【11.5校内测试】【倒计时5天】【DP】【二分+贪心check】【推式子化简+线段树】

Solution

非常巧妙的建立DP方程。

据dalao们说题目明显暗示根号复杂度？？（反正我是没看出来

因为每次分的块大小一定不超过$sqrt n$，要不然直接每个位置开一个块答案都才为$n$。

于是大佬们想到用一个非常巧妙的数组$pos[j]$，表示顺推到当前位置$i$时，以$i$作为右端点，区间出现了$j$个颜色的左端点的位置。

于是每次转移就变成了$dp[i]=min(dp[pos[j]-1]+j*j)$，而不需要把之前全部枚举。$j$的范围就是$<=sqrt n$的。

所以每次新到一个位置，就对于每个$j$看是否有新的贡献，记录$cnt[j]$表示$pos[j]$到$i$当前实际有多少个颜色。

如果$cnt[j]>j$表示当前$pos[j]$需要往后移动更新，那么每次往后一位一位暴力移动查询当前位置是否可以作为$pos[j]$，就是判断这个位置之后，$i$之前是否还出现了这个位置的颜色，如果出现了那么这个位置就不能作为$pos[j]$，因为它后面还有贡献。（据说均摊复杂度O(n)？？）

细节通过双向链表处理。

Code

#include<bits/stdc++.h>
#define LL long long
#define RG register
using namespace std;

LL dp[40005];
int nxt[40005], a[40005], pre[40005], las[40005], pos[40005], cnt[40006];
int n, m;

int main() {
    freopen("cleanup.in", "r", stdin);
    freopen("cleanup.out", "w", stdout);
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i ++) {
        scanf("%d", &a[i]);
        pre[i] = las[a[i]];
        nxt[las[a[i]]] = i;
        las[a[i]] = i;
        nxt[i] = n + 1;
    }
    memset(dp, 0x3f3f3f3f, sizeof(dp));    dp[0] = 0;
    int siz = sqrt(n);    for(int i = 1; i <= siz; i ++)    pos[i] = 1;
    for(RG int i = 1; i <= n; i ++) {
        for(RG int j = 1; j <= siz; j ++) {
            if(pre[i] < pos[j])    cnt[j] ++;
            if(cnt[j] > j) {
                cnt[j] --;
                while(nxt[pos[j]] < i)    pos[j] ++;
                pos[j] ++;
            }
            dp[i] = min(dp[pos[j] - 1] + j * j, dp[i]);
        }
    }
    printf("%lld", dp[n]);
    return 0;
}

Solution

二分性很明显，但是check竟然是用搜索回溯处理？？？真实震惊

搜索大概就是每次在剩下没有被覆盖的点中找到最靠左、最靠右、最靠下、最靠上四个极点位置，然后每次贪心往四个角（左上、左下、右上、右下）填矩阵，更新剩下没覆盖的点，搜三层回来即可.....

真的好玄学啊QAQ但是就是过了QAQ

Code

#include<bits/stdc++.h>
#define oo 0x3f3f3f3f
#define LL long long
using namespace std;

int n, xmi, xma, ymi, yma, MA;

struct Node {
    LL x, y;
} tr[20005];
bool cmp1(Node a, Node b) { return a.x < b.x; }
bool cmp2(Node a, Node b) { return a.y < b.y; }

bool flag, vis[20005];
LL now;

void dfs(int dep, int tot) {
    if(tot == n) { flag = 1;    return ; }
    if(dep == 3)    return ;
    bool used[20005] = {0};
    LL maxx = -oo, minx = oo, maxy = -oo, miny = oo;
    for(int i = 1; i <= n; i ++) {
        if(vis[i])    continue;
        maxx = max(maxx, tr[i].x);    minx = min(minx, tr[i].x);
        maxy = max(maxy, tr[i].y);    miny = min(miny, tr[i].y);
    }
    
    int cnt = 0;
    
    LL xl = minx, xr = minx + now, yl = miny, yr = miny + now;    cnt = 0;
    for(int i = 1; i <= n; i ++) {
        if(vis[i])    continue;
        if(tr[i].x <= xr && tr[i].x >= xl && tr[i].y >= yl && tr[i].y <= yr) {
            vis[i] = 1;    used[i] = 1; cnt ++;
        }
    }
    dfs(dep + 1, tot + cnt);    if(flag)    return ;
    for(int i = 1; i <= n; i ++)    if(used[i])    vis[i] = used[i] = 0;
    
    xl = minx, xr = minx + now, yl = maxy - now, yr = maxy;    cnt = 0;
    for(int i = 1; i <= n; i ++) {
        if(vis[i])    continue;
        if(tr[i].x <= xr && tr[i].x >= xl && tr[i].y >= yl && tr[i].y <= yr) {
            vis[i] = 1;    used[i] = 1; cnt ++;
        }
    }
    dfs(dep + 1, tot + cnt);    if(flag)    return ;
    for(int i = 1; i <= n; i ++)    if(used[i])    vis[i] = used[i] = 0;

    xl = maxx - now, xr = maxx, yl = maxy - now, yr = maxy;    cnt = 0;
    for(int i = 1; i <= n; i ++) {
        if(vis[i])    continue;
        if(tr[i].x <= xr && tr[i].x >= xl && tr[i].y >= yl && tr[i].y <= yr) {
            vis[i] = 1;    used[i] = 1; cnt ++;
        }
    }
    dfs(dep + 1, tot + cnt);    if(flag)    return ;
    for(int i = 1; i <= n; i ++)    if(used[i])    vis[i] = used[i] = 0;
    
    xl = maxx - now, xr = maxx, yl = miny, yr = miny + now;    cnt = 0;
    for(int i = 1; i <= n; i ++) {
        if(vis[i])    continue;
        if(tr[i].x <= xr && tr[i].x >= xl && tr[i].y >= yl && tr[i].y <= yr) {
            vis[i] = 1;    used[i] = 1; cnt ++;
        }
    }
    dfs(dep + 1, tot + cnt);    if(flag)    return ;
    for(int i = 1; i <= n; i ++)    if(used[i])    vis[i] = used[i] = 0;
}

bool check(LL mid) {
    now = mid;    flag = 0;
    for(int i = 1; i <= n; i ++)    vis[i] = 0;
    dfs(0, 0);
    return flag;
}

LL erfen() {
    LL l = 0, r = 100000000000, ans;
    while(l <= r) {
        LL mid = (l + r) >> 1;
        if(check(mid))    ans = mid, r = mid - 1;
        else    l = mid + 1;
    }
    return ans;
}

int main() {
    freopen("cover.in", "r", stdin);
    freopen("cover.out", "w", stdout);
    scanf("%d", &n);
    for(int i = 1; i <= n; i ++)
        scanf("%lld%lld", &tr[i].x, &tr[i].y);
    LL ans = erfen();
    printf("%lld", ans);
    return 0;
}

Solution

区间修改？？发现没那么简单QAQ，是求多个区间的和，没办法直接修改就快速求出答案。

所以推一推式子将$n^3$优化到$n^2$，要求的区间的区间和实际上可以由每条边的贡献算出来：$sum_{i=L}^{R}{v[i]*(i-L+1)*(R-i+1)}$，此处$R$是事先$--$了的，因为$R-R+1$的边不能被算入贡献。

化简得$sum{iv[i] * (L + R) - i^2v[i] + v[i] * (R - L * R + 1 - L)}$所以要维护的变量实际上只有$iv[i].i^2v[i].v[i]$三个的区间和即可，用一颗线段树结构体就好辣。

全都要开longlong才可以！！

Code

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

LL n, m;

LL gcd(LL a, LL b) {
    return b == 0 ? a : gcd(b, a % b);
}

struct Node {
    LL iv, v, iv2;
    Node operator + (const Node &a) const {
        Node c;
        c.iv = iv + a.iv;    c.v = v + a.v;    c.iv2 = iv2 + a.iv2;
        return c;
    }
} TR[400005];

LL tag[400005], iv[100005], iv2[100005];

void push_down(LL nd, LL l, LL r) {
    if(tag[nd]) {
        LL d = tag[nd], mid = (l + r) >> 1;
        TR[nd << 1].iv += d * (iv[mid] - iv[l - 1]);
        TR[nd << 1].iv2 += d * (iv2[mid] - iv2[l - 1]);
        TR[nd << 1].v += d * (mid - l + 1);
        TR[nd << 1 | 1].iv += d * (iv[r] - iv[mid]);
        TR[nd << 1 | 1].iv2 += d * (iv2[r] - iv2[mid]);
        TR[nd << 1 | 1].v += d * (r - mid);
        tag[nd << 1] += d, tag[nd << 1 | 1] += d;    tag[nd] = 0;
    }
}

void update(LL nd) {
    TR[nd] = TR[nd << 1] + TR[nd << 1 | 1];
}

void modify(LL nd, LL l, LL r, LL L, LL R, LL d) {
    if(l >= L && r <= R) {
        TR[nd].iv += d * (iv[r] - iv[l - 1]);
        TR[nd].iv2 += d * (iv2[r] - iv2[l - 1]);
        TR[nd].v += d * (r - l + 1);
        tag[nd] += d;
        return ;
    }
    push_down(nd, l, r);    LL mid = (l + r) >> 1;
    if(L <= mid)    modify(nd << 1, l, mid, L, R, d);
    if(R > mid)        modify(nd << 1 | 1, mid + 1, r, L, R, d);
    update(nd);
}

Node query(LL nd, LL l, LL r, LL L, LL R) {
    if(l >= L && r <= R)    return TR[nd];
    push_down(nd, l, r);
    LL mid = (l + r) >> 1;    Node ans;    ans.iv = ans.iv2 = ans.v = 0;
    if(L <= mid)    ans = ans + query(nd << 1, l, mid, L, R);
    if(R > mid)        ans = ans + query(nd << 1 | 1, mid + 1, r, L, R);
    return ans;
}

void init() {
    for(LL i = 1; i <= n; i ++) {
        iv[i] = iv[i - 1] + i;
        iv2[i] = iv2[i - 1] + i * i;
    }
}

int main() {
    freopen("roadxw.in", "r", stdin);
    freopen("roadxw.out", "w", stdout);
    scanf("%lld%lld", &n, &m);
    init();
    for(LL t = 1; t <= m; t ++) {
        char s[10];    LL L, R, V;
        scanf("%s", s);
        if(s[0] == 'C') {
            scanf("%lld%lld%lld", &L, &R, &V);
            modify(1, 1, n, L, R - 1, V);
        } else {
            scanf("%lld%lld", &L, &R);
            R --;
            Node a = query(1, 1, n, L, R);
            LL ans = a.iv * (L + R) - a.iv2 + a.v * (R - L * R + 1 - L);
            LL tot = R - L + 2;
            LL ans2 = tot * (tot - 1) / 2;
            LL d = gcd(ans, ans2);
            ans /= d, ans2 /= d;
            printf("%lld/%lld
", ans, ans2);
        }
    }
    return 0;
}