2024ccpc女生赛题解

考场上写的A，C，H，L，M
下来补一下剩下的

E

注意\(p[i]<i\)这个性质
和重心关系不大，一个简单的构造，手模几个样例就能发现规律。
倒着枚举：
\(c[i]=0\)的是叶子，不用处理
\(c[i]>0\)，这个点连到父亲所在连通块的根上即可。
并查集维护连通块以及连通块的根，根就是连通块中最小编号的点。

#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <cmath>
#include <vector>using namespace std;const int N=2e5+5;
#define LL long longint n;
int fa[N];
int find(int x) {return fa[x]==x?x:fa[x]=find(fa[x]);
}
int c[N];
vector<int>p[N];
void merge(int x,int y) {int fx=find(x),fy=find(y);if(fx>fy) swap(fx,fy);//id小的做根fa[fy]=fx;
}
bool vis[N];
int main() {int T;scanf("%d",&T);while(T--) {scanf("%d",&n);for(int i=1;i<=n;i++) {fa[i]=i;vis[i]=0;p[i].clear();}for(int i=1;i<=n;i++) {scanf("%d",&c[i]);if(c[i]==0)continue;for(int j=1,x;j<=c[i];j++) {scanf("%d",&x);p[i].push_back(x);}}for(int i=n;i;i--) {for(int j=0;j<c[i];j++) {int x=find(p[i][j]);printf("%d %d\n",i,x);merge(i,x);}}}return 0;
}

K 小凯的省奖之梦

（考场上根本没读这题）
大模拟，在暴力的基础上优化即可，就是考验码力

暴力想法是，两层循环，枚举买几个p，几个q，然后去check加上这些分后是否能拿到省奖，
算一下复杂度，
check是 nlogn，但由于要多次排序，常数挺大。
\(100*100*500*log(500)*const=5e7*const\)
t飞
第一次优化，二分q。在第22个点TLE。
第二次优化，发现可以分开处理两个学期的得奖情况，\(O(100*nlogn)\)处理好第一学期的，然后存在vector里，第二学期直接在第一学期基础上做就好。

AC代码：

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <vector>using namespace std;const int N = 505;
int rk[N];
int n;
struct Stu {int id;string nam;int a1, a2, b1, b2, c1, c2;//zhi ,de ,tiint sum1, sum2;int prize;//奖项分
} s[N];
vector<Stu> sav[101];
vector<Stu> v1;
bool cmp1(Stu x, Stu y) {return x.nam < y.nam;
}
char str[30];
Stu kk;
int fir, sec, thi;
int m, q, p;
void get_rk(vector<Stu> v) {v.push_back(kk);sort(v.begin(), v.end(), cmp1);int cnt = 0;for (auto x: v) {rk[x.id] = ++cnt;}
}bool cmp2(Stu x, Stu y) {if (x.sum1 == y.sum1) {return x.a1 == y.a1 ? rk[x.id] < rk[y.id] : x.a1 > y.a1;} else {return x.sum1 > y.sum1;}
}bool cmp3(Stu x, Stu y) {if (x.sum2 == y.sum2) {return x.a2 == y.a2 ? rk[x.id] < rk[y.id] : x.a2 > y.a2;} else {return x.sum2 > y.sum2;}
}bool cmp4(Stu x, Stu y) {if (x.prize == y.prize) {if (x.sum1 + x.sum2 == y.sum1 + y.sum2) {return x.a1 + x.a2 == y.a1 + y.a2 ? rk[x.id] < rk[y.id] : x.a1 + x.a2 > y.a1 + y.a2;} else {return x.sum1 + x.sum2 > y.sum1 + y.sum2;}} else {return x.prize > y.prize;}
}bool cmp_z1(Stu x, Stu y) {return x.a1 > y.a1;
}
bool cmp_z2(Stu x, Stu y) {return x.a2 > y.a2;
}
int rk_zhi[N];//智yu排序
int score1, score2, score3;void sinister1(vector<Stu> &v, int add, int fir, int sec, int thi) {kk.a1 += add;kk.sum1 += add;v.push_back(kk);sort(v.begin(), v.end(), cmp_z1);int nowrank = 1;for (int i = 0; i < n; i++, nowrank++) {if (i > 0 && v[i].a1 == v[i - 1].a1) {rk_zhi[v[i].id] = rk_zhi[v[i - 1].id];} elserk_zhi[v[i].id] = nowrank;}sort(v.begin(), v.end(), cmp2);nowrank = 0;for (auto &it: v) {nowrank++;it.prize = 0;if (fir && rk_zhi[it.id] <= score1) {it.prize += 15;fir--;continue;}if (sec && rk_zhi[it.id] <= score2) {it.prize += 10;sec--;continue;}if (thi && rk_zhi[it.id] <= score3) {it.prize += 5;thi--;continue;}}kk.a1 -= add;kk.sum1 -= add;
}
//拷贝
bool check(vector<Stu> v, int add, int fir, int sec, int thi) {for (int i = 0; i < n; i++) {if (kk.id == v[i].id) {v[i].a2 += add;v[i].sum2 += add;break;}}sort(v.begin(), v.end(), cmp_z2);int nowrank = 1;for (int i = 0; i < n; i++, nowrank++) {if (i > 0 && v[i].a2 == v[i - 1].a2) {rk_zhi[v[i].id] = rk_zhi[v[i - 1].id];} elserk_zhi[v[i].id] = nowrank;}sort(v.begin(), v.end(), cmp3);nowrank = 0;for (auto &it: v) {nowrank++;if (fir && rk_zhi[it.id] <= score1) {it.prize += 15;fir--;continue;}if (sec && rk_zhi[it.id] <= score2) {it.prize += 10;sec--;continue;}if (thi && rk_zhi[it.id] <= score3) {it.prize += 5;thi--;continue;}}sort(v.begin(), v.end(), cmp4);bool ok = 0;nowrank = 0;for (auto &it: v) {nowrank++;if (it.id == kk.id) {if (nowrank <= m) ok = 1;break;}}return ok;
}
int main() {scanf("%d", &n);kk.nam = "crazyzhk";for (int i = 1; i <= n; i++) {scanf("%s %d %d %d %d %d %d", str, &s[i].a1, &s[i].b1, &s[i].c1, &s[i].a2, &s[i].b2, &s[i].c2);s[i].nam = str;s[i].id = i;s[i].sum1 = s[i].a1 + s[i].b1 + s[i].c1;s[i].sum2 = s[i].a2 + s[i].b2 + s[i].c2;if (s[i].nam == kk.nam) {kk.a1 = s[i].a1;kk.a2 = s[i].a2;kk.b1 = s[i].b1;kk.b2 = s[i].b2;kk.c1 = s[i].c1;kk.c2 = s[i].c2;kk.id = i;kk.sum1 = s[i].sum1;kk.sum2 = s[i].sum2;} else {v1.push_back(s[i]);}}get_rk(v1);fir = floor(0.15 * (double) n);sec = floor(0.25 * (double) n);thi = floor(0.35 * (double) n);score1 = floor(0.25 * (double) n);score2 = floor(0.45 * (double) n);score3 = floor(0.75 * (double) n);bool flg = 0;int ans = 1e8;scanf("%d %d %d", &m, &p, &q);for (int i = 0; i + kk.a1 <= 100; i++) {sav[i] = v1;sinister1(sav[i], i, fir, sec, thi);int L = 0, R = 100 - kk.a2;while (L <= R) {int mid = (L + R) >> 1;if (check(sav[i], mid, fir, sec, thi)) {flg = 1;ans = min(ans, i * p + mid * q);R = mid - 1;} else {L = mid + 1;}}}if (flg) {printf("%d", ans);} else {puts("Surely next time");}return 0;
}