         puts ("");
     }

     return ;
 }