NOIP2018
評價:這年的NOIP感覺質量很屑,完全是在亂搞,聽說Day2T2是因爲漏題的原因直接換成了清華校內集訓的T1。分明就是我太菜了
T1
解答
水題,做個差分,把正的加起來就行了。
//
// Created by dhy on 18-11-10.
//
#include <cstdio>
#include <cstdlib>
const int MAXN = 100010;
int read(){
int x = 0,f = 1;
char c = getchar();
while(c>'9'||c<'0'){ if(c=='-')f = -1;c = getchar(); }
while(c>='0'&&c<='9'){ x = (x<<1)+(x<<3)+c-'0';c = getchar();}
return x*f;
}
int a[MAXN];
int d[MAXN];
int main(){
int n = read();
for(int i = 1;i<=n;i++){
a[i] = read();
d[i] = a[i]-a[i-1];
}
int ans = 0;
for(int i = 1;i<=n;i++){
if(d[i]>0)ans+=d[i];
}
printf("%d\n",ans);
return 0;
}
T2
解答
揹包,排個序,然後嘗試用小的更新大的。
代碼:
//
// Created by dhy on 18-11-10.
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cstdlib>
using namespace std;
int read(){
int x = 0,f = 1;
char c = getchar();
while(c>'9'||c<'0'){ if(c=='-')f = -1;c = getchar(); }
while(c>='0'&&c<='9'){ x = (x<<1)+(x<<3)+c-'0';c = getchar();}
return x*f;
}
const int MAXVAL = 25010;
int a[MAXVAL];
bool f[MAXVAL];
int main(){
int T = read();
while(T--){
memset(f,false,sizeof(f));
int n = read();
int ans = n;
for(int i = 1;i<=n;i++){
a[i] = read();
}
sort(a+1,a+1+n);
f[0] = true;
for(int i = 1;i<=n;i++){
if(f[a[i]]){
ans--;
continue;
}
for(int j = a[i];j<=a[n];j++)f[j] = f[j]|f[j-a[i]];
}
printf("%d\n",ans);
}
return 0;
}
T3
解答
首先不難看出是個二分加樹上的一個操作。考慮如何判斷解是否可行。記表示節點的子樹中,滿足條件的路徑有多少種,表示節點的子樹中,最長的一條鏈的長度,然後對於每個節點,直接合並子樹中的然後把全部丟到multiset
裏面,每次取最小的,再找一個剛好可以拼湊成的值拼湊,更新
代碼:
// luogu-judger-enable-o2
#include <iostream>
#include <cstring>
#include <cstdio>
#include <set>
#include <algorithm>
#define clean(x) memset(x,0,sizeof(x))
using namespace std;
const int MAXN = 101000;
struct edge{int t,w,next;}edges[MAXN<<1];
int head[MAXN],top;
void add(int f,int t,int w){
edges[++top].next = head[f];
edges[top].t = t;
edges[top].w = w;
head[f] = top;
}
int f[MAXN],g[MAXN];
multiset<int> st[MAXN];
int n,m;
void dfs(int x,int fa,int mid){
st[x].clear();f[x] = 0,g[x] = 0;
for(int i = head[x];i;i = edges[i].next){
int t = edges[i].t;
if(t==fa)continue;
dfs(t,x,mid);
g[x]+=g[t];
int v = f[t]+edges[i].w;
if(v>=mid)g[x]++;
else st[x].insert(v);
}
while(!st[x].empty()){
if(st[x].size()==1){
f[x] = max(f[x],*st[x].begin());
break;
}else{
multiset<int>::iterator it;
it = st[x].lower_bound(mid-*st[x].begin());
if(it==st[x].begin()&&st[x].count(*it)==1)it++;
if(it==st[x].end()){
f[x] = max(f[x],*st[x].begin());
st[x].erase(st[x].find(*st[x].begin()));
}else{
g[x]++;
st[x].erase(st[x].find(*it));
st[x].erase(st[x].find(*st[x].begin()));
}
}
}
}
bool check(int mid){
dfs(1,1,mid);
return g[1]>=m;
}
int main(){
scanf("%d%d",&n,&m);
int f,t,w;
int sum = 0;
for(int i = 1;i<n;i++){
scanf("%d%d%d",&f,&t,&w);
add(f,t,w);add(t,f,w);
sum+=w;
}
int l = 0,r = sum;
while(l<r){
int mid = l+r+1>>1;
if(check(mid))l = mid;
else r = mid-1;
}
printf("%d\n",l);
return 0;
}
T4
解答
暴力枚舉斷一條邊,然後貪心做就好了
代碼:
// luogu-judger-enable-o2
#include <iostream>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
vector<int> G[5010];
int vis[5010];
namespace tree{
vector<int> ans;
void dfs(int x){
vis[x] = true;
ans.push_back(x);
for(int i = 0;i<G[x].size();i++){
if(!vis[G[x][i]])dfs(G[x][i]);
}
}
void main(){
dfs(1);
for(int i = 0;i<ans.size();i++)cout<<ans[i]<<' ';
}
}
int n,m;
int edge[5010][2];
namespace ringBasedTree{
vector<int> ans,temp;
int del_f,del_t;
bool chect(int u,int v){
if((u==del_f&&v==del_t)||(u==del_t&&v==del_f)){
return false;
}
return true;
}
void dfs(int x){
vis[x] = true;
temp.push_back(x);
for(int i = 0;i<G[x].size();i++){
if(chect(x,G[x][i])&&!vis[G[x][i]]){
dfs(G[x][i]);
}
}
}
void main(){
for(int i = 1;i<=n;i++)ans.push_back(9999);
for(int i = 1;i<=m;i++){
memset(vis,false, sizeof(vis));
del_f = edge[i][0],del_t = edge[i][1];
temp.clear();
dfs(1);
if(temp<ans&&temp.size()==n)ans = temp;
}
for(int i = 0;i<ans.size();i++)cout<<ans[i]<<' ';
}
}
int main(void){
cin>>n>>m;
int x,y;
for(int i = 1;i<=m;i++){
cin>>x>>y;
edge[i][0] = x;
edge[i][1] = y;
G[x].push_back(y),G[y].push_back(x);
}
for(int i = 1;i<=n;i++)sort(G[i].begin(),G[i].end());
if(m==n-1){
tree::main();
}else{
ringBasedTree::main();
}
return 0;
}
T5
題面
我不會,據說可以dfs套dfs找規律得50分
T6
題面
我不會,DDP板子題