UVA 10564 Paths through the Hourglass DP

注意用long long。

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<sstream>
#include<cmath>
#include<climits>
#include<string>
#include<map>
#include<queue>
#include<vector>
#include<stack>
#include<set>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
#define pb(a) push_back(a)
#define INF 0x1f1f1f1f
#define lson idx<<1,l,mid
#define rson idx<<1|1,mid+1,r
#define PI  3.1415926535898
template<class T> T min(const T& a,const T& b,const T& c) {
    return min(min(a,b),min(a,c));
}
template<class T> T max(const T& a,const T& b,const T& c) {
    return max(max(a,b),max(a,c));
}
void debug() {
#ifdef ONLINE_JUDGE
#else

    freopen("d:\in.txt","r",stdin);
    freopen("d:\out1.txt","w",stdout);
#endif
}
int getch() {
    int ch;
    while((ch=getchar())!=EOF) {
        if(ch!=' '&&ch!='
')return ch;
    }
    return EOF;
}
int n,s;
int da[50][50];
ll dp[50][50][550];
ll f(int k,int pos,int s)
{
    if(dp[k][pos][s]>=0)return dp[k][pos][s];
    if(k==2*n-1)return dp[k][pos][s]=s==da[k][pos]?1:0;
    ll x=0;
    if(k<n)
    {
        if(pos!=1&&s>=da[k][pos])x+=f(k+1,pos-1,s-da[k][pos]);
        if(pos!=abs(k-n)+1&&s>=da[k][pos])x+=f(k+1,pos,s-da[k][pos]);
    }else
    {
        if(s>=da[k][pos])x+=f(k+1,pos,s-da[k][pos]);
        if(s>=da[k][pos])x+=f(k+1,pos+1,s-da[k][pos]);
    }
    return dp[k][pos][s]=x;
}
int main()
{
   // freopen("d:\in.txt","r",stdin);
    while(scanf("%d%d",&n,&s)!=EOF&&(n||s))
    {
        for(int i=1;i<=2*n-1;i++)
        {
            for(int j=1;j<=abs(i-n)+1;j++)
                scanf("%d",&da[i][j]);
        }
        ll num=0;
        memset(dp,-1,sizeof(dp));
        for(int i=1;i<=n;i++)
        {
            num+=f(1,i,s);
        }
        printf("%lld
",num);
        if(num==0)
            printf("
");
        else
        {
            for(int i=1;i<=n;i++)if(dp[1][i][s]>0)
            {
                printf("%d ",i-1);
                int pos=i;
                int ns=s;
                for(int k=1;k<2*n-1;k++)
                {
                    if(k<n)
                    {
                        if(pos!=1&&dp[k+1][pos-1][ns-da[k][pos]]>0)
                        {
                            printf("L");
                            ns-=da[k][pos];--pos;
                        }else
                        {
                            printf("R");
                            ns-=da[k][pos];
                        }
                    }else
                    {
                        if(dp[k+1][pos][ns-da[k][pos]]>0)
                        {
                            printf("L");
                            ns-=da[k][pos];
                        }else
                        {
                            printf("R");
                            ns-=da[k][pos];pos++;
                        }
                    }
                }
                break;
            }
            printf("
");
        }
    }
    return 0;
}