UFPE Starters Final Try-Outs 2020
作者：互联网
题目很简单。差点10题，非常可惜。
A：
签到题+4不应该。
solver：lzh
1 #include <bits/stdc++.h>
2 using namespace std;
3 typedef pair<int, int> pii;
4 typedef long long ll;
5 #define ff first
6 #define ss second
7
8 const int MAXSTRLEN = 500010;
9 int maxlen0 = 0, maxlen1 = 0;
10 char ch[2 * MAXSTRLEN];
11 int p[2 * MAXSTRLEN];
12 void manacher(char str[], int len) {
13 for (int i = 0; i < len; i++) {
14 ch[2 * i + 1] = '#';
15 ch[2 * i + 2] = str[i];
16 }
17 ch[0] = '$';
18 ch[2 * len + 1] = '#';
19 ch[2 * len + 2] = '\0';
20 len = len * 2 + 2;
21 p[0] = p[1] = 1;
22 int id = 1, mx = 2;
23 for (int i = 2; i < len; i++) {
24 int j = min(p[2 * id - i], mx - i);
25 while (ch[i - j] == ch[i + j])
26 j++;
27 p[i] = j;
28 if (i + p[i] > mx) {
29 id = i;
30 mx = i + p[i];
31 }
32 if ((p[i] - 1) % 2 == 0)
33 maxlen0 = max(maxlen0, p[i] - 1);
34 else
35 maxlen1 = max(maxlen1, p[i] - 1);
36 }
37 }
38 char s[MAXSTRLEN];
39 int main() {
40 int n, m;
41 scanf("%d%d%s", &n, &m, s);
42 manacher(s, n);
43 if ((m % 2 == 0 && maxlen0 >= m) || (m % 2 == 1 && maxlen1 >= m)) {
44 printf("Accept\n");
45 } else
46 printf("Reject\n");
47 }
View Code
B：
solver：zyh、czq
1 #include <bits/stdc++.h>
2 using namespace std;
3 struct Segment_Tree {
4 //constants and innerclass
5 struct node {
6 int l, r, lc, rc;
7 long long val;
8 };
9 static const int N = 500000;
10 static const int root = 0;
11
12 //variables
13 int num;
14 node tr[2 * N];
15
16 //methods
17 void setRange(int l, int r, int now = root) {
18 num = 0;
19 build(l, r, now);
20 }
21 void build(int l, int r, int now) {
22 tr[now].l = l; tr[now].r = r; tr[now].val = 0;
23 if (l < r) {
24 int mid = (l + r) >> 1;
25 ++num; tr[now].lc = num;
26 ++num; tr[now].rc = num;
27 build(l, mid, tr[now].lc);
28 build(mid + 1, r, tr[now].rc);
29 }
30 }
31
32 void pushup(node &f, node &lc, node &rc) {
33 f.val = lc.val + rc.val;
34 }
35
36 //warning: don't invoke both update and updateR in single segment tree, it may cause error
37 void update(int pos, long long val, int now = root) {
38 int mid = (tr[now].l + tr[now].r) >> 1;
39 if (tr[now].l == tr[now].r) {
40 tr[now].val = val;
41 return;
42 } else if (pos <= mid) update(pos, val, tr[now].lc);
43 else update(pos, val, tr[now].rc);
44 //write parent update here
45 pushup(tr[now], tr[tr[now].lc], tr[tr[now].rc]);
46 }
47
48 long long query(int l, int r, int now = root) {
49 if (tr[now].l == l && tr[now].r == r) return tr[now].val;
50 else {
51 //if (tr[now].cover!=0) pushdown(tr[now],tr[tr[now].lc],tr[tr[now].rc]);
52 int mid = (tr[now].l + tr[now].r) >> 1;
53 if (r <= mid) return query(l, r, tr[now].lc);
54 else if (l > mid) return query(l, r, tr[now].rc);
55 else return query(l, mid, tr[now].lc) + query(mid + 1, r, tr[now].rc);
56 }
57 }
58
59 };
60 Segment_Tree tr;
61 map<string, long long> Map;
62 string str[1000001];
63 char ch[10001];
64 int main() {
65 int n, m, q;
66 scanf("%d%d%d", &n, &m, &q);
67 for (int i = 1; i <= n; ++i) {
68 scanf("%s", ch);
69 str[i] = ch;
70 }
71 for (int i = 0; i < m; ++i) {
72 long long v;
73 scanf("%s%lld", ch, &v);
74 Map[string(ch)] = v;
75 }
76 tr.setRange(1, n);
77 for (int i = 1; i <= n; ++i) {
78 tr.update(i, Map[str[i]]);
79 }
80 while (q--) {
81 int op;
82 scanf("%d", &op);
83 if (op == 1) {
84 int x;
85 scanf("%d%s", &x, ch);
86 tr.update(x, Map[string(ch)]);
87 } else {
88 int l, r;
89 scanf("%d%d", &l, &r);
90 long long s = tr.query(l, r);
91 if (s <= 30 * (r - l + 1)) printf("NO\n");
92 else printf("YES\n");
93 }
94 }
95 }
View Code
D：
solver：lzh、czq
1 #include <bits/stdc++.h>
2 using namespace std;
3 typedef long long ll;
4 typedef pair<double, int> pii;
5 #define ff first
6 #define ss second
7 #define mp make_pair
8
9 pii p[100010];
10 int check(double x, int n) {
11 int cur = 1;
12 while (cur <= n) {
13 ll add = 0;
14 while (p[cur].ff - x <= 0)
15 add += p[cur++].ss;
16 if (add == 0)
17 return 0;
18 x += add;
19 }
20 return 1;
21 }
22 int main() {
23 int n, x, y;
24 scanf("%d%d%d", &n, &x, &y);
25 for (int i = 1; i <= n; i++) {
26 int u, v, w;
27 scanf("%d%d%d", &u, &v, &w);
28 p[i] = mp(sqrt(1ll * (u - x) * (u - x) + 1ll * (v - y) * (v - y)) - w, w);
29 }
30 sort(p + 1, p + 1 + n);
31
32 double l = max(0.0, p[1].ff), r = p[n].ff, eps = 1e-8;
33 double ans = r;
34 while (r - l > eps) {
35 double mid = (l + r) / 2.0;
36 if (check(mid, n)) {
37 r = mid;
38 ans = min(ans, mid);
39 } else
40 l = mid;
41 }
42 printf("%.10lf\n", ans);
43 }
View Code
E：
solver：zyh
1 #include <bits/stdc++.h>
2 using namespace std;
3 int li[] = {6, 28, 496, 8128};
4 int len = 4;
5 inline int read() {
6 int rnt = 0;
7 int sign = 0;
8 char ch = 0;
9 while (!isdigit(ch)) {
10 sign |= ch == '-';
11 ch = getchar();
12 }
13 while (isdigit(ch)) rnt = (rnt << 3) + (rnt << 1) + (ch ^ 48), ch = getchar();
14 return sign ? -rnt : rnt;
15 }
16 inline void print(int x) {
17 if (x < 0) {
18 putchar('-');
19 x = -x;
20 }
21 if (x > 9) print(x / 10);
22 putchar(x % 10 + '0');
23 }
24 int main() {
25 int n = read();
26 while (n--) {
27 int k = read();
28 int ans;
29 for (ans = len - 1; ans >= 0; --ans)
30 if (li[ans] <= k) break;
31 if (ans < 0) print(-1);
32 else print(li[ans]);
33 putchar('\n');
34 }
35 }
View Code
F：
solver：lzh
1 #include <bits/stdc++.h>
2 using namespace std;
3 typedef pair<int, int> pii;
4 typedef long long ll;
5 #define ff first
6 #define ss second
7
8 struct node {
9 int l, r, h;
10 int add;
11 node() {}
12 node(int _l, int _r, int _h, int _add): l(_l), r(_r), h(_h), add(_add) {}
13 };
14
15 const int N = 2e5 + 10;
16 int x[N];
17 ll sum[N << 2];
18 node q[N];
19 int mark[N << 2];
20
21 void pushup(int n, int left, int right) {
22 if (mark[n])sum[n] = x[right + 1] - x[left];
23 else if (left == right)sum[n] = 0;
24 else sum[n] = sum[n << 1] + sum[n << 1 | 1];
25 }
26
27 void pushdown(int n) {
28 mark[n << 1] += mark[n], mark[n << 1 | 1] += mark[n];
29 mark[n] = 0;
30 }
31
32 void update(int l, int r, int add, int n, int left, int right) {
33 if (l <= left && right <= r) {
34 mark[n] += add;
35 pushup(n, left, right);
36 return;
37 }
38 int mid = left + right >> 1;
39 if (l <= mid)update(l, r, add, n << 1, left, mid);
40 if (mid < r)update(l, r, add, n << 1 | 1, mid + 1, right);
41 pushup(n, left, right);
42 }
43
44 int main() {
45 int n;
46 scanf("%d", &n);
47 for (int i = 1; i <= n; i++) {
48 int l, r, h; scanf("%d%d%d", &l, &r, &h);
49 x[2 * i - 1] = l; x[2 * i] = r;
50 q[2 * i - 1] = node(l, r, 0, 1);
51 q[2 * i] = node(l, r, h, -1);
52 }
53 sort(x + 1, x + 1 + 2 * n);
54 sort(q + 1, q + 1 + 2 * n, [](node a, node b) {
55 return a.h < b.h;
56 });
57 int cnt = 1;
58 for (int i = 2; i <= 2 * n; i++)
59 if (x[i - 1] != x[i])x[++cnt] = x[i];
60 ll ans = 0;
61 q[2 * n + 1].h = q[2 * n].h;
62
63 for (int i = 1; i <= 2 * n; i++) {
64 int l = lower_bound(x + 1, x + 1 + cnt, q[i].l) - x;
65 int r = lower_bound(x + 1, x + 1 + cnt, q[i].r) - x - 1;
66 update(l, r, q[i].add, 1, 1, cnt);
67 ans += sum[1] * (q[i + 1].h - q[i].h);
68 }
69 printf("%lld\n", ans);
70 }
View Code
G：
solver：lzh、czq
1 #include <bits/stdc++.h>
2 using namespace std;
3 typedef pair<int, int> pii;
4 typedef long long ll;
5 #define ff first
6 #define ss second
7
8 map<string, int>name, prob;
9 char strname[50010][25];
10 int pts[50010];
11 set<int>submit[50010];
12 int main() {
13 int c, p, s; scanf("%d%d%d", &c, &p, &s);
14 int cname = 0, cprob = 0;
15 char str[50];
16 for (int i = 1; i <= c; i++) {
17 scanf("%s", strname[i]);
18 name[strname[i]] = ++cname;
19 }
20 for (int i = 1; i <= p; i++) {
21 int x;
22 scanf("%s%d", str, &x);
23 prob[str] = ++cprob; pts[cprob] = x;
24 }
25 for (int i = 1; i <= s; i++) {
26 int numname = -1, numprob = -1;
27 scanf("%s", str);
28 numname = name[str];
29 scanf("%s", str);
30 numprob = prob[str];
31 scanf("%s", str);
32 if (strlen(str) == 2 && str[0] == 'A' && str[1] == 'C') {
33 if (numname && numprob) {
34 submit[numname].insert(numprob);
35 }
36 }
37 }
38 for (int i = 1; i <= c; i++) {
39 int ans = 0;
40 for (auto j : submit[i])ans += pts[j];
41 printf("%s %d\n", strname[i], ans);
42 }
43 }
View Code
H：
solver：lzh、czq
1 #include <bits/stdc++.h>
2 using namespace std;
3 typedef long long ll;
4 typedef pair<ll, int> pii;
5 #define ff first
6 #define ss second
7 #define mp make_pair
8
9 const int N = 1e5 + 10;
10 vector<pii> v[N];
11 int T[N], vis[N];
12 ll ans[N];
13 int main() {
14 int n, m, k;
15 scanf("%d%d%d", &n, &m, &k);
16 for (int i = 1; i <= m; i++) {
17 int x, y, z;
18 scanf("%d%d%d", &x, &y, &z);
19 v[x].push_back(mp(y, z));
20 v[y].push_back(mp(x, z));
21 }
22 for (int i = 1; i <= n; i++)
23 scanf("%d", &T[i]);
24
25 priority_queue<pii, vector<pii>, greater<pii>> pq;
26 for (int i = 1; i <= k; i++) {
27 int x;
28 scanf("%d", &x);
29 pq.push(mp(T[x], x));
30 }
31
32 while (!pq.empty()) {
33 pii x = pq.top();
34 pq.pop();
35 if (vis[x.ss])
36 continue;
37 vis[x.ss]++;
38 ans[x.ss] = x.ff;
39
40 for (auto i : v[x.ss])
41 if (!vis[i.ff])
42 pq.push(mp(x.ff + i.ss + T[i.ff], i.ff));
43 }
44 for (int i = 1; i <= n; i++)
45 printf("%lld\n", ans[i]);
46 }
View Code
I：
solver：zyh、czq
1 #include <bits/stdc++.h>
2 using namespace std;
3 struct point {
4 int x, v;
5 point() {}
6 point(int _x, int _v) {
7 x = _x;
8 v = _v;
9 }
10 bool operator<(const point &b)const {
11 return v < b.v;
12 }
13 };
14 point pts[2000000];
15 int f[2000000];
16 int Size[2000000];
17 int a[2000000];
18 int find(int u) {
19 if (f[u] != u) f[u] = find(f[u]);
20 return f[u];
21 }
22 int Union(int u, int v) {
23 u = find(u);
24 v = find(v);
25 if (u != v) {
26 Size[u] += Size[v];
27 f[v] = u;
28 }
29 return u;
30 }
31 int main() {
32 int s, n, m;
33 scanf("%d%d%d", &s, &n, &m);
34 int len = n * m;
35 for (int i = 0; i < len; ++i) {
36 scanf("%d", &a[i]);
37 pts[i] = point(i, a[i]);
38 }
39 sort(pts, pts + len);
40 for (int i = 0; i < len; ++i) {
41 f[i] = i; Size[i] = 1;
42 }
43 int maxsize = 0;
44 for (int k = len - 1; k >= 0; --k) {
45 int i = pts[k].x;
46 int v = pts[k].v;
47 if (i - 1 >= 0 && (i / m) == ((i - 1) / m) && a[i - 1] >= v) Union(i, i - 1);
48 if (i + 1 < len && (i / m) == ((i + 1) / m) && a[i + 1] >= v) Union(i, i + 1);
49 if (i + m < len && a[i + m] >= v) Union(i, i + m);
50 if (i - m >= 0 && a[i - m] >= v) Union(i, i - m);
51 maxsize = max(maxsize, Size[find(i)]);
52 if (maxsize >= s) {
53 printf("%d\n", v);
54 return 0;
55 }
56 }
57 }
View Code
J：
solver：lzh
差一点就做出来了，非常难顶
1 #include <bits/stdc++.h>
2 using namespace std;
3 typedef pair<int, int> pii;
4 typedef long long ll;
5 #define ff first
6 #define ss second
7 #define mp make_pair
8
9 vector<pii> v[100010];
10 ll dp[100010][10];
11 const ll mod = 1e9 + 7;
12 int baoli2[20][2] = { 1, 2, 1, 3, 1, 4, 1, 5, 2, 1, 2, 3, 2, 4, 2, 5, 3, 1, 3, 2, 3, 4, 3, 5, 4, 1, 4, 2, 4, 3, 4, 5, 5, 1, 5, 2, 5, 3, 5, 4 };
13 int baoli3[60][3] = { 1, 2, 3, 1, 2, 4, 1, 2, 5, 1, 3, 2, 1, 3, 4, 1, 3, 5, 1, 4, 2, 1, 4, 3, 1, 4, 5, 1, 5, 2, 1, 5, 3, 1, 5, 4, 2, 1, 3, 2, 1, 4, 2, 1, 5, 2, 3, 1, 2, 3, 4, 2, 3, 5, 2, 4, 1, 2, 4, 3, 2, 4, 5, 2, 5, 1, 2, 5, 3, 2, 5, 4, 3, 1, 2, 3, 1, 4, 3, 1, 5, 3, 2, 1, 3, 2, 4, 3, 2, 5, 3, 4, 1, 3, 4, 2, 3, 4, 5, 3, 5, 1, 3, 5, 2, 3, 5, 4, 4, 1, 2, 4, 1, 3, 4, 1, 5, 4, 2, 1, 4, 2, 3, 4, 2, 5, 4, 3, 1, 4, 3, 2, 4, 3, 5, 4, 5, 1, 4, 5, 2, 4, 5, 3, 5, 1, 2, 5, 1, 3, 5, 1, 4, 5, 2, 1, 5, 2, 3, 5, 2, 4, 5, 3, 1, 5, 3, 2, 5, 3, 4, 5, 4, 1, 5, 4, 2, 5, 4, 3 };
14
15 void dfs(int x, int pre) {
16 int col = 0;
17 vector<int> nex;
18 for (auto i : v[x])
19 if (i.ff != pre) {
20 dfs(i.ff, x);
21 nex.push_back(i.ff);
22 } else
23 col = i.ss;
24 int sz = v[x].size();
25 if (pre == -1)
26 return;
27
28 if (sz == 1) {
29 if (v[x][0].ss)
30 dp[x][v[x][0].ss] = 1;
31 else
32 for (int i = 1; i <= 5; i++)
33 dp[x][i] = 1;
34 return;
35 } else if (sz == 2) {
36 for (int i = 0; i < 20; i++)
37 if (!col || col == baoli2[i][1]) {
38 dp[x][baoli2[i][1]] = (dp[x][baoli2[i][1]] + dp[nex[0]][baoli2[i][0]]) % mod;
39 }
40 } else if (sz == 3) {
41 for (int i = 0; i < 60; i++)
42 if (!col || col == baoli3[i][2]) {
43 dp[x][baoli3[i][2]] = (dp[x][baoli3[i][2]] + dp[nex[0]][baoli3[i][0]] * dp[nex[1]][baoli3[i][1]] % mod) % mod;
44 }
45 } else {
46 int a[10];
47 for (int i = 1; i <= 5; i++)
48 a[i] = i;
49 do {
50 if (!col || col == a[sz]) {
51 ll tmp = 1;
52 for (int i = 1; i <= sz - 1; i++)
53 tmp = tmp * dp[nex[i - 1]][a[i]] % mod;
54
55 dp[x][a[sz]] = (dp[x][a[sz]] + tmp) % mod;
56 }
57 } while (next_permutation(a + 1, a + 1 + 5));
58 }
59 }
60 int main() {
61 int n;
62 scanf("%d", &n);
63 for (int i = 1; i <= n - 1; i++) {
64 int a, b, c;
65 scanf("%d%d%d", &a, &b, &c);
66 v[a].push_back(mp(b, c));
67 v[b].push_back(mp(a, c));
68 }
69 for (int i = 1; i <= n; i++)
70 if (v[i].size() > 5) {
71 printf("0\n");
72 return 0;
73 }
74 if (n == 1) {
75 printf("1\n");
76 return 0;
77 }
78 if (n == 2) {
79 if (v[1][0].ss)
80 printf("1\n");
81 else
82 printf("5\n");
83 return 0;
84 }
85 dfs(1, -1);
86
87 ll ans = 0;
88 vector<int> nex;
89 for (auto i : v[1])
90 nex.push_back(i.ff);
91 if (v[1].size() == 1) {
92 for (int i = 1; i <= 5; i++) {
93 ans = (ans + dp[nex[0]][i]) % mod;
94 }
95 } else if (v[1].size() == 2) {
96 for (int i = 0; i < 20; i++) {
97 ll tmp = 1;
98 for (int j = 0; j < 2; j++)
99 tmp = tmp * dp[nex[j]][baoli2[i][j]] % mod;
100 ans = (ans + tmp) % mod;
101 }
102 } else if (v[1].size() == 3) {
103 for (int i = 0; i < 60; i++) {
104 ll tmp = 1;
105 for (int j = 0; j < 3; j++)
106 tmp = tmp * dp[nex[j]][baoli3[i][j]] % mod;
107 ans = (ans + tmp) % mod;
108 }
109 } else {
110 int a[10], sz = v[1].size();
111 for (int i = 1; i <= 5; i++)
112 a[i] = i;
113 do {
114 ll tmp = 1;
115 for (int i = 1; i <= sz; i++)
116 tmp = tmp * dp[nex[i - 1]][a[i]] % mod;
117 ans = (ans + tmp) % mod;
118 } while (next_permutation(a + 1, a + 1 + 5));
119 }
120
121 printf("%lld\n", ans);
122 }
View Code
K：
solver：lzh、czq
1 #include <bits/stdc++.h>
2 using namespace std;
3 typedef pair<int, int> pii;
4 typedef long long ll;
5 #define ff first
6 #define ss second
7
8 const int N = 1e6 + 10;
9 vector<ll> dp[N];
10 vector<int> a[N];
11 int main() {
12 int n, h; scanf("%d%d", &n, &h);
13 for (int i = 1; i <= n; i++) {
14 a[i].resize(h + 1);
15 dp[i].resize(h + 1);
16 for (int j = 1; j <= h; j++)
17 scanf("%d", &a[i][j]);
18 dp[i][1] = a[i][1];
19 }
20
21 a[0].resize(h + 1); a[n + 1].resize(h + 1);
22 dp[0].resize(h + 1); dp[n + 1].resize(h + 1);
23 for (int i = 1; i <= h; i++)a[0][i] = a[n + 1][i] = dp[0][i] = dp[n + 1][i] = 0;
24 for (int j = 2; j <= h; j++) {
25 for (int i = 1; i <= n; i++) {
26 dp[i][j] = a[i][j] + max(max(dp[i - 1][j - 1], dp[i + 1][j - 1]), dp[i][j - 1]);
27 }
28 }
29 ll maxx = 0;
30 for (int i = 1; i <= n; i++)maxx = max(maxx, dp[i][h]);
31 printf("%lld\n", maxx);
32 }
View Code
标签：UFPE,Outs,10,int,tr,long,Starters,now,define 来源： https://www.cnblogs.com/JHSeng/p/12193790.html