题解点我
Code
#include <bits/stdc++.h>
typedef long long LL;
typedef unsigned long long uLL;
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define MP(x, y) std::make_pair(x, y)
#define DE(x) cerr << x << endl;
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define GO cerr << "GO" << endl;
using namespace std;
inline void proc_status()
{
ifstream t("/proc/self/status");
cerr << string(istreambuf_iterator<char>(t), istreambuf_iterator<char>()) << endl;
}
inline int read()
{
register int x = 0; register int f = 1; register char c;
while (!isdigit(c = getchar())) if (c == '-') f = -1;
while (x = (x << 1) + (x << 3) + (c xor 48), isdigit(c = getchar()));
return x * f;
}
template<class T> inline void write(T x)
{
static char stk[30]; static int top = 0;
if (x < 0) { x = -x, putchar('-'); }
while (stk[++top] = x % 10 xor 48, x /= 10, x);
while (putchar(stk[top--]), top);
}
template<typename T> inline bool chkmin(T &a, T b) { return a > b ? a = b, 1 : 0; }
template<typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
const int maxN = 1000;
const int mod = 1e9 + 7;
void pls(int &x, int y)
{
x += y;
if (x >= mod) x -= mod;
if (x < 0) x += mod;
}
int n, K, to[maxN], sum[maxN];
int dp[maxN][maxN], g[maxN];
int main()
{
#ifndef ONLINE_JUDGE
freopen("xhc.in", "r", stdin);
freopen("xhc.out", "w", stdout);
#endif
cin >> n >> K;
for (int i = 1; i <= K; ++i)
{
int u, v;
cin >> u >> v;
to[u] = v;
to[v] = u;
sum[u]++;
sum[v]++;
}
n <<= 1;
for (int i = 1; i <= n; ++i)
sum[i] += sum[i - 1];
g[0] = 1;
for (int i = 2; i <= n; i += 2)
g[i] = 1ll * (i - 1) * g[i - 2] % mod;
for (int i = 1; i <= n; ++i)
{
for (int j = i + 1; j <= n; j += 2)
{
bool flag = 0;
for (int k = i; k <= j; ++k)
if (to[k] and (to[k] < i || to[k] > j))
{
flag = 1;
break;
}
if (flag)
continue;
dp[i][j] = g[j - i + 1 - sum[j] + sum[i - 1]];
for (int l = i + 1; l < j; l += 2)
pls(dp[i][j], -1ll * dp[i][l] * g[j - l- sum[j] + sum[l]] % mod);
}
}
K <<= 1;
int ans(0);
for (int i = 1; i <= n; ++i)
for (int j = i + 1; j <= n; j += 2)
pls(ans, 1ll * dp[i][j] * g[n - (j - i + 1) - (K - (sum[j] - sum[i - 1]))] % mod);
cout << ans << endl;
return 0;
}
转载于:https://www.cnblogs.com/cnyali-Tea/p/11439846.html
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