1. 程式人生 > >洛谷T47092 作業_簡單狀壓動歸

洛谷T47092 作業_簡單狀壓動歸

只要注意一下細節就毫無難點了。 Code:

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 20;
const long long mod = 4921057 ;
int a[maxn], b[maxn], pos[maxn], sumv[maxn];
long long  dp[1 << maxn][maxn];
inline int lowbit(int t) { return t & (-t); }
inline int dis(int x,int
y,int S){ if(x < y) swap(x, y); int ans = sumv[x - 1] - sumv[y]; for(int i = x; i > y; --i) if((S & pos[i]) != 0) --ans; return ans; } int main() { // freopen("input.in","r",stdin); int n; scanf("%d",&n); for(int i = 1;i <= n + 1; ++i) pos[i] = (1 <<
(i - 1)); for(int i = 1;i <= n; ++i) { scanf("%d",&a[i]); sumv[i] = sumv[i - 1] + 1; } for(int i = 1;i <= n; ++i) scanf("%d",&b[i]); for(int i = 1;i <= n; ++i) dp[pos[i]][i] = 1; for(int i = 1;i < pos[n + 1]; ++i) { for(int j = 1;j <=
n; ++j) { if((i & pos[j]) == 0) continue; int mx = (i ^ pos[j]); if(mx == 0) continue; for(int k = 1;k <= n; ++k) { if(k == j || (mx & pos[k]) == 0 || a[j] < a[k]) continue; if(dis(k, j,mx) > b[k]) continue; dp[i][j] += dp[mx][k]; dp[i][j] %= mod; } } } long long fin = 0; for(int i = 1;i <= n; ++i) fin += dp[pos[n + 1] - 1][i] , fin %= mod; printf("%lld",fin); return 0; }