从noip之后，自己的状态就非常不好。学文化课的时候心里总想着竞赛，真的停课了却又异常浮躁，简单题不想做，难题不会做，错了的题也不改，天天打游戏，静不下心。省选也是各种瞎打，day2T1竟然会因为没有memset掉了40分。虽说苟进了队，但毕竟进的不光彩，若是还以这样的状态下去，潦草离场，那当初选择这条路的意义又在哪里？
得开始好好学习了啊......
4.16
正式停课的第一天~
昨天的HNOI考的巨炸，今天就先改改题
（老吴一大早分析了一波局势，dalao_cxy被寄予厚望，Orz，要向他学习）
~~~~~
只改了Day2两道题 http://www.cnblogs.com/XYZinc/p/8854765.html
还做了一道@cxy出的dp
今天效率好低啊
4.17
今天就主要打一些基础算法调整一下状态
~~~~~
- 贪心
http://www.cnblogs.com/XYZinc/p/8862632.html
很容易想到贪心，对于所有未处理的点，在它的第k祖先处处理必然是最优的
f[i][j]表示节点i的子树中，距离为j且需要灭火器的点数
r[i][j]表示从节点i出发，可以向距离不大于j的节点分配的灭火器个数
1 #include <bits/stdc++.h> 2 using namespace std; 3 #define N 100005 4 #define ll long long 5 6 inline ll read() 7 { 8 ll x = 0, f = 1; char ch = getchar(); 9 while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();} 10 while(ch >= '0' && ch <= '9') {x = x * 10 + ch - '0'; ch = getchar();} 11 return x * f; 12 } 13 14 int n, s, k, ans; 15 vector<int> to[N]; 16 17 ll f[N][25], r[N][25]; 18 void dfs(int x, int fa) 19 { 20 f[x][0] = 1; 21 for(int i = 0; i < to[x].size(); i++) 22 { 23 int kk = to[x][i]; 24 if(fa == kk) continue; 25 dfs(kk, x);//从叶子节点向上dp 26 for(int j = 1; j <= k; j++) 27 f[x][j] += f[kk][j - 1], r[x][j - 1] += r[kk][j]; 28 } 29 if(f[x][k]) 30 { 31 int t = f[x][k] / s + (f[x][k] % s != 0); 32 ans += t; r[x][k] += t * s; 33 } 34 int p = k; 35 for(int i = k; i >= 0; i--)//从下到上处理能覆盖的点 36 while(f[x][i]) 37 { 38 while(!r[x][p] && p >= i) p--; 39 if(p < i) break;//不能继续处理，退出 40 int t = min(r[x][p], f[x][i]); 41 r[x][p] -= t; f[x][i] -= t; 42 } 43 } 44 45 int main() 46 { 47 n = read(); s = read(); k = read(); 48 for(int i = 1; i < n; i++) 49 { 50 int x = read(), y = read(); 51 to[x].push_back(y); to[y].push_back(x); 52 } 53 dfs(1, 0); 54 ll cnt = 0;//最后一起处理距离根k以内的节点 55 for(int i = 0; i <= k; i++) cnt += f[1][i]; 56 ans += cnt / s + (cnt % s != 0); 57 cout << ans; 58 return 0; 59 }
//在x上方给出n个点，要求用最少个数的半径为i，圆心在x轴上的圆覆盖所有点，输出最少需要的圆的个数，无法覆盖输出-1,多组测试数据 #include <iostream> #include <cstdio> #include <cmath> #include <algorithm> #include <cstdlib> #include <cstring> using namespace std; const int N = 1005; const int inf = 2000000000; int n, ans, T; int d, x[N], y[N]; struct node { double l, r; }s[N]; bool cmp(node a, node b) { if(a.l == b.l) return a.r < b.r; return a.l < b.l; } int main() { while(scanf("%d%d", &n, &d) && n && d) { T++; bool fail = false; ans = 0; for(int i = 1; i <= n; i++) { scanf("%d%d", &x[i], &y[i]); if(y[i] > d) fail = true; } if(fail) {printf("Case %d: -1 ", T);continue;} //问题转化为在x轴上选出尽量少的点，使所有区间都包含至少一个点} for(int i = 1; i <= n; i++) { double t = sqrt(1.0 * d * d - 1.0 * y[i] * y[i]); s[i].l = (double)x[i] - t; s[i].r = (double)x[i] + t; } sort(s + 1, s + 1 + n, cmp); double pos = -inf;//pos记录上一个点的位置 for(int i = 1; i <= n; i++) { if(s[i].l > pos)//若两个区间没有交集，点数+1 { ans++; pos = s[i].r; } else if(s[i].r < pos) pos = s[i].r; } printf("Case %d: %d ", T, ans); } return 0; }
- 分治
emmm......读入WA了2次
难受
因为X可以填入任意实数，所以填X可以改变逆序对的奇偶，即最后填X的人能直接获胜，判一下X的奇偶即可
若序列中不存在X，胜负已定，直接计算逆序对个数
特别地，对于 n = 1 的情况，qmqmqm 必输
#include <bits/stdc++.h>
using namespace std;
#define N 100005
int n,cnt;
int a[N], t[N];
void mergesort(int l, int r)
{
if(l == r) {t[l] = a[l]; return;}
int mid = (l + r) >> 1;
mergesort(l, mid); mergesort(mid + 1, r);
int i = l, j = mid + 1, k = l;
while(i <= mid && j <= r)
{
if(a[i] > a[j])
{
t[k++] = a[j++];
cnt += (mid - i + 1) % 2;
}
else t[k++] = a[i++];
}
while(i <= mid) t[k++] = a[i++];
while(j <= r) t[k++] = a[j++];
for(int i = l; i <= r; i++) a[i] = t[i];
}
int main()
{
scanf("%d", &n);
if(n == 1) {cout << "L"; return 0;}
for(int i = 1; i <= n; i++)
{
char s[15]; scanf("%s", s);
if(s[0] == 'X') cnt++;
else
{
int x = 0, f = 1, l = strlen(s);
for(int j = 0; j < l; j++)
{
if(s[j] == '-') f = -1;
else x = x * 10 + s[j] - '0';
}
a[++a[0]] = x * f;
}
}
if(!cnt) mergesort(1, a[0]);
if(cnt % 2) cout << "W"; else cout << "L";
return 0;
}
~~~~~
今天效率还是好低啊
4.18
这两天再水一下dp和数据结构，就得开始学习新东西了
~~~~~
#include <bits/stdc++.h> using namespace std; const int N = 100005; int a[N], f[N]; int main() { int x; while(scanf("%d", &x) != EOF) {a[++a[0]] = x;} for(int i = a[0]; i > 0; i--) { if(!f[0] || a[i] >= f[f[0]]) f[++f[0]] = a[i]; else { int p = upper_bound(f + 1, f + 1 + f[0], a[i]) - f; f[p] = a[i]; } } printf("%d ", f[0]); f[0] = 0; for(int i = 1; i <= a[0]; i++) { if(!f[0] || a[i] > f[f[0]]) f[++f[0]] = a[i]; else { int p = lower_bound(f + 1, f + 1 + f[0], a[i]) - f; f[p] = a[i]; } } printf("%d", f[0]); return 0; }
mn[i][j]表示合并完区间[i, j]的最小分数
很容易想到： mn[i][j] = min(mn[i][k] + mn[k + 1][j] + sum(i, j)); (i <= k < j)
考虑递推顺序，得到mn[i][j]的前提是已经提前算出[i, j]子区间的mn值，枚举 i 或 j 并不是很容易处理
可以想到枚举区间长度，再用 i 算出 j 值，这样便可以确保所需状态已经求出
最大值同理
#include <bits/stdc++.h>
using namespace std;
const int N = 205;
const int inf = 1e9;
int n, a[N], sum[N];
int mn[N][N], mx[N][N];//mn[i][j],mx[i][j]分别代表区间[i,j]合并的最 小/大 得分
int main()
{
scanf("%d", &n);
for(int i = 1; i <= n; i++) {scanf("%d", &a[i]); a[i + n] = a[i];}
for(int i = 1; i <= 2 * n; i++) sum[i] = sum[i - 1] + a[i];
for(int l = 1; l <= n; l++)
for(int i = 1, j = i + l; j < 2 * n; i++, j++)
{
mn[i][j] = inf;
for(int k = i; k < j; k++)
{
mn[i][j] = min(mn[i][j], mn[i][k] + mn[k + 1][j] + sum[j] - sum[i - 1]);
mx[i][j] = max(mx[i][j], mx[i][k] + mx[k + 1][j] + sum[j] - sum[i - 1]);
}
}
int ans[2] = {inf, 0};
for(int i = 1; i <= n; i++)
{
ans[0] = min(ans[0], mn[i][i + n - 1]);
ans[1] = max(ans[1], mx[i][i + n - 1]);
}
cout << ans[0] << endl << ans[1];
return 0;
}
#include <bits/stdc++.h>
using namespace std;
const int N = 105;
const int inf = 1e9;
const int sim[5][5] = { {5, -1, -2, -1, -3},
{-1, 5, -3, -2, -4},
{-2, -3, 5, -2, -2},
{-1, -2, -2, 5, -1},
{-3, -4, -2, -1, 0} };
int n, m, a[N], b[N];
int f[N][N];
char s[N];
void get(int l, int T[N])
{
for(int i = 0; i < l; i++)
{
if(s[i] == 'A') T[i + 1] = 0;
else if(s[i] == 'C') T[i + 1] = 1;
else if(s[i] == 'G') T[i + 1] = 2;
else if(s[i] == 'T') T[i + 1] = 3;
}
}
int main()
{
scanf("%d%s", &n, s); get(n, a);
scanf("%d%s", &m, s); get(m, b);
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++) f[i][j] = -inf;
for(int i = 1; i <= n; i++) f[i][0] = f[i - 1][0] + sim[a[i]][4];
for(int i = 1; i <= m; i++) f[0][i] = f[0][i - 1] + sim[b[i]][4];
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++)
{
f[i][j] = max(f[i][j], f[i][j - 1] + sim[b[j]][4]);
f[i][j] = max(f[i][j], f[i - 1][j] + sim[a[i]][4]);
f[i][j] = max(f[i][j], f[i - 1][j - 1] + sim[a[i]][b[j]]);
}
cout << f[n][m];
return 0;
}
f[i][j] 表示前 i 个纸牌，上下差值为 j 时的最小翻动次数
于是就有 f[i][j] = min(f[i - 1][j - (a[i] - b[i])], f[i - 1][j + (a[i] - b[i])])
注意数组别越界
#include <bits/stdc++.h>
using namespace std;
const int N = 1005;
int n, a[N], b[N];
int f[N][15 * N];
int main()
{
scanf("%d", &n);
for(int i = 1; i <= n; i++) scanf("%d%d", &a[i], &b[i]);
memset(f, 127, sizeof(f)); f[0][0 + 6000] = 0;
for(int i = 1; i <= n; i++)
for(int j = -6000; j <= 6000; j++)
{
j += 6000;
f[i][j] = min(f[i - 1][j - a[i] + b[i]], f[i - 1][j - b[i] + a[i]] + 1);
j -= 6000;
}
for(int i = 0; i <= 6000 ; i++)
{
int ans = min(f[n][6000 + i], f[n][6000 - i]);
if(ans <= n) {cout << ans; return 0;}
}
return 0;
}
从下向上跟从上到下没太大区别
f[i][j] = max(f[i - 1][j - 1], f[i - 1][j], f[i - 1][j + 1]);
注意宽为奇数，最后答案就是max(f[n][m / 2 ], f[n][m / 2 + 1], f[n][m / 2 + 2]);
#include <bits/stdc++.h>
using namespace std;
const int N = 205;
int n, m;
int mp[N][N], f[N][N];
int main()
{
scanf("%d%d", &n, &m);
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++) scanf("%d", &mp[i][j]);
memset(f, -11, sizeof(f));
for(int i = 0; i <= m + 1; i++) f[0][i] = 0;
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++)
{
f[i][j] = max(f[i - 1][j - 1], f[i - 1][j + 1]);
f[i][j] = max(f[i][j], f[i - 1][j]);
f[i][j] += mp[i][j];
}
cout << max(f[n][m / 2], max(f[n][m / 2 + 1], f[n][m / 2 + 2]));
return 0;
}
~~~~~
今天又被jushen鄙视了
我该如何委婉地提醒他我比他菜很多这一事实？
4.19
昨天的石子合并复杂度是O(n3)，四边形不等式优化一下可以达到O(n2)
#include <bits/stdc++.h>
using namespace std;
const int N = 205;
const int inf = 1e9;
int n, a[N], sum[N];
int mn[N][N], mx[N][N];//mn[i][j],mx[i][j]分别代表区间[i,j]合并的最 小/大 得分
int sn[N][N];//sn[i][j]代表区间[i,j]合并得到最小值时的k值
int main()
{
scanf("%d", &n);
for(int i = 1; i <= n; i++) {scanf("%d", &a[i]); a[i + n] = a[i];}
for(int i = 1; i <= 2 * n; i++)
{
sum[i] = sum[i - 1] + a[i];
sn[i][i] = i;
}
for(int l = 1; l <= n; l++)
for(int i = 1, j = i + l; j < 2 * n; i++, j++)
{
mx[i][j] = max(mx[i][j - 1], mx[i + 1][j]) + sum[j] - sum[i - 1];//最大值不满足单调性，不能用四边形不等式，但最大值总是在端点处取到???
mn[i][j] = inf;
for(int k = sn[i][j - 1]; k <= sn[i + 1][j]; k++)
if(mn[i][j] > mn[i][k] + mn[k + 1][j])
{mn[i][j] = mn[i][k] + mn[k + 1][j]; sn[i][j] = k;}
mn[i][j] += sum[j] - sum[i - 1];
}
int ans[2] = {inf, 0};
for(int i = 1; i < n; i++)
{
ans[0] = min(ans[0], mn[i][i + n - 1]);
ans[1] = max(ans[1], mx[i][i + n - 1]);
}
cout << ans[0] << endl << ans[1];
return 0;
}
其实还是有一点不太懂，先挖坑吧，以后会填的
#include <bits/stdc++.h>
using namespace std;
const int N = 105;
int n, m, mp[N][N];
int f[N][N];//f[i][j]表示以(i,j)为右下角的正方形边长
int main()
{
scanf("%d%d", &n, &m);
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++) scanf("%d", &mp[i][j]);
int ans = 0;
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++)
{
if(!mp[i][j]) f[i][j] = 0;
else f[i][j] = min(f[i - 1][j - 1], min(f[i][j - 1], f[i - 1][j])) + 1;
ans = max(ans, f[i][j]);
}
cout << ans;
return 0;
}
#include <bits/stdc++.h>
using namespace std;
const int N = 55;
#define ll long long
int T, n;
ll a[N], b[N], c[N];
ll f[100005];
struct node
{
ll a, b, c;
}s[N];
bool cmp(node x, node y)
{
return x.c * y.b < y.c * x.b;
}
int main()
{
scanf("%d%d", &T, &n);
for(int i = 1; i <= n; i++) scanf("%lld", &s[i].a);
for(int i = 1; i <= n; i++) scanf("%lld", &s[i].b);
for(int i = 1; i <= n; i++) scanf("%lld", &s[i].c);
sort(s + 1, s + 1 + n, cmp);
for(int i = 1; i <= n; i++)//前i件物品
for(ll j = T; j >= s[i].c; j--)//已经花费的时间
f[j] = max(f[j], f[j - s[i].c] + s[i].a - j * s[i].b);
ll ans = 0;
for(int i = 0; i <= T; i++) ans = max(ans, f[i]);
cout << ans;
return 0;
}
~~~~~
好烦啊，根本看不进去东西
4.20
#include <bits/stdc++.h>
using namespace std;
const int N = 2505;
int n, m, ans;
int l[N][N], r[N][N];
int mp[N][N], f[N][N][2];
int main()
{
while(scanf("%d%d", &n, &m) != EOF)
{
ans = 0;
memset(l, 0, sizeof(l)); memset(r, 0, sizeof(r));
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++)
{
scanf("%d", &mp[i][j]);
l[i][j] = l[i][j - 1] + mp[i][j];
r[i][j] = r[i - 1][j] + mp[i][j];
}
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++)
{
if(!mp[i][j]) f[i][j][0] = f[i][j][1] = 0;
else
{
for(int k = f[i - 1][j - 1][0]; k >= 0; k--)//这一层的常数我感觉很小，然而并不会证
{
if(r[i][j] - r[i - k - 1][j] != 1) continue;
if(l[i][j] - l[i][j - k - 1] != 1) continue;
f[i][j][0] = k + 1; break;
}
for(int k = f[i - 1][j + 1][1]; k >= 0; k--)
{
if(r[i][j] - r[i - k - 1][j] != 1) continue;
if(l[i][j + k] - l[i][j - 1] != 1) continue;
f[i][j][1] = k + 1; break;
}
}
ans = max(ans, max(f[i][j][0], f[i][j][1]));
}
cout << ans << endl;
}
return 0;
}
4.24
又浪了几天......
难受
~~~~~
http://www.cnblogs.com/XYZinc/p/8930911.html
4.25
dfs
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstdlib>
#include <cstring>
using namespace std;
const int N = 105;
int m, n;
struct node
{
int p, l, sz;
int t[N], v[N];
}s[N];
bool vis[N];
int dfs(int x, int mn, int mx)
{
int cnt = s[x].p;
for(int i = 1; i <= s[x].sz; i++)
{
int to = s[x].t[i]; if(vis[to]) continue;
if(abs(mn - s[to].l) <= m && abs(mx - s[to].l) <= m)
{
vis[to] = true;
cnt = min(cnt, dfs(to, min(mn, s[to].l), max(mx, s[to].l)) + s[x].v[i]);
vis[to] = false;
}
}
return cnt;
}
int main()
{
scanf("%d%d", &m, &n);
for(int i = 1; i <= n; i++)
{
scanf("%d%d%d", &s[i].p, &s[i].l, &s[i].sz);
for(int j = 1; j <= s[i].sz; j++) scanf("%d%d", &s[i].t[j], &s[i].v[j]);
}
vis[1] = true;
cout << dfs(1, s[1].l, s[1].l);
return 0;
}
kruskal
唉，注意输出格式
#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <algorithm> #include <queue> using namespace std; const int N = 1005; const int M = 1000005; int T, n, m, fa[N]; struct node{int x, y, w;}s[M]; bool cmp(node a, node b) {return a.w > b.w;} int findf(int x) { if(fa[x] == x) return x; return fa[x] = findf(fa[x]); } void merge(int x, int y) { x = findf(x); y = findf(y); if(x != y) fa[x] = y; } int main() { scanf("%d", &T); for(int id = 1; id <= T; id++) { scanf("%d%d", &n, &m); for(int i = 1; i <= m; i++) scanf("%d%d%d", &s[i].x, &s[i].y, &s[i].w); sort(s + 1, s + 1 + m, cmp); for(int i = 1; i <= n; i++) fa[i] = i; for(int i = 1; i <= m; i++) { merge(s[i].x, s[i].y); if(findf(1) == findf(n)) { printf("Scenario #%d: ", id); printf("%d ", s[i].w); if (id != T) printf(" "); break; } } } return 0; }
4.26
就水了场比赛
4.27
#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <algorithm> #include <vector> using namespace std; const int N = 1005; char s[30]; int n, m, ans, deg[N]; vector<int> to[N]; int t, dfn[N], low[N]; bool flag[N]; void tarjan(int x, int fa) { dfn[x] = low[x] = ++t; flag[x] = true; for(int i = 0; i < to[x].size(); i++) { int k = to[x][i]; if(k == fa) continue; if(!dfn[k]) { tarjan(k, x); low[x] = min(low[x], low[k]); } else if(flag[k]) low[x] = min(low[x], dfn[k]); } } void clear() { for(int i = 1; i <= n; i++) to[i].clear(); memset(low, 0, sizeof(low)); memset(dfn, 0, sizeof(dfn)); memset(flag, 0, sizeof(flag)); memset(deg, 0, sizeof(deg)); } int main() { while(scanf("%d%d", &n, &m) != EOF) { clear(); for(int i = 1; i <= m; i++) { int x, y; scanf("%d%d", &x, &y); to[x].push_back(y); to[y].push_back(x); } tarjan(1, 0); for(int i = 1; i <= n; i++) { for(int j = 0; j < to[i].size(); j++) { int k = to[i][j]; if(low[i] != low[k]) deg[low[i]]++; } } for(int i = 1; i <= n; i++) if(deg[i] == 1) ans++; cout << (ans + 1) / 2 << endl; } return 0; }
#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <algorithm> #include <vector> using namespace std; const int N = 105; int n, col[N], sum[3]; int a[N][N], b[N][N], num; bool mp[N][N], f[N], dp[N][N * 2], id[N][N * 2], fail; void dfs(int x, int c) { col[x] = c; if(c == 1) a[num][++sum[1]] = x; else b[num][++sum[2]] = x; for(int y = 1; y <= n; y++) { if(mp[x][y]) { if(col[y]) { if(col[y] != 3 - c) {fail = true; return;} } else dfs(y, 3 - c); } if(fail) return; } } int pos; vector<int> ans[2]; void print(int i, int dif) { if(!i) return; if(!id[i][dif]) { for(int k = 1; k <= a[i][0]; k++) ans[0].push_back(a[i][k]); for(int k = 1; k <= b[i][0]; k++) ans[1].push_back(b[i][k]); print(i - 1, dif - a[i][0] + b[i][0]); } else { for(int k = 1; k <= a[i][0]; k++) ans[1].push_back(a[i][k]); for(int k = 1; k <= b[i][0]; k++) ans[0].push_back(b[i][k]); print(i - 1, dif + a[i][0] - b[i][0]); } } int main() { scanf("%d", &n); for(int i = 1; i <= n; i++) { int x; while(scanf("%d", &x) && x) f[x] = 1; for(int j = 1; j <= n; j++) if(!f[j] && i != j) mp[i][j] = mp[j][i] = true; memset(f, 0, sizeof(f)); } for(int i = 1; i <= n; i++) if(!col[i]) { sum[1] = sum[2] = 0; num++; dfs(i, 1); if(fail) {cout << "No solution"; return 0;} a[num][0] = sum[1]; b[num][0] = sum[2]; } dp[0][n] = true; for(int i = 0; i < num; i++) { int x = a[i + 1][0] - b[i + 1][0]; for(int j = -n; j <= n; j++) { if(!dp[i][j + n]) continue; dp[i + 1][j + x + n] = true; id[i + 1][j + x + n] = 0; dp[i + 1][j - x + n] = true; id[i + 1][j - x + n] = 1; } } for(int i = 0; i <= n; i++) if(dp[num][n + i]) {pos = i; break;} else if(dp[num][n - i]) {pos = -i; break;} print(num, n + pos); sort(ans[0].begin(), ans[0].end()); sort(ans[1].begin(), ans[1].end()); cout << ans[0].size(); for(int i = 0; i < ans[0].size(); i++) cout << " " << ans[0][i]; cout << endl; cout << ans[1].size(); for(int i = 0; i < ans[1].size(); i++) cout << " " << ans[1][i]; return 0; }
4.28
打了一下fft的板子
~~~~~
被抓去中午给萌新讲动态规划。。。
我其实很菜的啊。。。
先预习一波吧。。。
4.29
kd-tree
5.3
树链剖分，边权转化为点权
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int N = 100005;
const int M = N * 2;
const int inf = 1e9;
inline int read()
{
int x = 0, f = 1; char ch = getchar();
while(!isdigit(ch)) {if(ch == '-') f = -1; ch = getchar();}
while(isdigit(ch)) {x = x * 10 + ch - '0'; ch = getchar();}
return x * f;
}
struct edge
{
int fr, to, w;
}E[N];
struct node
{
int l, r, f, mn, mx;
node(){}
node(int ll, int rr, int ff, int nn, int xx) {l = ll; r = rr; f = ff; mn = nn; mx = xx;}
}T[N << 3];
int num, to[M], w[M], nt[M], pt[N];
void add(int x, int y, int v) {to[++num] = y; w[num] = v; nt[num] = pt[x]; pt[x] = num;}
int n, sz[N], fa[N], deep[N], son[N];
void dfs1(int x)
{
sz[x] = 1;
for(int e = pt[x]; e; e = nt[e])
{
int k = to[e];
if(k == fa[x]) continue;
fa[k] = x;
deep[k] = deep[x] + 1;
dfs1(k);
sz[x] += sz[k];
if(sz[k] > sz[son[x]]) son[x] = k;
}
}
int t, dfn[N], top[N];
void dfs2(int x, int tp)
{
top[x] = tp; dfn[x] = ++t;
if(!son[x]) return;
dfs2(son[x], tp);
for(int e = pt[x]; e; e = nt[e])
if(to[e] != son[x] && to[e] != fa[x]) dfs2(to[e], to[e]);
}
void build(int p, int x, int y)
{
T[p] = node(x, y, 0, inf, -inf);
if(x == y) return;
int mid = (x + y) >> 1;
build(p << 1, x, mid); build(p << 1 | 1, mid + 1, y);
}
void neg(int p)
{
T[p].mn *= -1; T[p].mx *= -1;
swap(T[p].mn, T[p].mx);
}
void pushup(int p)
{
T[p].mn = min(T[p << 1].mn, T[p << 1 | 1].mn);
T[p].mx = max(T[p << 1].mx, T[p << 1 | 1].mx);
}
void pushdown(int p)
{
if(T[p].l == T[p].r || !T[p].f) return;
T[p << 1].f ^= 1; T[p << 1 | 1].f ^= 1;
neg(p << 1); neg(p << 1 | 1);
T[p].f = 0;
}
void update(int p, int x, int y, int v)
{
pushdown(p);
int pl = T[p].l, pr = T[p].r;
if(pl == x && pr == y)
{
if(v == inf) {T[p].f = 1; neg(p);}
else T[p].mn = T[p].mx = v;
return;
}
int mid = (pl + pr) >> 1;
if(y <= mid) update(p << 1, x, y, v);
else if(x > mid) update(p << 1 | 1, x, y, v);
else {update(p << 1, x, mid, v); update(p << 1 | 1, mid + 1, y, v);}
pushup(p);
}
void Update(int x, int y, int v)
{
while(top[x] != top[y])
{
if(deep[top[x]] < deep[top[y]]) swap(x, y);
update(1, dfn[top[x]], dfn[x], v);
x = fa[top[x]];
}
if(x == y) return;
if(deep[x] < deep[y]) swap(x, y);
update(1, dfn[y] + 1, dfn[x], v);
}
int query(int p, int x, int y)
{
pushdown(p);
int pl = T[p].l, pr = T[p].r;
if(pl == x && pr == y) return T[p].mx;
int mid = (pl + pr) >> 1;
if(y <= mid) return query(p << 1, x, y);
else if(x > mid) return query(p << 1 | 1, x, y);
else return max(query(p << 1, x, mid), query(p << 1 | 1, mid + 1, y));
}
int Query(int x, int y)
{
int cnt = -inf;
while(top[x] != top[y])
{
if(deep[top[x]] < deep[top[y]]) swap(x, y);
cnt = max(cnt, query(1, dfn[top[x]], dfn[x]));
x = fa[top[x]];
}
if(x == y) return cnt;
if(deep[x] < deep[y]) swap(x, y);
cnt = max(cnt, query(1, dfn[y] + 1, dfn[x]));
return cnt;
}
void clear()
{
num = t = 0;
memset(to, 0, sizeof(to)); memset(w, 0, sizeof(w));
memset(nt, 0, sizeof(nt)); memset(pt, 0, sizeof(pt));
memset(deep, 0, sizeof(deep)); memset(son, 0, sizeof(son));
memset(fa, 0, sizeof(fa)); memset(sz, 0, sizeof(sz));
}
int main()
{
int op = read();
while(op--)
{
clear();
n = read();
for(int i = 1; i < n; i++)
{
E[i].fr = read(); E[i].to = read(); E[i].w = read();
add(E[i].fr, E[i].to, E[i].w); add(E[i].to, E[i].fr, E[i].w);
}
dfs1(1);
dfs2(1, 1);
build(1, 1, t);
for(int i = 1; i < n; i++)
{
if(deep[E[i].fr] > deep[E[i].to]) swap(E[i].fr, E[i].to);
update(1, dfn[E[i].to], dfn[E[i].to], E[i].w);
}
while(true)
{
char s[10]; scanf("%s", s);
if(s[0] == 'C')
{
int id = read(), v = read();
update(1, dfn[E[id].to], dfn[E[id].to], v);
}
else if(s[0] == 'N')
{
int x = read(), y = read();
Update(x, y, inf);
}
else if(s[0] == 'Q')
{
int x = read(), y = read();
printf("%d
", Query(x, y));
}
else break;
}
}
return 0;
}
5.4
tarjan
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;
const int N = 200005;
inline int read()
{
int x = 0, f = 1; char ch = getchar();
while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
while(ch >= '0' && ch <= '9') {x = x * 10 + ch - '0'; ch = getchar();}
return x * f;
}
int n, m, cas, Q, ans;
vector<int> to[N];
int t, dfn[N], low[N], fa[N], deep[N]; bool flag[N], br[N];
void tarjan(int x)
{
dfn[x] = low[x] = ++t; flag[x] = true;
for(int i = 0; i < to[x].size(); i++)
{
int k = to[x][i];
if(!dfn[k])
{
fa[k] = x;
deep[k] = deep[x] + 1;
tarjan(k);
low[x] = min(low[x], low[k]);
if(dfn[x] < low[k]) ans++, br[k] = true;//br[i]记录i与fa[i]的连边是否为桥
}
else if(k != fa[x])low[x] = min(low[x], dfn[k]);
}
}
void lca(int x, int y)
{
if(deep[x] < deep[y]) swap(x, y);
while(deep[x] > deep[y])
{
if(br[x]) {ans--; br[x] = false;}
x = fa[x];
}
while(x != y)
{
if(br[x]) {ans--; br[x] = false;}
if(br[y]) {ans--; br[y] = false;}
x = fa[x]; y = fa[y];
}
}
void clear()
{
ans = t = 0;
for(int i = 1; i <= n; i++) to[i].clear();
memset(flag, 0, sizeof(flag)); memset(br, 0, sizeof(br));
memset(deep, 0, sizeof(deep)); memset(fa, 0, sizeof(fa));
memset(dfn, 0, sizeof(dfn));
}
int main()
{
n = read(); m = read();
while(n + m)
{
clear();
printf("Case %d:
", ++cas);
for(int i = 1; i <= m; i++)
{
int x = read(), y = read();
to[x].push_back(y); to[y].push_back(x);
}
tarjan(1);
Q = read();
while(Q--)
{
int x = read(), y = read();
lca(x, y);
printf("%d
", ans);
}
printf("
");
n = read(); m = read();
}
return 0;
}
5.5
又水了一天
5.10 - 5.14
丝滑拿铁
除了意识到自己到底多菜，什么收获也没有
5.15
早上做了一下 NOI2014 DAY1
接下来的时间打算就这样，早上做题感受一下自己多菜，下午改题，晚上学点新东西
也不打算学什么太难的东西了，学了也打不出来
emmm......
难受
5.18 - 5.28
考试
5.28
两边都没过，没学上了啊
这段时间真的是完全荒废了，什么都没学，考试的题也没改
怕是要痛改前非，重新做人啊
5.29
-
分治
luogu1429（优美一点的分治，或者旋转坐标轴乱搞）
- 矩阵快速幂
6.12
- 树的直径
6.13
- 树的直径
6.15
gcd && lcm
#include <bits/stdc++.h> using namespace std; int gcd(int a, int b) { if(!b) return a; return gcd(b, a % b); } int main() { int T; scanf("%d", &T); while(T--) { int a[2], b[2]; scanf("%d%d%d%d", &a[0], &a[1], &b[0], &b[1]); int k = a[0] / a[1], t = b[1] / b[0], ans = 0; for(int x = 1; x * x <= b[1]; x++) { if(b[1] % x != 0) continue; if(x % a[1] == 0 && gcd(x / a[1], k) == 1 && gcd(t, b[1] / x) == 1) ans++; int y = b[1] / x; if(x == y) continue; if(y % a[1] == 0 && gcd(y / a[1], k) == 1 && gcd(t, b[1] / y) == 1) ans++; } printf("%d ", ans); } return 0; }