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[BZOJ3925][[Zjoi2015]地震后的幻想乡][期望+状压DP]

2017-03-29 19:12 525 查看

[BZOJ3925][[Zjoi2015]地震后的幻想乡][期望+状压DP]

]http://blog.csdn.net/skywalkert/article/details/47792065]

这位神犇写得很好啊QAQ

代码:

#include <bits/stdc++.h>
const int N = 11, M = 50;
typedef long long ll;

inline void read(ll &x) {
scanf("%lld", &x);
}
ll n, m, e[1 << N], f[1 << N][M], g[1 << N][M], sz[1 << N], cnt[1 << N], C[M][M];
double ans;

inline void Pre(int m) {
C[0][0] = 1;
for (int i = 1; i <= m; i++) {
C[i][0] = C[i][i] = 1;
for (int j = 1; j < i; j++) C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
}
}
int main(void) {
//freopen("in.txt", "r", stdin);
read(n), read(m); Pre(m);
for (int i = 1; i <= m; i++) {
ll u, v; read(u), read(v);
u--; v--; e[u] |= 1 << v; e[v] |= 1 << u;
}
for (int s = 0; s < 1 << n; s++) {
sz[s] = sz[s >> 1] + (s & 1);
if (sz[s] == 1) {
g[s][0] = 1;
continue;
}
for (int i = 0; i < n; i++)
if (s & (1 << i)) {
cnt[s] += sz[e[i] & s];
}
cnt[s] >>= 1;
int p = s & -s;
for (int t = (s - 1) & s; t; t = (t - 1) & s)
if (t & p) {
for (int i = 0; i <= cnt[t]; i++)
for (int j = 0; j <= cnt[s ^ t]; j++)
f[s][i + j] += g[t][i] * C[cnt[s ^ t]][j];
}
for (int i = 0; i <= cnt[s]; i++)
g[s][i] = C[cnt[s]][i] - f[s][i];
}
for (int i = 0; i <= m; i++)
ans += (double)f[(1 << n) - 1][i] / C[cnt[(1 << n) - 1]][i];
ans /= (m + 1);
printf("%.6lf\n", ans);
return 0;
}


完。

By g1n0st
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