UVA 10564 Paths through the Hourglass[DP 打印]

Posted 2020-08-17 Candy?

 

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=25,M=505,INF=1e9;
typedef long long ll;
inline int read(){
    char c=getchar();int x=0,f=1;
    while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1;c=getchar();}
    while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘;c=getchar();}
    return x*f;
}
int n,s,w[N<<1][N];
ll f[N<<1][N][M];
void dp(){
    memset(f,0,sizeof(f));
    for(int j=1;j<=n;j++)
        f[n+n-1][j][w[n+n-1][j]]=1;
    
    //xia
    for(int i=n+n-2;i>=n;i--)
        for(int j=1;j<=i-n+1;j++)
            for(int k=w[i][j];k<=s;k++)
                f[i][j][k]=f[i+1][j][k-w[i][j]]+f[i+1][j+1][k-w[i][j]];
    //shang
    for(int i=n-1;i>=1;i--)
        for(int j=1;j<=n-i+1;j++)
            for(int k=w[i][j];k<=s;k++)
                f[i][j][k]=f[i+1][j][k-w[i][j]]+f[i+1][j-1][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-1) return;
    if(i<n){
        if(j>1&&f[i+1][j-1][k-w[i][j]]){
            putchar(‘L‘);
            print(i+1,j-1,k-w[i][j]);
        }else{
            putchar(‘R‘);
            print(i+1,j,k-w[i][j]);
        }
    }else{
        if(f[i+1][j][k-w[i][j]]){
            putchar(‘L‘);
            print(i+1,j,k-w[i][j]);
        }else{
            putchar(‘R‘);
            print(i+1,j+1,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,0,sizeof(w));
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n-i+1;j++) w[i][j]=read();
        for(int i=n+1;i<=n+n-1;i++)
            for(int j=1;j<=i-n+1;j++) w[i][j]=read();
        dp();
        int p=0;
        ll sum=0;
        for(int i=n;i>=1;i--) if(f[1][i][s]){p=i;sum+=f[1][i][s];}
        
        printf("%lld\n",sum);
        if(p) printf("%d ",p-1),print(1,p,s);
        putchar(‘\n‘);
    }
}