51NOD 算法马拉松12

OTZ做出题目的神犇。。断断续续改完了在这里存一下思路吧

A题:第K大区间
题意:
定义一个区间的值为其众数出现的次数
现给出n个数,求将所有区间的值排序后,第K大的值为多少。

分析:
二分答案mid,任务就是判定有多少个区间的众数≥mid。
#include <bits/stdc++.h>
 
using namespace std;
 
typedef long long ll;
 
const int maxn = 100010;
 
int n, A[maxn], c[maxn], cnt[maxn], mx, tmp[maxn];
ll k;
 
void Add(int x){
    cnt[c[x]] --;
    c[x] ++;
    cnt[c[x]] ++;
    mx = max(mx, c[x]);
}
 
void Del(int x){
    cnt[c[x]] --;
    c[x] --;
    cnt[c[x]] ++;
    while(!cnt[mx])mx --;
}
 
int check(int x){
    ll ans = 0, l = 1; mx = 0;
    memset(cnt, 0, sizeof cnt);
    memset(c, 0, sizeof c);
    cnt[0] = n;
    for(int i = 1; i <= n; i ++){
        Add(A[i]);
        while(true){
            Del(A[l]);
            if(mx < x) {Add(A[l]); break;}
            l ++;
        }
        if(mx >= x) ans += l;
    }
    return ans >= k;
}
 
int main(){
    scanf("%d%d", &n, &k);
    for(int i = 1; i <= n; i ++){
        scanf("%d", &A[i]);
        tmp[i] = A[i];
    }
    sort(tmp+1, tmp+1+n);
    int p = unique(tmp+1, tmp+1+n) - tmp - 1;
    for(int i = 1; i <= n; i ++)
        A[i] = lower_bound(tmp+1, tmp+1+p, A[i]) - tmp;
    int l = 1, r = n+1;
    while(l < r){
        int mid = l + (r - l + 1) / 2;
        if(check(mid))l = mid;
        else r = mid - 1;
    }
    printf("%d\n", l);
    return 0;
}

  

 
 
B题戳这里

C题:逛街

分析:
对于喜欢的店的个数一定要大于等于k,这个可以用一个堆来贪心,维护堆中的元素等于k个,对于其他的离散化一下扔进线段树(splay也不介意,但是好像会t)
然后在线段树上二分
注意val有负值a。。然后不能加a
#include <bits/stdc++.h>
#define maxn 200010
using namespace std;
typedef long long ll;
int n, T, k, root, smz;

int A[maxn], B[maxn], C[maxn];
long long sum[maxn];


priority_queue<pair<ll, int> > Q;

int b[maxn];

struct Hash{
    int id, p;
    ll val;
    bool operator<(const Hash& k)const{return val < k.val;}
}h[maxn];

bool cmp(const Hash& a, const Hash& b){return a.p < b.p;}

struct Node{
    int l, r, size; ll sum;
}t[maxn << 2];
#define lc id<<1
#define rc id<<1|1
void build(int id, int l, int r){
    t[id].l = l, t[id].r = r;
    if(l == r)return;
    int mid = l+r >> 1;
    build(lc, l, mid);
    build(rc, mid+1, r);
}

void pushup(int id){
    t[id].sum = t[lc].sum + t[rc].sum;
    t[id].size = t[lc].size + t[rc].size;
}

void update(int id, int pos, ll val){
    if(val < 0)return;
    if(t[id].l == t[id].r){
        t[id].sum = val;
        t[id].size ++;
        return;
    }
    int mid = t[id].l + t[id].r >> 1;
    if(pos <= mid)update(lc, pos, val);
    else update(rc, pos, val);
    pushup(id);
}

int ask(int id, ll T){
    if(t[id].l == t[id].r)return min(t[id].size, (int)(T >= t[id].sum));
    if(T >= t[lc].sum)return t[lc].size + ask(rc, T - t[lc].sum);
    return ask(lc, T);
}

int vis[maxn];

int main(){
    scanf("%d%d%d", &n, &T, &k);
    for(int i = 1; i <= n; i ++)scanf("%d", &A[i]);
    for(int i = 1; i <= n; i ++)scanf("%d", &B[i]), h[i].val = B[i], h[i].p = i;
    for(int i = 1; i <= n; i ++)scanf("%d", &C[i]);

    sort(h+1, h+1+n);
    
    for(int i = 1; i <= n; i ++)
        h[i].id = i;
    sort(h+1, h+1+n, cmp);
    for(int i = 1; i <= n; i ++)b[i] = h[i].id;
    build(1, 1, n);
    int ans = 0;
    ll Sum = 0;
    for(int i = 1; i <= n; i ++){
        if(C[i]){
            if(Q.size() < k)Q.push(make_pair(B[i], i)), Sum += B[i];
            else if(!Q.empty() && (Q.size() == k && B[i] < Q.top().first)){
                ll t = Q.top().first;
                int t1 = Q.top().second;Q.pop();
                Q.push(make_pair(B[i], i));Sum = Sum - t + B[i];
                update(1, b[t1], t);
            }else update(1, b[i], B[i]);
        }
        else update(1, b[i], B[i]);
        if(Q.size() >= k && Sum + A[i] <= T)ans = max(ans, k + ask(1, T - Sum - A[i]));
    }
    if(ans < k)puts("-1");
    else printf("%d\n", ans);
    return 0;
}

  

 
D题:Rikka with Sequences
题目大意:给定一个序列,有更改操作,每次查询一段区间的历史区间和最小值。
n, m ≤ 10^5, Ai ≤ 10^9
分析:
这个在线好难做a。所以就要离线la。所以就要用KD-tree啦。(QAQ)
对于每一个操作我们可以计算它对询问的贡献。把一个询问看成二维平面上的一个点(l, r),每一次操作都是将(pos, pos)左上角(l <= pos, r >= pos)的询问更改一下,通过打标记来实现历史最小值的询问。时光倒流,如果是每次给一个位置添加数字,询问历史最小值是个经典问题la
一定不要忘记pushdown。。。一定不要忘记mn[i]和mx[i]
#include <bits/stdc++.h>
#define maxn 100010
using namespace std;

typedef long long ll;

const int inf = 0x7fffffff;

int n, m, dfs_clock, D;
//----------------------KD-tree--------------------------//
struct Node{
    int l, r, mn[2], mx[2], d[2], id;
    ll sum, add, ans, Minadd;
    int& operator[](const int& k){return d[k];}
    bool operator<(Node k)const{return d[D] < k[D];}
    Node(int x = 0, int y = 0, int i = 0, ll s = 0){
        sum = ans = s, id = i;
        mn[0] = mx[0] = d[0] = x;
        mn[1] = mx[1] = d[1] = y;
        l = r = add = Minadd = 0;
    }
}t[maxn], dfn[maxn], P;
int tot, root;

void update(int o){
    for(int i = 0; i < 2; i ++){
        t[o].mn[i] = t[o].mx[i] = t[o][i];
        if(t[o].l){
            t[o].mn[i] = min(t[o].mn[i], t[t[o].l].mn[i]);
            t[o].mx[i] = max(t[o].mx[i], t[t[o].l].mx[i]);
        }
        if(t[o].r){
            t[o].mn[i] = min(t[o].mn[i], t[t[o].r].mn[i]);
            t[o].mx[i] = max(t[o].mx[i], t[t[o].r].mx[i]);
        }
    }
}

int Newnode(){
    t[++ tot] = P;
    return tot;
}

inline void Down1(int o, ll val){
    if(!o)return;
    t[o].ans = min(t[o].ans, t[o].sum + val);
    t[o].Minadd = min(t[o].Minadd, t[o].add + val);
}

inline void Down2(int o, ll val){
    if(!o)return;
    t[o].ans = min(t[o].ans, t[o].sum += val);
    t[o].Minadd = min(t[o].Minadd, t[o].add += val);
}

inline void pushdown(int o){
    if(t[o].Minadd){
        Down1(t[o].l, t[o].Minadd), Down1(t[o].r, t[o].Minadd);
        t[o].Minadd = 0;
    }
    
    if(t[o].add){
        Down2(t[o].l, t[o].add), Down2(t[o].r, t[o].add);
        t[o].add = 0;
    }
}

void Modify(int o, int pos, ll val){
    pushdown(o);
    if(t[o].mx[0] <= pos && t[o].mn[1] >= pos) {Down2(o, val); return;}
    if(t[o][0] <= pos && t[o][1] >= pos)t[o].sum += val, t[o].ans = min(t[o].ans, t[o].sum);
    if(t[o].l && t[t[o].l].mn[0] <= pos && t[t[o].l].mx[1] >= pos)Modify(t[o].l, pos, val);
    if(t[o].r && t[t[o].r].mn[0] <= pos && t[t[o].r].mx[1] >= pos)Modify(t[o].r, pos, val);
}

void Insert(int o, int type){
    pushdown(o), D = type;
    if(P < t[o]){
        if(t[o].l)Insert(t[o].l, type^1);
        else t[o].l = Newnode();
    }
    else{
        if(t[o].r)Insert(t[o].r, type^1);
        else t[o].r = Newnode();
    }
    update(o);
}

ll ans[maxn];
void dfs(int o){
    pushdown(o);
    dfn[++ dfs_clock] = t[o];
    ans[t[o].id] = t[o].ans;
    if(t[o].l)dfs(t[o].l);
    if(t[o].r)dfs(t[o].r);
}

int build(int l, int r, int type){
    if(l > r)return 0; D = type;
    int mid = l + r >> 1;
    nth_element(dfn+l, dfn+mid, dfn+r+1);
    t[mid] = dfn[mid];
    t[mid].l = build(l, mid-1, type^1);
    t[mid].r = build(mid+1, r, type^1);
    update(mid); return mid;
}

//----------------------query---------------------------//
int tp[maxn], l[maxn], r[maxn];
//----------------------BIT-----------------------------//
ll A[maxn], Bit[maxn];
#define lowbit(i) i&(~i+1)
void add(int pos, ll val){
    for(int i = pos; i <= n; i += lowbit(i))
        Bit[i] += val;
}
ll ask(int pos){
    if(!pos)return 0;ll ret = 0;
    for(int i = pos; i; i -= lowbit(i))
        ret += Bit[i];
    return ret;
}
//----------------------solve---------------------------//
int main(){
    root = Newnode();
    t[root].mn[0] = t[root].mn[1] = inf;
    t[root].mx[0] = t[root].mx[1] = -inf;
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i ++)
        scanf("%lld", &A[i]), add(i, A[i]);
    for(int i = 1; i <= m; i ++){
        scanf("%d%d%d", &tp[i], &l[i], &r[i]);
        if(tp[i] == 1){
            r[i] -= A[l[i]];//增量
            A[l[i]] += r[i];
            add(l[i], r[i]);
        }
    }
    
    int Buf = 2000;
    for(int i = m; i >= 1; i --){
        if(tp[i] == 1){
            add(l[i], -r[i]);
            Modify(root, l[i], -r[i]);
        }
        else{
            P = Node(l[i], r[i], i, ask(r[i]) - ask(l[i]-1));
            Insert(root, 0);
        }
        if(i % Buf == 0)dfs_clock = 0, dfs(root), root = build(1, dfs_clock, 0);
    }
    
    dfs_clock = 0, dfs(root);
    for(int i = 1; i <= m; i ++)
        if(tp[i] == 2)printf("%lld\n", ans[i]);
    return 0;
}

  

 
E题:小Z的Tire

广义后缀自动机。倍增找到位置输出parent树中子树节点个数

#include <bits/stdc++.h>
#define maxn 2000010
using namespace std;

struct Node{int len, link, nxt[26];}st[maxn];

int root, size, last;

int s[maxn];

void init(){
    root = size = last = 0;
    st[root].len = 0;
    st[root].link = -1;
}

void Extend(char ch){
    int c = ch - 'a', p = last, q = st[p].nxt[c];
    if(q){
        if(st[q].len == st[p].len + 1)
            last = q;
        else{
            int clone = ++ size;
            st[clone] = st[q];
            st[clone].len = st[p].len + 1;
            for(; ~p && st[p].nxt[c] == q; p = st[p].link)
                st[p].nxt[c] = clone;
            st[q].link = clone;
            last = clone;
        }
    }
    else{
        int cur = ++ size;
        st[cur].len = st[p].len + 1;
        for(; ~p && !st[p].nxt[c]; p = st[p].link)
            st[p].nxt[c] = cur;
        if(p == -1)
            st[cur].link = root;
        else{
            q = st[p].nxt[c];
            if(st[q].len == st[p].len + 1)
                st[cur].link = q;
            else{
                int clone = ++ size;
                st[clone] = st[q];
                st[clone].len = st[p].len + 1;
                for(; ~p && st[p].nxt[c] == q; p = st[p].link)
                    st[p].nxt[c] = clone;
                st[q].link = st[cur].link = clone;
            }
        }
        last = cur;
    }
    s[last] = 1;
}

int n;

char c[maxn];

int anc[maxn][22];

int t[maxn], w[maxn];


void pre_anc(){
    for(int i = 1; i <= size; i ++)w[st[i].len] ++;
    for(int i = 1; i <= size; i ++)w[i] += w[i-1];
    for(int i = 1; i <= size; i ++)t[w[st[i].len] --] = i;
    for(int i = size; i >= 1; i --)s[st[t[i]].link] += s[t[i]];

    memset(anc, -1, sizeof anc);
    for(int i = 1; i <= size; i ++)
        anc[i][0] = st[i].link;
    for(int j = 1; 1 << j <= size; j ++){
        for(int i = 1; i <= size; i ++){
            int a = anc[i][j-1];
            if(~a)anc[i][j] = anc[a][j-1];
        }
    }
}

vector<int>V[100010];

int ask_pos(int r, int len){
    int now = r;
    for(int i = 21; i >= 0; i --){
        int t = anc[now][i];
        if(t == -1)continue;
        if(st[t].len >= len)now = t;
    }
    return now;
}

int main(){
    init();
    scanf("%d", &n);
    for(int i = 1; i <= n; i ++){
        scanf("%s", c+1);
        int N = strlen(c+1);
        last = root;
        for(int j = 1; j <= N; j ++){
            Extend(c[j]);
            V[i].push_back(last);
        }
    }
    
    pre_anc();

    int m, u, v, d;
    scanf("%d", &m);
    for(int i = 1; i <= m; i ++){
        scanf("%d%d%d", &u, &v, &d);
        printf("%d\n", s[ask_pos(V[u][d-1], d-v+1)]);
    }
    return 0;
}

 

  

 
posted @ 2016-03-29 17:02  _Horizon  阅读(468)  评论(0编辑  收藏  举报