8.8集训
上午
考试
下午
讲图论，改题
第一题
给定一个长度为\(n\)的序列，求这个序列所有长度为\(k\)的子序列的最大值之和，答案对\(1e9+7\)取模
显然，观察题面发现是“子序列”，也就是可以不连续
所以我们首先排序，然后考虑\(a[i]\)对于答案造成的贡献，发现以\(a[i]\)为最大值的序列，\(i-1\)之前的东东他都可以造成贡献，变相的说，是在前\(i-1\)个数里面，选\(k-1\)个，然后这么多次方案，\(a[i]\)对于每个方案都有贡献，
于是答案呼之欲出啊：\(Ans = \sum_\limits {i = k} ^nC_{i-1}^{k-1}\times a[i]\)，可惜在考场上没想出来，一直在想数据结构单调栈之类的去维护他，最后没办法，打了二十分暴力走人
PS:估计只有我不太清楚\(C_n^m = C_{n-1}^{m} + C_{n-1}^{m-1}\)吧
小贴士:搞不清楚组合数的规律的时候，试试杨辉三角？
#include <bits/stdc++.h>
#define LL long long
#define debug
using namespace std;
const int N = 1e5+66, mod = 1e9+7;
int n, k;
int a[N];
LL c[N][66], res;
inline int thestars() {
cin >> n >> k;
for (int i = 1; i <= n; ++ i) scanf ("%d", &a[i]);
sort (a + 1, a + n + 1);
for (int i = 0; i <= n; ++ i) c[i][0] = 1;
for (int i = 1; i <= n; ++ i) {
for (int j = 1; j < k; ++ j) {
c[i][j] = (c[i - 1][j] + c[i - 1][j - 1])%mod;
}
}
for (int i = k; i <= n; ++ i) res = (res + c[i - 1][k - 1]*a[i])%mod;
cout << res;
return 0;
}
int youngore = thestars();
signed main() {;}
第二题
题目大意：在一个有\(n\)个站点的路上，对于\(1\leq i < n\)，站点\(i\)可以一票到达站点\(i+1,i+2, i+3......Min(i+a_i, n)\)
求任意两点\(i,j\)从\(i\text~j\)的最少票数和，即求\(\sum_\limits{i=1}^{n-1}\sum_\limits{j=i}^{n} p(i,j)\)其中\(p(i,j)\)表示\(i\text~j\)的最少票数
状态：设\(f[i]\)表示\(\sum_\limits{j = i+1}^np(i,j)\)，即\(i\text~j\)的最少票数和，
转移：
- 对于\((i,i+a_i\rbrack\)，我们可以一票卖完
- 对于\((i+a_i,n\rbrack\)，选择一个合适的\(j\)作为跳板，跳到终点，显然选择一个\({j+a[j]}_{max}\)是最合适不过的了
- 故当\(i+a_i=n\)时，\(f[i] = n-i\)，表示\(i+1 \text~n\)都可以一票走
- 而当\(i+a_i<n\)时，用树状数组或线段树查询\((i,i+a_i \rbrack\)中，\(j+a_i\)最大的\(j\)
- 转移呼之欲出：\(f[i] = a_i+f[j]-(a_i -j)+(n-a_i)\)
PS
\(a[i] = min(a[i]+i, n)\)
\(f[i] = a[i] - i\)
结果：\(Ans = \sum_\limits{i=1}^{n-1}f[i]\)
给出AC代码：
#include <bits/stdc++.h>
#define lowbit(x) (x&-x)
#define LL long long
#define debug
using namespace std;
const int N = 1e5+66;
int n, a[N], mp[N], t;
LL f[N], res;
inline void chenge(int x) {
for (int i = x; i <= n; i += lowbit(i)) {
if (a[mp[i]] < a[x]) {
mp[i] = x;
}
}
}
inline int ask(int x) {
int yhm(0);
for (int i = x; i; i -= lowbit(i)) {
if (a[yhm] < a[mp[i]]) {
yhm = mp[i];
}
}
return yhm;
}
inline int thestars() {
scanf ("%d", &n);
for (int i = 1; i <= n-1; ++ i) {
scanf ("%d", &a[i]);
a[i] = min (a[i]+i, n);
}
for (int i = n-1; i; -- i) {
f[i] = a[i] - i;
if (a[i] < n) t = ask(a[i]), f[i] += f[t] - (a[i] - t) + (n - a[i]);
chenge(i);
}
for (int i = 1; i <= n; ++ i) res += f[i];
cout << res;
return 0;
}
int youngore = thestars();
signed main() {;}
第三题
没时间了，贴上题解与代码吧
一句话题意：给定一个\(1\text~n\)的序列，\(m\)次询问，每次询问\(\lfloor l_i, r_i \rfloor\)中选出两个不同的数所得的\(gcd\)的最大值
其中\(1 \leq n,m \leq 1e5\)
给出代码：
#include <bits/stdc++.h>
#define LL long long
#define debug
using namespace std;
const int N = 1e5+66;
vector<int>lv[N], lp[N];
int n, m, p(1), tot;
int a[N], d[N], f[N];
int pos[N], ans[N];
struct node {
int l, r, id;
bool operator<(const node &a) const {return r < a.r;}
}q[N];
inline void add(int x, int val) {
for (int i = x; i; i -= (i&-i))
f[i] = max(f[i], val);
}
inline int chaxun(int x) {
int res(0);
for (int i = x; i <= n; i += (i&-x))
res = max(res, f[i]);
return res;
}
inline int thestars() {
cin >> n >> m;
for (int i = 1; i <= n; ++ i) {
cin >> a[i];
pos[a[i]] = i;
}
for (int i = 1; i <= m; ++ i) {
cin >> q[i].l >> q[i].r;
q[i].id = i;
}
for (int i = 1; i <= n; ++ i) {
int tot(0);
for (int j = i; j <= n; j += i) d[++ tot] = pos[j];
sort(d + 1, d + tot + 1);
for (int j = 2; j <= tot; ++ j) {
lp[d[j]].push_back(d[j - 1]);
lv[d[j]].push_back(i);
}
}
sort(q + 1, q + m + 1);
for (int i = 1; i <= n; ++ i) {
for (int j = 0; j < (int)lp[i].size(); ++ j) add(lp[i][j], lv[i][j]);
while (p <= m && q[p].r <= i) {
ans[q[p].id] = chaxun(q[p].l);
++ p;
}
}
for (int i = 1; i <= m; ++ i) printf("%d\n", ans[i]);
return 0;
}
int youngore = thestars();
signed main() {;}
晚上
做例题呗.....，还有一道题没有D出来
一个黄题这么操蛋.....
明星牛
近乎板子题，缩点之后，找出度为零的点\(x\)，答案就是\(x\)的\(size\)
PS：如果发现出度为零的\(x\)不止一个，那么\(res = 0\)
#include <bits/stdc++.h>
#define LL long long
#define debug
using namespace std;
const int N = 1e5+66;
int n, m, res;
struct node {int to, nxt;}e[N]; int head[N], num;
inline void add_edge(int u, int v) {
++ num;
e[num].to = v, e[num].nxt = head[u];
head[u] = num;
}
vector<int>scc[N];
int tp, tot, cnt;
int dfn[N], low[N], sta[N], in[N], tar[N], d[N];
//tar[i] has showed the i belong which SCC
inline void tarjan (int x) {
dfn[x] = low[x] = ++ tot; in[x] = 1, sta[++ tp] = x;
for (int i = head[x]; i; i = e[i].nxt) {
int y = e[i].to;
if (!dfn[y]) {
tarjan(y);
low[x] = min(low[x], low[y]);
} else if (in[y]) low[x] = min(low[x], dfn[y]);
}
if (dfn[x] == low[x]) {
int y; ++ cnt;
do {
y = sta[tp --]; in[y] = 0;
tar[y] = cnt;
scc[cnt].push_back(y);
} while (x != y);
}
}
inline int thestars() {
cin >> n >> m;
for (int i = 1; i <= m; ++ i) {
int u, v;
scanf ("%d%d", &u, &v);
add_edge(u, v);
}
for (int i = 1; i <= n; ++ i) {
if (!dfn[i]) {
tarjan(i);
}
}
for (int x = 1; x <= n; ++ x) {
for (int i = head[x]; i; i = e[i].nxt) {
int y = e[i].to;
if (tar[x] == tar[y]) continue;
else d[tar[x]] = 1;
}
}
for (int i = 1; i <= cnt; ++ i) {
if (!d[i]) {
if (res) {res = 0; break;}
res = scc[i].size();
}
}
cout << res;
return 0;
}
//the topic aim:find all the T.size(), and the T = ("chudu" == 0)
//how can I do?
int youngore = thestars();
signed main() {;}
佳肴
这个题是一个拓扑排序，我们只需要把图倒着连一遍，然后跑一遍最大拓扑排序即可，把序列倒着输出就好了
正常的排序用队列，这里用优先队列
证明：
1 尽量靠前，然后2尽量靠前... 在反图反序列中就相当于1尽量靠后，然后2尽量靠后...这就相当于：能不选1就不选1，然后能不选2就不选2...
等价于：选最大的，然后再选接下来的最大的...
PS：需要特判两个地方，第一个是是否有入度为零的点，第二个是拓扑排序完成之后，判断点的个数是否等于\(n\)
#include <bits/stdc++.h>
#define LL long long
#define debug
using namespace std;
const int N = 1e5+66;
int T, n, m, num, tp, flag;
int a[N], z[N];
int deg[N];
struct edge{int to, nxt;}e[N]; int head[N], cnt;
inline void add_edge(int u, int v) {
++ cnt;
e[cnt].to = v, e[cnt].nxt = head[u];
head[u] = cnt; ++ deg[v];
}
inline void topsort() {
priority_queue<int>q;
for (int i = 1; i <= n; ++ i) {
if (!deg[i]) {
q.push(i);
}
}
while (q.size()) {
flag = 1;
int x = q.top(); q.pop();
z[++ num] = x;
for (int i = head[x]; i; i = e[i].nxt) {
int y = e[i].to;
if (-- deg[y] == 0) {
q.push(y);
}
}
}
}
inline int thestars() {
cin >> T;
while (T --) {
flag = 0, num = 0, tp = 0, cnt = 0;
memset(a, 0, sizeof a);
memset(z, 0, sizeof z);
memset(head, 0, sizeof head);
memset(deg, 0, sizeof deg);
scanf ("%d%d", &n, &m);
for (int i = 1; i <= m; ++ i) {
int x, y;
scanf ("%d%d", &x, &y);
add_edge(y, x);
}
topsort();
if (num == n) {
for (int i = 1; i <= num; ++ i) a[num-i+1] = z[i];
for (int i = 1; i <= num; ++ i) cout << a[i] << ' ';
puts("");
} else puts("Impossible!");
}
return 0;
}
int youngore = thestars();
signed main() {;}
