KSGT

上午，强开女神，瘫软在地

下午，怒写体积交

晚上，试水匈牙利

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区间合并的问题，只是要维护两组的变量，这两组的变量有优先级的差异，才用的是两颗线段树各维护一组变量的技巧，减少了代码规模，其实想写一个基类，两个子类，继承重载的方法

注意！
1.在区间合并中，用query把长度转换成具体的[l, r]区间，可以大大较少update里面的思维复杂度
2.push_down 与 push_up只有全空与全满的情况，因为在几种update中只有全空、全满的更新，没有其他的
（正是因为陷入了半满的怪圈，才导致强上女神而不得）

#define _CRT_SECURE_NO_WARNINGS#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <vector>using namespace std;const int maxn = 100000 + 10;struct SegTree{    int l, r;    int left_time, max_time, right_time;    void init(int l, int r)    {        this->l = l, this->r = r;        left_time = max_time = right_time = r - l + 1;    }}goddess_segTree[maxn << 2], friend_segTree[maxn << 2];int t, n;void build(int k, int l, int r){    goddess_segTree[k].init(l, r);    friend_segTree[k].init(l, r);    if (l == r) return;    int m = (l + r) >> 1;    build(k << 1, l, m);    build(k << 1 | 1, m + 1, r); }int query(int k, SegTree * segTree, int len){    if (segTree[k].left_time >= len) return segTree[k].l;    if (segTree[k << 1].max_time >= len) return query(k << 1, segTree, len);    if (segTree[k << 1].right_time + segTree[k << 1 | 1].left_time >= len)        return segTree[k << 1].r - segTree[k << 1].right_time + 1;    return query(k << 1 | 1, segTree, len);}void push_down(int k){    if (friend_segTree[k].l == friend_segTree[k].r) return;    if (friend_segTree[k].max_time == 0)    {        friend_segTree[k << 1].left_time = friend_segTree[k << 1].max_time            = friend_segTree[k << 1].right_time = 0;        friend_segTree[k << 1 | 1].left_time = friend_segTree[k << 1 | 1].max_time            = friend_segTree[k << 1 | 1].right_time = 0;    }    if (friend_segTree[k].max_time == friend_segTree[k].r - friend_segTree[k].l + 1)    {        friend_segTree[k << 1].left_time = friend_segTree[k << 1].max_time = friend_segTree[k << 1].right_time            = friend_segTree[k << 1].r - friend_segTree[k << 1].l + 1;        friend_segTree[k << 1 | 1].left_time = friend_segTree[k << 1 | 1].max_time = friend_segTree[k << 1 | 1].right_time            = friend_segTree[k << 1 | 1].r - friend_segTree[k << 1 | 1].l + 1;    }    //_________________________________________________________________________    if (goddess_segTree[k].max_time == 0)    {        goddess_segTree[k << 1].left_time = goddess_segTree[k << 1].max_time            = goddess_segTree[k << 1].right_time = 0;        goddess_segTree[k << 1 | 1].left_time = goddess_segTree[k << 1 | 1].max_time            = goddess_segTree[k << 1 | 1].right_time = 0;    }    if (goddess_segTree[k].max_time == goddess_segTree[k].r - goddess_segTree[k].l + 1)    {        goddess_segTree[k << 1].left_time = goddess_segTree[k << 1].max_time = goddess_segTree[k << 1].right_time            = goddess_segTree[k << 1].r - goddess_segTree[k << 1].l + 1;        goddess_segTree[k << 1 | 1].left_time = goddess_segTree[k << 1 | 1].max_time = goddess_segTree[k << 1 | 1].right_time            = goddess_segTree[k << 1 | 1].r - goddess_segTree[k << 1 | 1].l + 1;    }}void push_up(int k){    if (friend_segTree[k].l == friend_segTree[k].r) return;    friend_segTree[k].left_time = friend_segTree[k << 1].left_time;    if (friend_segTree[k << 1].l + friend_segTree[k << 1].left_time - 1 == friend_segTree[k << 1].r)        friend_segTree[k].left_time += friend_segTree[k << 1 | 1].left_time;    friend_segTree[k].right_time = friend_segTree[k << 1 | 1].right_time;    if (friend_segTree[k << 1 | 1].r - friend_segTree[k << 1 | 1].right_time + 1 == friend_segTree[k << 1 | 1].l)        friend_segTree[k].right_time += friend_segTree[k << 1].right_time;    friend_segTree[k].max_time = max(friend_segTree[k << 1].max_time, friend_segTree[k << 1 | 1].max_time);    friend_segTree[k].max_time = max(friend_segTree[k].max_time,         friend_segTree[k << 1].right_time + friend_segTree[k << 1 | 1].left_time);    //_________________________________________________________________________    goddess_segTree[k].left_time = goddess_segTree[k << 1].left_time;    if (goddess_segTree[k << 1].l + goddess_segTree[k << 1].left_time - 1 == goddess_segTree[k << 1].r)        goddess_segTree[k].left_time += goddess_segTree[k << 1 | 1].left_time;    goddess_segTree[k].right_time = goddess_segTree[k << 1 | 1].right_time;    if (goddess_segTree[k << 1 | 1].r - goddess_segTree[k << 1 | 1].right_time + 1 == goddess_segTree[k << 1 | 1].l)        goddess_segTree[k].right_time += goddess_segTree[k << 1].right_time;    goddess_segTree[k].max_time = max(goddess_segTree[k << 1].max_time, goddess_segTree[k << 1 | 1].max_time);    goddess_segTree[k].max_time = max(goddess_segTree[k].max_time,        goddess_segTree[k << 1].right_time + goddess_segTree[k << 1 | 1].left_time);}void update(int k, int l, int r, int flag){    if (l > friend_segTree[k].r || r < friend_segTree[k].l) return;    if (l <= friend_segTree[k].l && friend_segTree[k].r <= r)    {        friend_segTree[k].left_time = friend_segTree[k].max_time            = friend_segTree[k].right_time = 0;        if (flag)        {            goddess_segTree[k].left_time = goddess_segTree[k].max_time                = goddess_segTree[k].right_time = 0;        }        return;    }    push_down(k);    update(k << 1, l, r, flag);    update(k << 1 | 1, l, r, flag);    push_up(k);}void solve_ds(int len){    if (friend_segTree[1].max_time >= len)    {        int beg = query(1, friend_segTree, len);        printf("%d,let's fly\n", beg);        update(1, beg, beg + len - 1, 0);        return;    }    puts("fly with yourself");}void solve_ns(int len){    if (friend_segTree[1].max_time >= len)    {        int beg = query(1, friend_segTree, len);        printf("%d,don't put my gezi\n", beg);        update(1, beg, beg + len - 1, 1);        return;    }    if (goddess_segTree[1].max_time >= len)    {        int beg = query(1, goddess_segTree, len);        printf("%d,don't put my gezi\n", beg);        update(1, beg, beg + len - 1, 1);        return;    }    puts("wait for me");}void update_study(int k, int l, int r){    if (l > friend_segTree[k].r || r < friend_segTree[k].l) return;    if (l <= friend_segTree[k].l && friend_segTree[k].r <= r)    {        friend_segTree[k].left_time = friend_segTree[k].max_time = friend_segTree[k].right_time            = friend_segTree[k].r - friend_segTree[k].l + 1;        goddess_segTree[k].left_time = goddess_segTree[k].max_time = goddess_segTree[k].right_time            = goddess_segTree[k].r - goddess_segTree[k].l + 1;        return;    }    push_down(k);    update_study(k << 1, l, r);    update_study(k << 1 | 1, l, r);    push_up(k);}void solve_study(int l, int r){    puts("I am the hope of chinese chengxuyuan!!");    update_study(1, l, r);}int main(){    int ttt; scanf("%d", &ttt);    int kase = 0;    while (ttt--)    {        printf("Case %d:\n", ++kase);        scanf("%d%d", &t, &n);        build(1, 1, t);        for (int i = 0; i < n; i++)        {            char cmd[10];            int a, b;            scanf("%s%d", cmd, &a);            if (cmd[0] == 'D') solve_ds(a);            else if (cmd[0] == 'N') solve_ns(a);            else if (cmd[0] == 'S')            {                scanf("%d", &b);                solve_study(a, b);            }        }    }    return 0;}

这种代码仿佛是在写项目….

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周长交 -> 面积交 -> 体积交
一维 二维 三维

一维的周长中，差值来计算

二维的面积中，pre_len * deta_h

三维的体积中，以体积层为枚举单位，利用当前层内的面积 * deta_z

主要难点在于枚举体积层，其他的，比如重合三次，因为刚刚完成了面积交，所以很顺畅

#define _CRT_SECURE_NO_WARNINGS#include <iostream>#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace std;typedef long long ll;const int maxn = 3000 + 10;struct Line{    ll l, r, h;    ll z1, z2;    ll flag;    Line(ll l = 0, ll r = 0, ll h = 0, ll z1 = 0, ll z2 = 0, ll flag = 0)    {        this->l = l, this->r = r, this->h = h, this->z1 = z1, this->z2 = z2, this->flag = flag;    }    bool operator < (const Line & rhs) const    {        return h < rhs.h;    }}line[maxn];struct SegTree{    ll l, r;    ll cnt;    ll sum1, sum2, sum3;    void init(ll l, ll r)    {        this->l = l, this->r = r;        sum1 = sum2 = sum3 = 0;        cnt = 0;    }}segTree[maxn << 2];ll new_x[maxn], new_z[maxn];ll cur1;void build(int k, ll l, ll r){    segTree[k].init(l, r);    if (l + 1 >= r) return;    ll m = (l + r) >> 1;    build(k << 1, l, m);    build(k << 1 | 1, m, r);}ll depress_x(){    vector<ll> pool;    for (int i = 0; i < cur1; i++)    {        pool.push_back(line[i].l);        pool.push_back(line[i].r);    }    sort(pool.begin(), pool.end());    pool.erase(unique(pool.begin(), pool.end()), pool.end());    for (int i = 0; i < cur1; i++)    {        ll cur;        cur = find(pool.begin(), pool.end(), line[i].l) - pool.begin();        new_x[cur] = line[i].l; line[i].l = cur;        cur = find(pool.begin(), pool.end(), line[i].r) - pool.begin();        new_x[cur] = line[i].r; line[i].r = cur;    }    return pool.size() - 1;}ll depress_z(){    vector<ll> pool;    for (int i = 0; i < cur1; i++)    {        pool.push_back(line[i].z1);        pool.push_back(line[i].z2);    }    sort(pool.begin(), pool.end());    pool.erase(unique(pool.begin(), pool.end()), pool.end());    for (int i = 0; i < cur1; i++)    {        ll cur;        cur = find(pool.begin(), pool.end(), line[i].z1) - pool.begin();        new_z[cur] = line[i].z1;        cur = find(pool.begin(), pool.end(), line[i].z2) - pool.begin();        new_z[cur] = line[i].z2;    }    return pool.size() - 1;}void chan_sum(int k){    if (segTree[k].cnt >= 3)    {        segTree[k].sum3 = new_x[segTree[k].r] - new_x[segTree[k].l];        segTree[k].sum2 = new_x[segTree[k].r] - new_x[segTree[k].l];        segTree[k].sum1 = new_x[segTree[k].r] - new_x[segTree[k].l];    }    else if (segTree[k].cnt == 2)    {        segTree[k].sum3 = segTree[k << 1].sum1 + segTree[k << 1 | 1].sum1;        segTree[k].sum2 = new_x[segTree[k].r] - new_x[segTree[k].l];        segTree[k].sum1 = new_x[segTree[k].r] - new_x[segTree[k].l];    }    else if (segTree[k].cnt == 1)    {        segTree[k].sum3 = segTree[k << 1].sum2 + segTree[k << 1 | 1].sum2;        segTree[k].sum2 = segTree[k << 1].sum1 + segTree[k << 1 | 1].sum1;        segTree[k].sum1 = new_x[segTree[k].r] - new_x[segTree[k].l];    }    else if (segTree[k].cnt == 0)    {        segTree[k].sum3 = segTree[k << 1].sum3 + segTree[k << 1 | 1].sum3;        segTree[k].sum2 = segTree[k << 1].sum2 + segTree[k << 1 | 1].sum2;        segTree[k].sum1 = segTree[k << 1].sum1 + segTree[k << 1 | 1].sum1;    }}void update(int k, ll l, ll r, ll c){    if (l > segTree[k].r || r < segTree[k].l) return;    if (l <= segTree[k].l && segTree[k].r <= r)    {        segTree[k].cnt += c;        chan_sum(k);        return;    }    if (segTree[k].l + 1 == segTree[k].r)    {        if (segTree[k].cnt >= 3)        {            segTree[k].sum3 = new_x[segTree[k].r] - new_x[segTree[k].l];            segTree[k].sum2 = new_x[segTree[k].r] - new_x[segTree[k].l];            segTree[k].sum1 = new_x[segTree[k].r] - new_x[segTree[k].l];        }        else if (segTree[k].cnt == 2)        {            segTree[k].sum3 = 0;            segTree[k].sum2 = new_x[segTree[k].r] - new_x[segTree[k].l];            segTree[k].sum1 = new_x[segTree[k].r] - new_x[segTree[k].l];        }        else if (segTree[k].cnt == 1)        {            segTree[k].sum3 = 0;            segTree[k].sum2 = 0;            segTree[k].sum1 = new_x[segTree[k].r] - new_x[segTree[k].l];        }        else if (segTree[k].cnt == 0)        {            segTree[k].sum3 = 0;            segTree[k].sum2 = 0;            segTree[k].sum1 = 0;        }        return;    }    update(k << 1, l, r, c);    update(k << 1 | 1, l, r, c);    chan_sum(k);}ll square(ll down_lim, ll up_lim){    ll pre_h = 0, pre_len = 0;    ll res = 0;    for (int i = 0; i < cur1; i++)    {        if (line[i].z1 <= down_lim && up_lim <= line[i].z2)        {            update(1, line[i].l, line[i].r, line[i].flag);            res += (line[i].h - pre_h) * pre_len;            pre_h = line[i].h;            pre_len = segTree[1].sum3;        }    }    return res;}ll solve(){    ll w_x = depress_x();    ll w_z = depress_z();    ll res = 0;    for (int i = 1; i <= w_z; i++)    {        build(1, 0, w_x);        ll down_lim = new_z[i - 1];        ll up_lim = new_z[i];        res += (up_lim - down_lim) * square(down_lim, up_lim);    }    return res;}int main(){    int t; scanf("%d", &t);    int kase = 0;    while (t--)    {        cur1 = 0;        int n; scanf("%d", &n);        for (int i = 0; i < n; i++)        {            ll xx1, yy1, zz1, xx2, yy2, zz2;            scanf("%lld%lld%lld%lld%lld%lld", &xx1, &yy1, &zz1, &xx2, &yy2, &zz2);            line[cur1++] = Line(xx1, xx2, yy1, zz1, zz2, 1);            line[cur1++] = Line(xx1, xx2, yy2, zz1, zz2, -1);        }        sort(line, line + cur1);        printf("Case %d: %lld\n", ++kase, solve());    }    return 0;}

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复习了匈牙利算法，难为我还能记得，然而建模上面缺乏经验，被队友随意摩擦