[hihocoder][Offer收割]编程练习赛57

1-偏差排列

斐波那契数列

#pragma comment(linker, "/STACK:102400000,102400000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<algorithm>
#include<iostream>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<functional>
#include<math.h>
//#include<bits/stdc++.h>
using namespace std;
typedef long long lint;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef queue<int> QI;

void makedata() {
    freopen("input.txt", "w", stdout);
    fclose(stdout);
}

lint f[];

int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    //makedata();
    std::ios::sync_with_stdio(), cin.tie();
    int n;
    cin >> n;
    f[] = , f[] = ;
    for (int i = ; i <= n; i++) f[i] = f[i - ] + f[i - ];
    cout << f[n] << endl;
    return ;
}

递增N元组

先把每一行排序

dp[i][j]表示从第i行开始，且第i行选择第j个及以后的数字的方案数

#pragma comment(linker, "/STACK:102400000,102400000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<algorithm>
#include<iostream>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<functional>
#include<math.h>
//#include<bits/stdc++.h>
using namespace std;
typedef long long lint;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef queue<int> QI;
typedef map<int, int> MII;

void makedata() {
    freopen("input.txt", "w", stdout);
    fclose(stdout);
}

int mp[][];
lint dp[][];

int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    //makedata();
    std::ios::sync_with_stdio(), cin.tie();
    int n, m, x;
    cin >> n >> m;
    for (int i = ; i <= n; i++) {
        for (int j = ; j < m; j++) {
            cin >> mp[i][j];
        }
        sort(mp[i], mp[i] + m);
    }
    for (int i = ; i < m; i++) dp[n][i] = m - i;
    for (int i = n - ; i >= ; i--) {
        int p = m - ;
        for (int j = m - ; j >= ; j--) {
            dp[i][j] = dp[i][j + ];
            while (mp[i + ][p - ] > mp[i][j] && p) p--;
            if (mp[i][j] >= mp[i + ][p]) continue;
            dp[i][j] += dp[i + ][p];
            dp[i][j] %= ;
        }
    }
    cout << dp[][] << endl;
    return ;
}

逃离迷宫5

bfs...

#pragma comment(linker, "/STACK:102400000,102400000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<algorithm>
#include<iostream>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<functional>
#include<math.h>
//#include<bits/stdc++.h>
using namespace std;
typedef long long lint;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef queue<int> QI;
typedef map<int, int> MII;

void makedata() {
    freopen("input.txt", "w", stdout);
    fclose(stdout);
}

struct node {
    int x, y, k;
    node(int xx, int yy, int kk) : x(xx), y(yy), k(kk) {}
};
queue<node> q;
int d[][][];
string mp[];
const int dx[] = { -, , ,  };
const int dy[] = { , , , - };

int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    //makedata();
    std::ios::sync_with_stdio(), cin.tie();
    int n;
    cin >> n;
    for (int i = ; i < n; i++) cin >> mp[i];
    memset(d, 0x3F, sizeof(d));
    while (!q.empty()) q.pop();
    if (mp[][] == '.') d[][][] = , q.push(node(, , ));
    else d[][][] = , q.push(node(, , ));
    while (!q.empty()) {
        node p = q.front();
        q.pop();
        for (int i = ; i < ; i++) {
            int px = p.x + dx[i], py = p.y + dy[i];
            if (px <  || px >= n || py <  || py >= n) continue;
            if (mp[px][py] == '.') {
                if (d[px][py][p.k] > d[p.x][p.y][p.k] + ) {
                    d[px][py][p.k] = d[p.x][p.y][p.k] + ;
                    q.push(node(px, py, p.k));
                }
            } else {
                if (p.k == ) {
                    if (d[px][py][] > d[p.x][p.y][] + ) {
                        d[px][py][] = d[p.x][p.y][] + ;
                        q.push(node(px, py, ));
                    }
                }
            }
        }
    }
    if (d[n - ][n - ][] != 0x3F3F3F3F || d[n - ][n - ][] != 0x3F3F3F3F) cout << min(d[n - ][n - ][], d[n - ][n - ][]) << endl;
    else cout << - << endl;
    return ;
}

最大割集

算出每个点相关的边的异或和，割<S,T>中边的异或和就是S集合中点的边异或和的异或和，因为点集内部的边在这些异或和中被计算了两次变成0，只剩下S到T的边。然后按w排序从大到小添加线性基、

#pragma comment(linker, "/STACK:102400000,102400000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<algorithm>
#include<iostream>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<functional>
#include<math.h>
#include<iomanip>
//#include<bits/stdc++.h>
using namespace std;
typedef long long lint;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef queue<int> QI;
typedef map<int, int> MII;

void makedata() {
    freopen("input.txt", "w", stdout);
    fclose(stdout);
}

template <typename T> class LnBase {
public:
    int sz, szc;
    T *x;
    int *y;
    LnBase() {
        x = , sz = sizeof(T) << ;
        szc = -, resize(sz);
    }
    void resize(int size) {
        sz = size;
        if (!x) delete (x);
        x = new T[sz + ];
        y = new int[sz + ];
        memset(x, , sz * sizeof(T));
        memset(y, , sz << );
    }
    T operator[](int h) {
        return x[h];
    }
    //增加一个向量，若该向量能被已有向量线性表示，返回-1;否则返回该向量增加的是哪一个维度
    int add(T v) {
        for (int i = sz - ; i >= ; i--)
            if (v & (T) << i) {
                if (!x[i]) {
                    x[i] = v;
                    szc = -;
                    return i;
                }
                v ^= x[i];
            }
        return -;
    }
    //若该向量能被已有向量线性表示，返回1;否则返回0
    int find(T v) {
        for (int i = sz - ; i >= ; i--) {
            if (v & (T) << i && x[i]) v ^= x[i];
            if (!v) return ;
        }
        return ;
    }
    //线性基能表示出的最大向量
    T max() {
        T s = ;
        for (int i = sz - ; i >= ; i--) {
            if ((s ^ x[i]) > s) s ^= x[i];
        }
        return s;
    }
    //线性基能表示出的最小向量，为空返回-1
    T min() {
        for (int i = ; i < sz; i++) if (x[i]) return x[i];
        return -;
    }
    //矩阵标准化
    void canonicity() {
        int i, j;
        for (i = sz - ; i > ; i--)
            for (j = i - ; j >= ; j--) if (x[i] & (T) << j) x[i] ^= x[j];
        for (szc = i = ; i < sz; i++) if (x[i]) y[szc++] = i;
    }
    //线性基能表示出的第k大的向量
    T kth(long long K) {
        if (szc < ) canonicity();
        if (K >= 1ll << szc) return -;
        T s = ;
        for (int i = szc - ; i >= ; i--) if (K & 1ll << i) s ^= x[y[i]];
        return s;
    }
};
LnBase<lint> lb;
pair<lint, lint> p[];

int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    //makedata();
    std::ios::sync_with_stdio(), cin.tie();
    int n, m, u, v;
    lint t;
    long double sum = ;
    cin >> n >> m;
    memset(p, , sizeof(p));
    for (int i = ; i < n; i++) {
        cin >> p[i].first;
        sum += p[i].first;
    }
    for (int i = ; i < m; i++) {
        cin >> u >> v >> t;
        u--, v--;
        p[u].second ^= t, p[v].second ^= t;
    }
    sort(p, p + n);
    long double sel = ;
    for (int i = n - ; i >= ; i--) {
        if (lb.add(p[i].second) != -) sel += p[i].first;
    }
    cout << setprecision() << fixed;
    cout << ( * sel - sum) << endl;
    return ;
}