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