UVA 10564 Paths through the Hourglass .

#include <bits/stdc++.h>
using namespace std;
#define REP(i,a,b)  for(int i=a;i<=(int)(b);++i)
#define REPD(i,a,b) for(int i=a;i>=(int)(b);--i)
long long d[50][25][500];
int a[50][25],rest,n;
void DFS1(int i,int j,int w){
if(i==n) {rest=w; return;}
if(j>1&&d[i+1][j-1][w-a[i][j]]) {
printf("L");
DFS1(i+1,j-1,w-a[i][j]);
}
else if(j<=n-i&&d[i+1][j][w-a[i][j]]) {
printf("R");
DFS1(i+1,j,w-a[i][j]);
}
}
void DFS2(int i,int j,int w){
if(i==2*n-1) return;
if(d[i+1][j][w-a[i][j]]) {
printf("L");
DFS2(i+1,j,w-a[i][j]);
}
else if(d[i+1][j+1][w-a[i][j]]) {
printf("R");
DFS2(i+1,j+1,w-a[i][j]);
}
}
int main(int argc, char const *argv[])
{
int S;
while(scanf("%d%d",&n,&S)==2&&n+S){
REP(i,1,n)       REP(j,1,n-i+1) scanf("%d",&a[i][j]);
REP(i,n+1,n+n-1) REP(j,1,i-n+1)   scanf("%d",&a[i][j]);

memset(d,0,sizeof(d));
REP(i,1,n) d[2*n-1][i][a[2*n-1][i]]=1;
REPD(i,2*n-2,n) REP(j,1,i-n+1) {
int w=a[i][j];
REP(k,w,S) {
d[i][j][k]+=d[i+1][j][k-w];
d[i][j][k]+=d[i+1][j+1][k-w];
}
}

long long ans=0;
REPD(i,n-1,1) REP(j,1,n-i+1) {
int w=a[i][j];
REP(k,w,S) {
if(j>1) d[i][j][k]+=d[i+1][j-1][k-w];
if(j<=n-i) d[i][j][k]+=d[i+1][j][k-w];
}
if(i==1) ans+=d[i][j][S];
}
printf("%lld\n", ans);
REP(i,1,n) if(d[1][i][S]) {
printf("%d ", i-1);
DFS1(1,i,S); DFS2(n,1,rest);
break;
}
printf("\n");
}
return 0;
}