UVA 10564 Paths through the Hourglass[DP 打印]

UVA - 10564

Paths through the Hourglass

题意：

要求从第一层走到最下面一层，只能往左下或右下走

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。

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f[i][j][k]从下往上到第i层第j个和为k的方案数

上下转移不一样，分开处理

没必要判断走出沙漏

打印方案倒着找下去行了，尽量往左走

 

沙茶的忘注释掉文件WA好多次

 

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int N=,M=,INF=1e9;
typedef long long ll;
inline int read(){
    char c=getchar();int x=,f=;
    while(c<''||c>''){if(c=='-')f=-;c=getchar();}
    while(c>=''&&c<=''){x=x*+c-'';c=getchar();}
    return x*f;
}
int n,s,w[N<<][N];
ll f[N<<][N][M];
void dp(){
    memset(f,,sizeof(f));
    for(int j=;j<=n;j++)
        f[n+n-][j][w[n+n-][j]]=;

    //xia
    for(int i=n+n-;i>=n;i--)
        for(int j=;j<=i-n+;j++)
            for(int k=w[i][j];k<=s;k++)
                f[i][j][k]=f[i+][j][k-w[i][j]]+f[i+][j+][k-w[i][j]];
    //shang
    for(int i=n-;i>=;i--)
        for(int j=;j<=n-i+;j++)
            for(int k=w[i][j];k<=s;k++)
                f[i][j][k]=f[i+][j][k-w[i][j]]+f[i+][j-][k-w[i][j]];

}
void print(int i,int j,ll k){//printf("print %d %d %d\n",i,j,k);
    if(i==n+n-) return;
    if(i<n){
        if(j>&&f[i+][j-][k-w[i][j]]){
            putchar('L');
            print(i+,j-,k-w[i][j]);
        }else{
            putchar('R');
            print(i+,j,k-w[i][j]);
        }
    }else{
        if(f[i+][j][k-w[i][j]]){
            putchar('L');
            print(i+,j,k-w[i][j]);
        }else{
            putchar('R');
            print(i+,j+,k-w[i][j]);
        }
    }
}
int main(){
    //freopen("1.in","r",stdin);
    //freopen("1.out","w",stdout);
    while(scanf("%d%d",&n,&s)!=EOF&&(n||s)){
        memset(w,,sizeof(w));
        for(int i=;i<=n;i++)
            for(int j=;j<=n-i+;j++) w[i][j]=read();
        for(int i=n+;i<=n+n-;i++)
            for(int j=;j<=i-n+;j++) w[i][j]=read();
        dp();
        int p=;
        ll sum=;
        for(int i=n;i>=;i--) if(f[][i][s]){p=i;sum+=f[][i][s];}

        printf("%lld\n",sum);
        if(p) printf("%d ",p-),print(,p,s);
        putchar('\n');
    }
}