UVA 10564 Paths through the Hourglass

练习打印路径。。本题居然没有PE，搞死我了。。

后来还是参考了一位名为“加速！！！！！！！！！！”的好人的BLOG，自己改了半天才过了。

思路就是dp[i][j][k]代表从(i,j)出发，到达最底层的权值之和为k的路径条数，细节方面小心点。哎~~

贴代码：

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
typedef long long int llg;
llg dp[50][50][501];
int left1[50][50];
int right1[50][50];
int h[50][50];
char answer[50];
int nlines;
bool ok(int x,int y)
{
    return y > 0 && y <= h[x][0];
}
llg DP(int i,int j,int k)
{
    llg &ans = dp[i][j][k];
    if(ans != -1)
        return ans;
    ans = 0;
    int x,y;
    x = i + 1;
    y = left1[i][j];
    if(ok(x,y))
    {
        dp[i][j][k] += DP(x,y,k - h[i][j]);
    }
    y = right1[i][j];
    if(ok(x,y))
    {
        dp[i][j][k] += DP(x,y,k - h[i][j]);
    }
    return ans;
}
void printans(int i,int j,int k)
{
    if(i == nlines * 2 - 2)
        return;
    bool flag = false;
    int x,y;
    x = i + 1;
    y = left1[i][j];
    if(ok(x,y) && dp[x][y][k - h[i][j]] > 0)
    {
        answer[i] = 'L';
        flag = true;
        printans(x,y,k - h[i][j]);
    }
    x = i + 1;
    y = right1[i][j];
    if(!flag && ok(x,y) && dp[x][y][k - h[i][j]] > 0)
    {
        answer[i] = 'R';
        printans(x,y,k - h[i][j]);
    }
}
int main()
{
    int num,i,j,temp;
    llg total;
    while(cin>>nlines>>num)
    {
        if(nlines == 0 && num == 0)
            break;
        for(i = 0;i < nlines;i++)
        {
            temp = nlines - i;//本行数字的个数
            h[i][0] = temp;
            for(j = 1;j <= h[i][0];j++)
            {
                left1[i][j] = j - 1;
                right1[i][j] = j;
                cin>>h[i][j];
            }
        }
        left1[nlines - 1][1] = 1;
        right1[nlines - 1][1] = 2;
        for(i = nlines;i < nlines * 2 - 1;i++)
        {
            temp = 2 + i - nlines;
            h[i][0] = temp;
            for(j = 1;j <= h[i][0];j++)
            {
                left1[i][j] = j;
                right1[i][j] = j + 1;
                cin>>h[i][j];
            }
        }
        memset(dp,-1,sizeof(dp));
        for(i = 1;i <= h[nlines * 2 - 2][0];i++)
        {
            temp = h[nlines * 2 - 2][i];
            dp[nlines * 2 - 2][i][temp] = 1;
        }
        int start = -1;
        total = 0;
        for(i = 1;i <= h[0][0];i++)
        {
            total += DP(0,i,num);
            if(start == -1 && dp[0][i][num] > 0)
            {
                start = i;
                printans(0,i,num);
            }
        }
        if(total)
        {
            cout<<total<<endl;
            cout<<start - 1<<" ";
            for(i = 0;i < 2 * nlines - 2;i++)
                cout<<answer[i];
            cout<<endl;
        }
        else
        {
            cout<<"0"<<endl;
            cout<<endl;
        }
    }
    return 0;
}