hdu 5555 Immortality of Frog（状压dp）

题目链接：hdu 5555 Immortality of Frog

解题思路

因为每只青蛙最多碰到10个bad的membrane，所以可以用一个二进制状态表示到第i只青蛙还有哪些membrane可以选，但是二进制的状态的转移必须预处理，因为每只青蛙对应的bad membrane是不一样的。注意对于第i个位置来说，有membrane以这个位置终止，但是并没有被用掉则是非法的。剩下的good membrane就可以任意选了。

代码

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>

using namespace std;
typedef long long ll;
const int maxn = 1050;
const int mod = 105225319;

int N, L[maxn], R[maxn];
int dp[maxn][maxn], G[maxn][maxn];
vector<int> B[maxn];

int find(int id, int x) {
for (int i = 0; i < B[id].size(); i++)
if (B[id][i] == x) return i;
return -1;
}

bool init () {
memset(G, 0, sizeof(G));
for (int i = 1; i < N; i++) {
int n = B[i].size(), m = B[i+1].size();
if (n > 10) return false;

for (int j = 0; j < m; j++) {
int mv = find(i, B[i+1][j]);
for (int s = 0; s < (1<<n); s++) {
if (mv == -1 || (s&(1<<mv)))
G[i][s] |= (1<<j);
}
}

}
for (int i = 1; i <= N; i++) {
int n = B[i].size();
for (int s = 0; s < (1<<n); s++) {
int c = 0;
for (int j = 0; j < n; j++) {
int v = B[i][j];
if (s&(1<<j) && R[v] == i) c++;
}
G[i][s] = (c ? -1 : G[i][s]);
}
}
return B
.size() <= 10;;
}

int solve (int k) {
if (!init()) return 0;
memset(dp, 0, sizeof(dp));
dp[1][(1<<B[1].size())-1] = 1;

for (int i = 1; i <= N; i++) {
int n = B[i].size();
for (int s = 0; s < (1<<n); s++) {
if (dp[i][s] == 0) continue;

for (int j = 0; j < n; j++) if (s&(1<<j)) {
int v = G[i][s^(1<<j)];
if (v != -1)
dp[i+1][v] = (dp[i+1][v] + dp[i][s]) % mod;
}
int u = G[i][s];
if (u != -1)
dp[i+1][u] = (dp[i+1][u] + dp[i][s]) % mod;
//      printf("%d ", dp[i][s]);
}
//  printf("\n");
}
int ans = dp[N+1][0];
for (int i = 1; i <= k; i++)
ans = (1LL * ans * i) % mod;
return ans;
}

int main () {
int cas;
scanf("%d", &cas);
for (int kcas = 1; kcas <= cas; kcas++) {
scanf("%d", &N);
for (int i = 1; i <= N; i++) scanf("%d", &L[i]);
for (int i = 1; i <= N; i++) scanf("%d", &R[i]);

int cnt = 0;
for (int i = 1; i <= N; i++) B[i].clear();
for (int i = 1; i <= N; i++) if (R[i] - L[i] + 1 != N) {
for (int j = L[i]; j <= R[i]; j++) B[j].push_back(i);
cnt++;
}

printf("Case #%d: %d\n", kcas, solve(N - cnt));
}
return 0;
}