UVA 10564 Paths through the Hourglass

UVA_10564

    我们可以用f[i][j][k]表示到第i行第j个格子时路径上值得和是否可以为k，为了保证记录的路径字典序最小的，我们可以将i由大向小循环。

    状态转移的思路还是很好想的，但是写起来有些蛋疼。

#include<stdio.h>
#include<string.h>
#define MAXD 50
#define MAXS 520
int N, S, f[MAXD][MAXD][MAXS], p[MAXD][MAXD][MAXS], num[MAXD], a[MAXD][MAXD];
long long int d[MAXD][MAXD][MAXS];
int init()
{
    int i, j, k;
    scanf("%d%d", &N, &S);
    if(!N && !S)
        return 0;
    for(i = 0; i < N; i ++)
        for(j = 0; j < N - i; j ++)
        {
            scanf("%d", &a[i][j]);
            num[i] = N - i;
        }
    for(; i < 2 * N - 1; i ++)
        for(j = 0; j < i - N + 2; j ++)
        {
            scanf("%d", &a[i][j]);
            num[i] = i - N + 2;
        }
    return 1;
}
void printpath(int k)
{
    int i, j, t;
    printf("%d ", k);
    for(i = 0, j = k, t = S; i < N - 1; i ++)
    {
        if(p[i][j][t] == 0)
        {
            printf("L");
            t -= a[i][j --];
        }
        else
        {
            printf("R");
            t -= a[i][j];
        }
    }
    for(; i < 2 * N - 2; i ++)
    {
        if(p[i][j][t] == 0)
        {
            printf("L");
            t -= a[i][j];
        }
        else
        {
            printf("R");
            t -= a[i][j ++];
        }
    }
}
void solve()
{
    int i, j, k;
    long long int res;
    memset(p, -1, sizeof(p));
    memset(f, 0, sizeof(f));
    memset(d, 0, sizeof(d));
    for(i = 2 * N - 2, j = 0; j < num[i]; j ++)
    {
        f[i][j][a[i][j]] = 1;
        d[i][j][a[i][j]] = 1;
    }
    for(i = 2 * N - 3; i >= N - 1; i --)
        for(j = 0; j < num[i]; j ++)
        {
            for(k = 0; k <= S; k ++)
                if(f[i + 1][j][k])
                {
                    f[i][j][k + a[i][j]] = 1;
                    d[i][j][k + a[i][j]] += d[i + 1][j][k];
                    p[i][j][k + a[i][j]] = 0;
                }
            for(k = 0; k <= S; k ++)
                if(f[i + 1][j + 1][k])
                {
                    d[i][j][k + a[i][j]] += d[i + 1][j + 1][k];
                    if(!f[i][j][k + a[i][j]])
                    {
                        f[i][j][k + a[i][j]] = 1;
                        p[i][j][k + a[i][j]] = 1;
                    }
                }
        }
    for(i = N - 2; i >= 0; i --)
        for(j = 0; j < num[i]; j ++)
        {
            if(j)
                for(k = 0; k <= S; k ++)
                    if(f[i + 1][j - 1][k])
                    {
                        f[i][j][k + a[i][j]] = 1;
                        d[i][j][k + a[i][j]] += d[i + 1][j - 1][k];
                        p[i][j][k + a[i][j]] = 0;
                    }
            if(j < num[i + 1])
                for(k = 0; k <= S; k ++)
                    if(f[i + 1][j][k])
                    {
                        d[i][j][k + a[i][j]] += d[i + 1][j][k];
                        if(!f[i][j][k + a[i][j]])
                        {
                            f[i][j][k + a[i][j]] = 1;
                            p[i][j][k + a[i][j]] = 1;
                        }
                    }
        }
    res = 0;
    for(i = 0; i < num[0]; i ++)
        if(f[0][i][S])
        {
            if(!res)
                k = i;
            res += d[0][i][S];
        }
    printf("%lld\n", res);
    if(res)
        printpath(k);
    printf("\n");
}
int main()
{
    while(init())
        solve();
    return 0;
}