【2005-2006 ACM-ICPC, NEERC, Moscow Subregional Contest】Problem J. Jack-pot

简单dfs，差分一下A数组和建出字典树能写得更方便，若不这么做代码时就会像我一样难受。

#include
#include
#include
using namespace std;
typedef long long ll;

const int N = 100003;

int a[13][N], n, m, k, A[13], pos[13][N];

ll dfs(int tmp, int l, int r, int f) {
    if (tmp > m) {
        return 1ll * (r - l + 1) * A[m];
    }
    
    int und = l;
    ll ans = 1000000000000000ll;
    if (r - l + 1 < k || (a[tmp][l] != 0) || (a[tmp][r] != k - 1)) {
        ans = 1ll * A[tmp - 1] * (r - l + 1);
        if (a[tmp][l] != 0) pos[tmp - 1][f] = 0;
        if (a[tmp][r] != k - 1) pos[tmp - 1][f] = -(k - 1);
    }
    for (int i = l + 1; i <= r; ++i)
        if (a[tmp][i] != a[tmp][i - 1] && a[tmp][i] != a[tmp][i - 1] + 1) {
            ans = 1ll * A[tmp - 1] * (r - l + 1);
            pos[tmp - 1][f] = -(a[tmp][i - 1] + 1);
            break;
        }
    
    for (int i = l + 1; i <= r + 1; ++i)
        if (i > r || a[tmp][i] != a[tmp][und]) {
            int num = dfs(tmp + 1, und, i - 1, und) + 1ll * A[tmp - 1] * (r - l + 1 - (i - und));
            if (num < ans) {
                ans = num;
                pos[tmp - 1][f] = und;
            }
            
            und = i;
        }
    return ans;
}

char s[13];

int main() {
    scanf("%d%d%d", &n, &m, &k);
    for (int i = 1; i <= m; ++i) scanf("%d", &A[i]);
    for (int i = 1; i <= n; ++i) {
        scanf("%s", s + 1);
        for (int j = 1; j <= m; ++j)
            a[j][i] = s[j] - '0';
    }
    ll ans = dfs(1, 1, n, 1); int tmp = 1;
    for (int i = 1; i <= m; ++i) {
        tmp = pos[i - 1][tmp];
        if (tmp <= 0) {
            printf("%d", -tmp);
            for (int j = i + 1; j <= m; ++j) putchar('0');
            break;
        }
        printf("%d", a[i][tmp]);
    }
    printf("\n%lld\n", ans);
    return 0;
}

转载于:https://www.cnblogs.com/abclzr/p/9210699.html