UVA 10564 Paths through the Hourglass（背包）

为了方便打印路径，考虑从下往上转移。dp[i][j][S]表示在i行j列总和为S的方案，

dp[i][j][S] = dp[i+1][left][S-x]+dp[i+1][right][S-x]

方案O(2^2*n-1)，结果要用long long保存。

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
const int maxn = ,maxs = ;
int hg[maxn<<][maxn];
ll dp[maxn<<][maxn][maxs];

//#define LOCAL
int main()
{
#ifdef LOCAL
    freopen("in.txt","r",stdin);
#endif
    int n, S;
    while(scanf("%d%d", &n, &S) ,n){
        int rc = ;
        for(int i = n; i > ; i--){
            int *h = hg[rc++];
            for(int j = ; j <= i; j++){
                scanf("%d",h+j);
            }
        }
        for(int i = ; i <= n; i++){
            int *h = hg[rc++];
            for(int j = ; j <= i; j++){
                scanf("%d",h+j);
            }
        }
        rc--;
        memset(dp,,sizeof(dp));
        for(int j = ; j <= n; j++){
            dp[rc][j][hg[rc][j]] = ;
        }
        while(rc>=n){
            for(int j = , mj = (rc-- + ) - n; j <= mj; j++){
                int x = hg[rc][j];
                ll *f = dp[rc][j];
                for(int s = S-x; s >= ; s--){
                    f[s+x] = dp[rc+][j][s]+dp[rc+][j+][s];
                }
            }
        }
        for(int i = ; rc--,i <= n; i++){
            for(int j = ; j <= i; j++){
                int x = hg[rc][j];
                ll *f = dp[rc][j];
                for(int s = S-x; s >= ; s--){
                    f[s+x] = dp[rc+][j-][s]+dp[rc+][j][s];
                }
            }
        }
        ll ans = ;
        int index = ;
        for(int j = ; j <= n; j++){
            if(dp[][j][S]){
                ans += dp[][j][S];
                if(!index) index = j;
            }
        }
        printf("%lld\n",ans);
        if(ans){
            int j = index, s = S ;
            printf("%d ",j-);
            rc = n;
            for(int i = ; i < rc; i++){
                s -= hg[i-][j];
                if(dp[i][j-][s]){
                    putchar('L');
                    j--;
                }else {
                    putchar('R');
                }

            }
            rc = *n-;
            for(int i = n; i < rc; i++){
                s -= hg[i-][j];
                if(dp[i][j][s]){
                    putchar('L');
                }else {
                    j++;
                    putchar('R');
                }
            }
        }
        puts("");
    }
    return ;
}