hdu 6001 容斥 + dfs

题解不如代码

代码:

#include <bits/stdc++.h>using namespace std;typedef vector<int> vi;typedef long long ll;const int max_m = 1 << 15;const int max_n = 100001;const ll mod = 1e9 + 7;int n, m;char s[20];ll p[max_n];int cnt[max_m];ll v[max_m];int bitcnt[max_m];bool vis[max_m];vi vec[max_m];template <typename T>T read(){    T x = 0, f = 1; char ch = getchar();    while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}    while(ch >= '0' && ch <= '9') {x = 10 * x + ch - '0'; ch = getchar();}    return x * f;}inline ll Mod(ll x){return (x % mod + mod) % mod;}inline int lowbit(int x){return x & (-x);}void init(){    bitcnt[0] = 0; p[0] = 1;    for(int i = 1; i < max_n; ++i) p[i] = p[i - 1] * 2 % mod;    for(int i = 1; i < max_m; ++i) bitcnt[i] = bitcnt[i ^ lowbit(i)] + 1;    for(int i = 1; i < max_m; ++i)    {        vi bit_pos;        for(int j = 0; j < 15; ++j) if(i & (1 << j)) bit_pos.push_back(j);        for(int j = 0; j < 1 << bitcnt[i]; ++j)        {            int x = 0;            for(int k = 0; k < bitcnt[i]; ++k)                if(j & (1 << k)) x |= 1 << bit_pos[k];            vec[i].push_back(x);        }    }}void dfs(int x, int m){    vis[x] = 1;    for(int k = 0; k < vec[x].size(); ++k)    {        int u = vec[x][k];                       //非空子集        v[x] = Mod(v[x] + p[cnt[u]] - 1);            //2 ^ cnt - 1    }    v[x] = Mod(v[x] + 1);                       //空集                                                //更新cnt    for(int i = 0; i < m; ++i)        if(x & (1 << i) && !vis[x ^ (1 << i)])        {            for(int k = 0; k < vec[x].size(); ++k)            {                int u = vec[x][k];                if(u & (1 << i)) cnt[u ^ (1 << i)] += cnt[u];//x子集中有bit i就更新            }            dfs(x ^ (1 << i), m);            for(int k = 0; k < vec[x].size(); ++k)            {                int u = vec[x][k];             //恢复                if(u & (1 << i)) cnt[u ^ (1 << i)] -= cnt[u];            }        }}void in(){    memset(cnt, 0, sizeof cnt);    memset(vis, 0, sizeof vis);    memset(v, 0, sizeof v);    n = read<int>(), m = read<int>();    for(int i = 0; i < n; ++i)    {        scanf("%s", s);        int x = 0;        for(int j = 0; j < m; ++j)            if(s[j] == 'Y') x |= 1 << j;        ++cnt[x];    }   // v[0] = 1;    vis[0] = 1;    dfs((1 << m) - 1, m);}ll solve(){    in();    ll res = 0;                 // res    for(int i = 1; i < 1 << m; ++i)    {        ll ans = 0;             //the ans for bit i                                //for i : v1 && v2 && ... vn = 2^n - (!v1 || !v2 || .. !vn)                                // = 2^n - (bitcnt[j] & 1 ? v[j] : -v[j]) for j bit of i(j是i的非空子集)容斥        for(int k = 0; k < vec[i].size(); ++k)        {            int j = vec[i][k];            if(j == 0) continue;            ans += bitcnt[j] & 1 ? v[j] : -v[j];//v[j]是bit j的贡献(j bit位各子集数量分别贡献)            ans = Mod(ans);        }        ans = Mod(p[n] - ans);        res ^= ans;    }    return res;}int main(){    init();    int T = read<int>();    for(int kas = 1; kas <= T; ++kas) printf("Case #%d: %lld\n", kas,solve());    return 0;}