题目传送门

题意：

给一个相上面的图。要求从第一层走到最下面一层，只能往左下或右下走，经过的数字之和为sum。

问有多少条路径之和刚好等于S？ 如果有的话，输出字典序最小的路径。

思路：

f[i][j][k] 代表从(i,j)点往下走到最后一层和为k的方案数
那么，显然可以得到状态转移：
f[i][j][k] = f[i+1][left][k-val] + f[i+1][right][k-val],  val=(i,j)格上的数字，left是往坐下走的坐标，right往右下走的坐标

代码：

/**==========================================*   This is a solution for ACM/ICPC problem**   @author: shuangde*   @blog: blog.csdn.net/shuangde800*   @email: zengshuangde@gmail.com*===========================================*/#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
#include<cmath>
#include<cstring>
using namespace std;typedef long long int64;
const int INF = 0x3f3f3f3f;
const double PI  = acos(-1.0);int n, s;
int hourGlass[50][22];
int64 f[50][22][510];void input(){for(int i=1; i<=n; ++i)for(int j=1; j<=n-i+1; ++j) scanf("%d", &hourGlass[i][j]);for(int i=n+1; i<=2*n-1; ++i)for(int j=1; j<=i+1-n; ++j)scanf("%d", &hourGlass[i][j]); }void print_path(int i, int j, int sum){if(i >= 2*n-1) return;int val = hourGlass[i][j];if(i<n){ if(j>1 && f[i+1][j-1][sum-val]){printf("L");print_path(i+1, j-1, sum-val);return ;} printf("R");print_path(i+1, j, sum-val);}else{if(f[i+1][j][sum-val]){printf("L"); print_path(i+1, j, sum-val);return;} printf("R"); print_path(i+1, j+1, sum-val);}
}int main(){while(~scanf("%d%d", &n, &s) && n+s){input();memset(f, 0, sizeof(f));// 初始化最下面一行for(int i=1; i<=n; ++i)f[2*n-1][i][hourGlass[2*n-1][i]] = 1;// 下半部分dpfor(int i=2*n-2; i>=n; --i){for(int j=1; j<=i+1-n; ++j){for(int v=hourGlass[i][j]; v<=s; ++v){int w = hourGlass[i][j];f[i][j][v] = f[i+1][j][v-w] + f[i+1][j+1][v-w]; }} }// 上半部分dpint64 ans = 0;for(int i=n-1; i>=1; --i){for(int j=1; j<=n-i+1; ++j){for(int v=hourGlass[i][j]; v<=s; ++v){int w = hourGlass[i][j];if(j>1) f[i][j][v] += f[i+1][j-1][v-w];if(j<n-i+1) f[i][j][v] += f[i+1][j][v-w];}if(i==1) ans += f[1][j][s];} }cout << ans << endl;for(int i=1; i<=n; ++i){if(f[1][i][s]){printf("%d ", i-1); print_path(1, i, s);break;}}puts("");}return 0;
}