UVA - 12633 Super Rooks on Chessboard (FFT)

标签：统计   eof   lld   分析   编号   ace   print   多少   struct   

题意:一个N*M的棋盘上,放置N个皇后,皇后(i,j)可以攻击整行整列和两条对角线.求放完这N个皇后,棋盘上还有多少个点不会被攻击到.
分析:除了行和列之外,还要考虑对角线.对于每一个格点\((x,y)\),都有其对应的主对角线\(x+M-j\)(保证编号>0).如果行i和列j都不会被占据,那么点\((i,j)\)就会对其主对角线做出贡献,当然,需要保证这条对角线上没有皇后.将主对角线视作\(A^k\),则\(A^k的系数为\sum_{i+M-j=k}R^i*C^i\).若i行没有被占据,则\(R^i = 1\),否则为0, 若j列没被占据,\(C^{M-j}\)为1,否则为0.
求出卷积后,若对角线k上没有皇后,则该对角线的系数对答案做出贡献.

#include <bits/stdc++.h>    
using namespace std;
typedef long long LL;
const int MAXN = 4e5 + 10;
const double PI = acos(-1.0);
struct Complex{
    double x, y;
    inline Complex operator+(const Complex b) const {
        return (Complex){x +b.x,y + b.y};
    }
    inline Complex operator-(const Complex b) const {
        return (Complex){x -b.x,y - b.y};
    }
    inline Complex operator*(const Complex b) const {
        return (Complex){x *b.x -y * b.y,x * b.y + y * b.x};
    }
} va[MAXN * 2 + MAXN / 2], vb[MAXN * 2 + MAXN / 2];
int lenth = 1, rev[MAXN * 2 + MAXN / 2];
int N, M;   // f 和 g 的数量
    // f g和 的系数
    // 卷积结果
    // 大数乘积
int f[MAXN],g[MAXN];
vector<LL> conv;
vector<LL> multi;
//f g
void init()
{
    int tim = 0;
    lenth = 1;
    conv.clear(), multi.clear();
    memset(va, 0, sizeof va);
    memset(vb, 0, sizeof vb);
    while (lenth <= N + M - 2)
        lenth <<= 1, tim++;
    for (int i = 0; i < lenth; i++)
        rev[i] = (rev[i >> 1] >> 1) + ((i & 1) << (tim - 1));
}

void FFT(Complex *A, const int fla)
{
    for (int i = 0; i < lenth; i++){
        if (i < rev[i]){
            swap(A[i], A[rev[i]]);
        }
    }
    for (int i = 1; i < lenth; i <<= 1){
        const Complex w = (Complex){cos(PI / i), fla * sin(PI / i)};
        for (int j = 0; j < lenth; j += (i << 1)){
            Complex K = (Complex){1, 0};
            for (int k = 0; k < i; k++, K = K * w){
                const Complex x = A[j + k], y = K * A[j + k + i];
                A[j + k] = x + y;
                A[j + k + i] = x - y;
            }
        }
    }
}
void getConv(){             //求多项式
    init();
    for (int i = 0; i < N; i++)
        va[i].x = f[i];
    for (int i = 0; i < M; i++)
        vb[i].x = g[i];
    FFT(va, 1), FFT(vb, 1);
    for (int i = 0; i < lenth; i++)
        va[i] = va[i] * vb[i];
    FFT(va, -1);
    for (int i = 0; i <= N + M - 2; i++)
        conv.push_back((LL)(va[i].x / lenth + 0.5));
}

int row[MAXN], col[MAXN];
int vis[MAXN];

int main()
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt","r",stdin);
        freopen("out.txt","w",stdout);
    #endif
    int T,cas=1; scanf("%d",&T);
    while(T--){
        int R,C,n; scanf("%d %d %d",&R, &C, &n);
        int x,y;
        memset(vis,0,sizeof(vis));
        for(int i=1;i<=R;++i) row[i] = 1;
        for(int i=1;i<=C;++i) col[C-i] = 1;
        for(int i=1;i<=n;++i){
            scanf("%d %d",&x, &y);
            vis[x+C-y] = true;                  //表示该对角线被占据
            row[x] = 0, col[C-y] = 0;           //行列被占据
        }
        N = R+1;
        M = C;                                 //没被占据的行列做卷积,求合法的对角线
        for(int i=0;i<N;++i) f[i] = row[i];
        for(int i=0;i<M;++i) g[i] = col[i];
        getConv();
        int sz = conv.size();
        LL ans=0;
        for(int i=0;i<sz;++i){
            if(!vis[i])
                ans+= conv[i];          //若该对角线没被占据,则统计
        }
        printf("Case %d: %lld\n",cas++,ans);
    }
    return 0;
}

UVA - 12633 Super Rooks on Chessboard (FFT)

标签：统计   eof   lld   分析   编号   ace   print   多少   struct   

原文地址：https://www.cnblogs.com/xiuwenli/p/9729120.html