19.9.21 考试总结

  

今天的题比较水... 班上的大佬们都AK了 我这种蒟蒻就完全不会矩阵快速幂

我永远恨矩阵快速幂

然后这道题还是比较简单的 就是维护区间连续最大和

线段树维护一下就可以了

代码

#include <bits/stdc++.h>
#define oo 2 * 1e9
using namespace std;

const int N = 5 * 1e5 + 5; 
int n,m,lmax[4 * N],rmax[4 * N],f[4 * N],a[N],g[4 * N],sum[4 * N];
struct data{
    int lma,rma,s,sm;
    data(int lma = 0,int rma = 0,int s = 0,int sm = 0): lma(lma),rma(rma),s(s),sm(sm) { }
};

void update(int o) {
    
    g[o] = max(g[2 * o], g[2 * o + 1]); sum[o] = sum[2 * o] + sum[2 * o + 1];
    lmax[o] = lmax[2 * o]; rmax[o] = rmax[2 * o + 1];
    f[o] = max(f[2 * o], f[2 * o + 1]); f[o] = max(f[o], rmax[2 * o] + lmax[2 * o + 1]);
    lmax[o] = max(lmax[o], sum[2 * o] + lmax[2 * o + 1]);
    rmax[o] = max(rmax[o], sum[2 * o + 1] + rmax[2 * o]);
    f[o] = max(f[o], max(lmax[o], rmax[o]));
}

void build(int o,int l,int r) {
    
    if(l == r) {
        sum[o] = a[l];
        g[o] = a[l];
        rmax[o] = max(0, a[l]); lmax[o] = max(0, a[l]);
        f[o] = max(0,a[l]);
        return ;
    }
    int mid = (l + r) >> 1;
    build(2 * o, l, mid);
    build(2 * o + 1, mid + 1, r);
    update(o);
}

void Init( ) {
    
    scanf("%d%d",& n,& m);
    for(int i = 1;i <= n;i ++) scanf("%d",& a[i]);
    build(1, 1, n);
}

data query(int o,int l,int r,int L,int R) {
    
    if(l >= L && r <= R) return data(lmax[o], rmax[o], f[o], sum[o]);
    int mid = (l + r) >> 1;
    data q1,q2,q3; q1 = data(0, 0, 0, 0); q2 = data(0, 0, 0, 0); q3 = data(0, 0, 0, 0);
    int t1 = 0,t2 = 0;
     if(L <= mid) {
         q1 = query(2 * o, l, mid, L, R); t1 = 1;
     }
    if(mid < R) {
        q2 = query(2 * o + 1, mid + 1, r, L, R); t2 = 1;
    }
    if(! t1 && t2) return q2;
    if(t1 && ! t2) return q1;
    if(t1 && t2) {
        q3.lma = max(q1.lma, q3.lma); q3.rma = max(q3.rma, q2.rma);
        q3.s = max(q3.s, max(q1.s, q2.s)); q3.s = max(q3.s, q1.rma + q2.lma);
        q3.lma = max(q3.lma, q1.sm + q2.lma);
        q3.rma = max(q3.rma, q2.sm + q1.rma);
        q3.s = max(q3.s, max(q3.lma, q3.rma));
    } 
    return q3;
}

void modify(int o,int l,int r,int pos,int del) {
    
    if(l == r) {
        sum[o] = del;
        rmax[o] = max(0, del); lmax[o] = max(0, del);
        f[o] = max(0,del); g[o] = del;
        return ;
    }
    int mid = (l + r) >> 1;
    if(pos <= mid) modify(2 * o, l, mid, pos, del);
    else modify(2 * o + 1, mid + 1, r, pos, del);
    update(o);
}

int query_max(int o,int l,int r,int L,int R) {
    
    if(l >= L && r <= R) return g[o];
    int mid = (l + r) >> 1;
    int ans = -oo;
    if(L <= mid) ans = max(ans, query_max(2 * o, l, mid, L, R));
    if(mid < R) ans = max(ans, query_max(2 * o + 1, mid + 1, r, L, R));
    return ans;
}

inline int read( ) {
    
    int t = 1,ans = 0;
    char x; x = getchar( );
    while(x < '0' || x > '9') {
        if(x == '-') t = -1;
        x = getchar( );
    }
    while(x >= '0' && x <= '9') {
        ans = ans * 10 + x - '0';
        x = getchar( );
    }
    return ans * t;
}

void Solve( ) {
    
    while(m --) {
        int opt,l,r;
        opt = read( );
        if(opt == 1) {
            l = read( ); r = read( );
            data ans = query(1, 1, n, l, r);
            if(ans.s == 0) ans.s = query_max(1, 1, n, l, r);
            printf("%d
",ans.s);
        }
        else {
            l = read( ); r = read( );
            a[l] = r;
            modify(1, 1, n, l, r);
        }
    }
}

int main( ) {
    
    freopen("BRS.in","r",stdin);
    freopen("BRS.out","w",stdout);
    Init( );
    Solve( );
}

 

这道题就是矩阵快速幂 然后借着这道题的机会我重新学习了一波矩阵快速幂...

方程还是很简单的 就不说了 最重要的就是构造转移矩阵和初始矩阵 

完了发现这是一道水水题 先把$dp[1],dp[2] ... dp[k]$预处理出来 

然后转移矩阵 $1111$   初始矩阵 $dp[k]$

 　　　　　　$1000$　　　　　 $dp[k - 1]$

　　　　　　 $0100$                 $dp[...]$

　　　　　　 $0010$                 $dp[2]$

　　　　　　 $0001$                 $dp[1]$          类似于这种

代码

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const int N = 1e3 + 5;
const ll MOD = 7777777;
ll dp[N],s[20][5];
ll K,n;
struct Matrix {
    
    ll a[15][15];
}h;

void Moc(ll & a,ll b) {
    
    a = (a + b) % MOD;
}

Matrix pl(Matrix A,Matrix B) {
    
    Matrix c;
    for(int i = 1;i <= K;i ++) 
        for(int j = 1;j <= K;j ++) c.a[i][j] = 0;
    for(int i = 1;i <= K;i ++)
        for(int j = 1;j <= K;j ++) {
            for(int k = 1;k <= K;k ++)
                Moc(c.a[i][j], 1ll * A.a[i][k] * B.a[k][j] % MOD);
        }
    return c;
}

void Init( ) {
    
    scanf("%d%d",& K,& n);
    dp[0] = 1;
    for(int i = 1;i <= K;i ++) {
        for(int k = 0;k < i;k ++) {
            Moc(dp[i], dp[k]);
        }
    }
    for(int i = 1;i <= K;i ++) {
        h.a[1][i] = 1;    
    }
    for(int j = 2;j <= K;j ++) h.a[j][j-1] = 1;
    for(int i = 1;i <= K;i ++) s[i][1] = dp[i];
}

Matrix fast_pow(Matrix A,int del) {
    
    Matrix ans;
    for(int i = 1;i <= K;i ++) 
        for(int j = 1;j <= K;j ++) {
            ans.a[i][j] = 0;
            if(i == j) ans.a[i][j] = 1;
        }
    for(;del;del >>= 1,A = pl(A, A))
        if(del & 1) ans = pl(ans, A);
    return ans;
}

void Solve( ) {
    
    h = fast_pow(h, n - K);
    ll ans = 0;
    for(int i = 1;i <= K;i ++) Moc(ans, h.a[1][i] * s[K + 1 - i][1] % MOD);
    printf("%lld
",ans);
}

int main( ) {
    
    freopen("fyfy.in","r",stdin);
    freopen("fyfy.out","w",stdout);
    Init( );
    Solve( );
}

 ...这道题是扫描线的板子啊 而且数据范围这么小 就乱搞都可以过得

代码

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const int M = 10005;
int n,q;
ll ans = 0;
int yy[M * 2];
struct Event{
    int x,y1,y2;
    int d;
};
vector<Event>e;

struct Node {
    
    int sum,cnt;
    Node * ls,* rs;
    
    void up(int l, int r){
        if(cnt) sum = yy[r] - yy[l - 1];
        else if(l != r)
            sum = ls -> sum + rs -> sum;
        else sum = 0;
    }
}pool[M],* root,* tail = pool;

Node *build(int l = 1, int r = q) {
    
    Node * nd = ++ tail;
    if(l == r) nd -> sum = 0, nd -> cnt = 0;
    else {
        int mid = (l + r) >> 1;
        nd -> ls = build(l, mid);
        nd -> rs = build(mid + 1, r);
        nd -> up(l, r);
    }
    return nd;
}

#define Ls l, mid, nd -> ls
#define Rs mid+1, r, nd -> rs

void modify(int L, int R, int d, int l = 1, int r = q, Node * nd = root) {
    
    if(L <= l && R >= r) nd -> cnt += d, nd -> up(l, r);
    else {
        int mid = (l + r) >> 1;
        if(L <= mid)  modify(L, R, d, Ls);
        if(R > mid)   modify(L, R, d, Rs);
        nd -> up(l, r);
    }
}

bool operator < (const Event & a, const Event & b) {
    
    return a.x < b.x;
}

inline int read( ) {
    
    int t = 1,ans = 0;
    char x; x = getchar( );
    while(x < '0' || x > '9') {
        if(x == '-') t = -1;
        x = getchar( );
    }
    while(x >= '0' && x <= '9') {
        ans = ans * 10 + x - '0';
        x = getchar( );
    }
    return ans * t;
}

void Init( ) {
    
    scanf("%d",& n);
    int cnt = 0,k = 0;
    for(int i = 1;i <= n;i ++){
        int x1,x2,y1,y2;
        x1 = read( ); y1 = read( ); 
        x2 = read( ); y2 = read( );
        e.push_back((Event){x1, y1, y2, 1});
        e.push_back((Event){x2, y1, y2, -1});
        yy[++ cnt] = y1; yy[++ cnt] = y2;
    }
    sort(yy + 1, yy + 1 + cnt);
    sort(e.begin( ), e.end( ));
    q = unique(yy + 1, yy + 1 + cnt) - yy - 1;
    root = build( );
}

void Solve( ) {
    
    for(int i = 0;i < 2 * n;i ++) {
        ll dx = i == 0 ? 0 : e[i].x - e[i - 1].x;
        ans += 1ll * root -> sum * dx;
        int p1 = find(yy + 1, yy + 1 + q, e[i].y1) - yy;
        int p2 = find(yy + 1, yy + 1 + q, e[i].y2) - yy;
        modify(p1 + 1, p2, e[i].d);
    }
    printf("%lld
", ans);
}

int main( ) {
    
    freopen("olddriver.in","r",stdin);
    freopen("olddriver.out","w",stdout);
    Init( );
    Solve( );
}