今天又考的奇差
题解：
第一题；
这么简单一道题我想了好久，智商实在是下线了；
#include<bits/stdc++.h>
using namespace std;
int main(){
freopen("shortway.in","r",stdin);
freopen("shortway.out","w",stdout);
int n, k;
scanf("%d%d", &n, &k);
if((n - 1) % k == 0)printf("%d\n", (n - 1) / k * 2);
if((n - 1) % k == 1)printf("%d\n", (n - 1) / k * 2 + 1);
if((n - 1) % k >= 2)printf("%d\n", (n - 1) / k * 2 + 2);
}
第二题：我们发现e的长度很小，我们可以在上面做文章，其实每个位置对应的%strlen(e)都是一样的；我们用树状数组维护rt[pos][len][yu][id]表示1到POS每次跳len，当前位置%len = yu的id字符有多少个；对于一个字符串，查询就是rt[R][len][(L + i - 1) % len][a[i]],修改就把长度枚举一下；
这道题我开始想到每次跳len, 维护一个前缀和，但没有想到树状数组，暴力修改和我直接暴力复杂度差不多；如果想到余数这一维就好了
#include<bits/stdc++.h>
using namespace std;
const int M = 1e5 + 10;
char s[M], ch[M];
int rt[M][11][10][4], n, a[M], b[M];
inline int get(char x){
switch(x){
case ‘A‘:return 0;
case ‘T‘:return 1;
case ‘C‘:return 2;
case ‘G‘:return 3;
}
}
inline int lowbit(int x){return x & (-x);}
void add(int pos, int len, int yu, int id, int d){
for(int x = pos; x <= n; x += lowbit(x)){
rt[x][len][yu][id] += d;
}
}
int query(int pos, int id, int k, int yu){
int ret = 0;
for(int x = pos; x; x -= lowbit(x)){
ret += rt[x][k][yu][id];
}
return ret;
}
void init(){
for(int i = 1; i <= n; i++)
for(int k = 1; k <= 10; k++){
add(i, k, i % k, a[i], +1);
}
}
void add(int pos, int id, int d){
for(int k = 1; k <= 10; k++)
add(pos, k, pos % k, id, d);
}
int main(){
freopen("evolution.in","r",stdin);
freopen("evolution.out","w",stdout);
scanf("%s", s);
n = strlen(s);
for(int i = 1; i <= n; i++)a[i] = get(s[i - 1]);
init();
int q, opt, l, r;
scanf("%d", &q);
while(q--){
scanf("%d", &opt);
if(opt == 1){
scanf("%d%s", &l, ch);
int t = get(ch[0]);
add(l, a[l], -1);
add(l, t, +1);
a[l] = t;
}
else {
scanf("%d%d%s", &l, &r, ch );
int ans = 0, len = strlen(ch);
for(int i = 1; i <= len; i++){
b[i] = get(ch[i - 1]);
ans += query(r, b[i], len, (l + i - 1) % len);
ans -= query(l - 1, b[i], len, (l + i - 1) % len);
}
printf("%d\n", ans);
}
}
}
第三题：
线段树存树边，维护最大值更新非树遍，用非树边更新线段树上的限制；
#include<bits/stdc++.h>
using namespace std;
const int M = 2e5 + 10;
const int oo = 0x3f3f3f3f;
#define ex(i, u) for(int i = h[u]; i; i = G[i].nxt)
#define Ls nd->ls, lf, mid
#define Rs nd->rs, mid + 1, rg
int fa[M], ans[M], n, m, h[M], top[M], siz[M], son[M], dep[M], in[M];
int idc, tot, seq[M], line[M];
bool intree[M], fg;
struct edge{int u, v, w, id;}g[M];
struct Edge{int nxt, v, id;}G[M << 1];
void add(int u, int v, int i){
G[++tot].v = v; G[tot].nxt = h[u]; h[u] = tot; G[tot].id = g[i].id;
}
bool cmp(edge a, edge b){
return a.w < b.w;
}
bool cmp_id(edge a, edge b){
return a.id < b.id;
}
int find(int x){
if(x == fa[x])return x;
return fa[x] = find(fa[x]);
}
void uni(int u, int v){
fa[find(u)] = find(v);
}
struct Node;
void modify(int L, int R, int d, Node * nd, int lf, int rg);
struct Node{
Node *ls, *rs;
int mx, mi, tag;
inline void down(int lf, int rg){
if(tag != oo){
int mid = (lf + rg) >> 1;
modify(lf, mid, tag, ls, lf, mid);
modify(mid + 1, rg, tag, rs, mid + 1, rg);
tag = oo;
}
}
inline void up(){
mx = max(ls->mx, rs->mx);
}
}pool[M << 2], *tail = pool, *root;
Node * build(int lf = 1, int rg = n){
Node *nd = ++tail;
if(lf == rg) {
nd->mi = nd->tag = oo;
nd->mx = g[line[seq[lf]]].w;
}
else {
int mid = (lf + rg) >> 1;
nd->ls = build(lf, mid);
nd->rs = build(mid + 1, rg);
nd->tag = nd->mi = oo;
nd->up();
}
return nd;
}
int query(int L, int R, Node * nd = root, int lf = 1, int rg = n){
if(L <= lf && rg <= R) return nd->mx;
else {
nd->down(lf, rg);
int mid = (lf + rg) >> 1;
int ans = 0;
if(L <= mid)ans = max(ans, query(L, R, Ls));
if(R > mid)ans = max(ans, query(L, R, Rs));
return ans;
}
}
void modify(int L, int R, int d, Node * nd = root, int lf = 1, int rg = n){
if(L <= lf && rg <= R) {
nd->tag = min(nd->tag, d);
nd->mi = min(nd->mi, d);
return ;
}
else {
nd->down(lf, rg);
int mid = (lf + rg) >> 1;
if(L <= mid)modify(L, R, d, Ls);
if(R > mid)modify(L, R, d, Rs);
//nd->up();// ???
}
}
void getdown(Node * nd = root, int lf = 1, int rg = n){
if(lf == rg){
ans[line[seq[lf]]] = nd->mi;
return ;
}
nd->down(lf, rg);
int mid = (lf + rg) >> 1;
getdown(Ls);
getdown(Rs);
}
void dfs(int u, int f){
dep[u] = dep[f] + 1;
siz[u] = 1;
fa[u] = f;
ex(i, u){
int v = G[i].v;
if(v == f)continue;
dfs(v, u);
line[v] = G[i].id;
siz[u] += siz[v];
if(siz[v] > siz[son[u]])son[u] = v;
}
}
void dfs2(int u, int tp){
in[u] = ++idc;
seq[idc] = u;
top[u] = tp;
if(son[u])dfs2(son[u], tp);
ex(i, u){
int v = G[i].v;
if(v == fa[u] || v == son[u])continue;
dfs2(v, v);
}
}
void Modify(int u, int v, int d){
while(top[u] != top[v]){
if(dep[top[u]] < dep[top[v]])swap(u, v);
modify(in[top[u]], in[u], d);
u = fa[top[u]];
}
if(dep[u] < dep[v])swap(u, v);
if(in[u] != in[v])modify(in[v] + 1, in[u], d);
}
int Query(int u, int v){
int ans = 0;
while(top[u] != top[v]){
if(dep[top[u]] < dep[top[v]])swap(u, v);
ans = max(ans, query(in[top[u]], in[u]));
//if(fg)printf("%d %d %d %d %d %d %d\n", u, v, dep[top[u]], dep[top[v]], in[top[u]], in[u], ans);
u = fa[top[u]];
}
if(dep[u] < dep[v])swap(u, v);
if(in[u] != in[v])ans = max(ans, query(in[v] + 1, in[u]));
//if(fg)printf("%d %d %d\n", u, v, ans);
return ans;
}
int main(){
freopen("mst.in","r",stdin);
freopen("mst.out","w",stdout);
int u, v, w;
scanf("%d%d", &n, &m);
for(int i = 1; i <= m; i++){
scanf("%d%d%d", &u, &v, &w);
g[i] = (edge){u, v, w, i};
}
sort(g + 1, g + 1 + m, cmp);
for(int i = 1; i <= n; i++)fa[i] = i;
for(int i = 1; i <= m; i++){
int u = g[i].u, v = g[i].v;
if(find(u) == find(v))continue;
uni(u, v);
intree[g[i].id] = 1;
add(u, v, i); add(v, u, i);
}
line[1] = 0;
dfs(1, 0);
dfs2(1, 1);
//for(int i = 1; i <= n; i++)printf("%d %d %d\n", i, dep[i], top[i]);puts("");
sort(g + 1, g + 1 + m, cmp_id);
root = build();
for(int i = 1; i <= m; i++){
if(intree[i])continue;
int u = g[i].u, v = g[i].v;
if(i == 473)fg = 1;
int t = Query(u, v);
ans[i] = t - 1;
Modify(u, v, g[i].w - 1);
fg = 0;
}
getdown();
for(int i = 1; i <= m; i++)
printf("%d ",ans[i] == oo ? -1 : ans[i]);
}
原文地址：https://www.cnblogs.com/EdSheeran/p/9543925.html
时间： 2024-10-31 15:48:59