UVa 10564 - Paths Through the Hourglass

链接

传送门

题意

给出一个沙漏,求从顶至底经过数字的和为定值的路径数目。如果数目不为零,输出字典序最小的一条。

思路

类似于背包的dp,对于每个点保存从当前点到沙漏底部不同值的路径数目。

代码

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#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
typedef long long LL;
const int maxn = 21;
const int maxs = 505;
int a[maxn << 1][maxn];
LL dp[maxn << 1][maxn][maxs];
int main() {
int n, s;
while (~scanf("%d%d", &n, &s) && (n || s)) {
for (int i = 1; i < n; ++i) {
for (int j = 1; j <= n - i + 1; ++j) {
scanf("%d", &a[i][j]);
}
}
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= i; ++j) {
scanf("%d", &a[n + i - 1][j]);
}
}
memset(dp, 0, sizeof dp);
int n2 = (n << 1) - 1;
for (int i = 1; i <= n; ++i) {
dp[n2][i][a[n2][i]] = 1;
}
for (int i = n2 - 1; i >= n; --i) {
for (int j = 1; j <= i - n + 1; ++j) {
for (int k = a[i][j]; k <= s; ++k) {
dp[i][j][k] = dp[i + 1][j][k - a[i][j]] + dp[i + 1][j + 1][k - a[i][j]];
}
}
}
LL ans = 0;
for (int i = n - 1; i >= 1; --i) {
for (int j = 1; j <= n - i + 1; ++j) {
for (int k = a[i][j]; k <= s; ++k) {
dp[i][j][k] = 0;
if (j > 1) {
dp[i][j][k] += dp[i + 1][j - 1][k - a[i][j]];
}
if (j < n - i + 1) {
dp[i][j][k] += dp[i + 1][j][k - a[i][j]];
}
}
if (i == 1) {
ans += dp[i][j][s];
}
}
}
printf("%lld\n", ans);
if (ans > 0) {
for (int i = 1; i <= n; ++i) {
if (dp[1][i][s] > 0) {
printf("%d ", i - 1);
int p = i, ss = s;
for (int j = 1; j < n; ++j) {
ss -= a[j][p];
if (p > 1 && dp[j + 1][p - 1][ss] > 0) {
putchar('L');
--p;
} else {
putchar('R');
}
}
for (int j = n; j < n2; ++j) {
ss -= a[j][p];
if (dp[j + 1][p][ss] > 0) {
putchar('L');
} else {
++p;
putchar('R');
}
}
break;
}
}
}
puts("");
}
return 0;
}