洛谷P1896 互不侵犯king

来源：互联网发布：云计算项目编辑：程序博客网时间：2024/06/05 01:14
#include <iostream>#include <vector>#include <cstdio>#include <cstdlib>#define ll long longusing namespace std;const int MAXN = 2000;const int MAXK = 1100;ll f[2][MAXN][MAXK],Ans;int N,K;int s[MAXN],num[MAXN];int Next[MAXN*10],h[MAXN*10],st[MAXN],flag;inline int getNum(int x){    int t = 0;    while (x>0){        x = x&(x-1);//每执行一次将最右边的1置为0        t++;    }    return t;}void DFS(int len,int now){    if (len == N){        ++s[0];        s[s[0]] = now;        num[s[0]] = getNum(now);        return;    }    DFS(len+1,now << 1);//不放猛兽,最右位放置0    if((now & 1) != 1){//当前最右为为0时可放置猛兽        DFS(len+1,(now << 1)|1);//右移一位并把最后一位变成1    }}void add(int one,int two){    Next[++flag] = st[one];    h[flag] = two;    st[one] = flag;}vector < vector <int> > G;void init(){    cin >> N >> K;    s[0] = 0;    //枚举出总状态数    for (int i = 0; i < (1 << N); ++i) {        if(i & (i << 1))            continue;        s[0]++;        int t = i;        num[s[0]] = 0;        while (t){            num[s[0]] += (t&1);            t = t >> 1;        }        s[s[0]] = i;    }    int s1,s2;    G.clear();    G.resize(MAXN*10);    for (int j = 1; j <= s[0]; ++j) {        for (int i = j+1; i <= s[0]; ++i) {//为什么不从j开始循环,因为0到0时会push两次1-1,会造成错误哒            s1 = s[j],s2 = s[i];            if( ((s2 << 1) & s1) || ((s2 >> 1) & s1) || (s1 & s2) )//判断s1状态与s2状态是否冲突                continue;            G[i].push_back(j);//状态号            G[j].push_back(i);        }    }    G[1].push_back(1);//0到0,也就是说可以连续出现两次第一个状态,也就是两行都不放}void DP(){    f[0][1][0] = 1;//f(i,j,k) 考虑前i行,是第k个状态时,共放了j只猛兽//第一个状态就是0000    int m = 0,s1;    for (int i = 1; i <= N; ++i) {        m ^= 1;        //初始化数组        for (int j = 1; j <= s[0]; ++j) {            for (int k = 0; k <= K; ++k) {                f[m][j][k] = 0;            }        }        for (int j = 1; j <= s[0]; ++j) {            for (int k = 0; k <= K; ++k) {                for (int l = 0; l < G[j].size(); ++l) {                    s1 = G[j][l];                    if(k - num[j] >= 0){                        f[m][j][k] += f[m^1][s1][k-num[j]];                    }                //冻柜sa                }            }        }    }    Ans = 0;    for (int n = 1; n <= s[0]; ++n) {        Ans += f[m][n][K];    }    cout << Ans << endl;}int main() {    init();    DP();    return 0;}
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