contest 1.18

D.装饰

列方程

#include <iostream>
#include<cstdio>
#include<algorithm>
typedef long long ll;
using namespace std;
 
ll a[5];
 
int main()
{
    int t;scanf("%d",&t);
    while(t--){
        scanf("%lld%lld%lld",&a[0],&a[1],&a[2]);
        sort(a,a+3);
        ll tmp=2*(a[0]+a[1])-a[2];
        if(tmp<=0){
            ll ans=a[0]+a[1];
            printf("%lld
",ans);
        }
        else if(tmp%3==0){
            ll ans=tmp/3+(a[2]-tmp/3)/2;
            printf("%lld
",ans);
        }
        else{
            ll ans1=tmp/3+(a[2]-tmp/3)/2;
            ll ans2=(tmp/3+1)+(a[0]+a[1]-2*(tmp/3+1));
            printf("%lld
",max(ans1,ans2));
        }
    }
    return 0;
}

F.Convention

二分

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define inf 0x3f3f3f3f3f3f3f3f
typedef long long ll;
using namespace std;
 
ll n,m,c;
ll t[100005];
ll bus_time[100005];
ll bus_num[100005];
 
bool check(ll x){
  for(ll i=1;i<=m;i++){
    bus_num[i]=0,bus_time[i]=inf;
  }
  ll now=1;
  for(ll i=1;i<=n;i++){
    if(bus_num[now]==c){
        if(now+1>m) return 0;
        else now++,bus_num[now]=1,bus_time[now]=t[i];
    }
    else{
        if(t[i]-bus_time[now]<=x) bus_num[now]++;
        else{
            if(now+1>m) return 0;
            else now++,bus_num[now]=1,bus_time[now]=t[i];
        }
    }
  }
  return 1;
}
 
int main()
{
    scanf("%lld%lld%lld",&n,&m,&c);
    for(ll i=1;i<=n;i++) scanf("%lld",&t[i]);
    sort(t+1,t+1+n);
    ll l=0,r=t[n];
    while(l<=r){
        ll mid=(l+r)/2;
        //printf("mid=%lld %d
",mid,check(mid));
        if(check(mid)) r=mid-1;
        else l=mid+1;
    }
    printf("%lld
",l);
    return 0;
}

G.Convention II

模拟+优先队列

#include <iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
typedef long long ll;
using namespace std;
 
struct Node{
  ll s,t,id;
  bool operator<(const Node& b) const{
    return id>b.id;
  }
}cow[100005];
priority_queue<Node> q;
 
bool cmp(Node a,Node b){
  if(a.s!=b.s) return a.s<b.s;
  else return a.id<b.id;
}
 
int main()
{
    ll n;scanf("%lld",&n);
    for(ll i=1;i<=n;i++){
        scanf("%lld%lld",&cow[i].s,&cow[i].t);
        cow[i].id=i;
    }
    sort(cow+1,cow+1+n,cmp);
    ll p=1,last=0,ans=0;
    while(p<=n){
        while(cow[p].s<=last) {
            q.push(cow[p]);
            p++;
            if(p>n) break;
        }
        if(q.empty()){
            last=cow[p].s+cow[p].t;
            p++;
        }
        else{
            Node top=q.top();
            //printf("last=%d top.s=%d top.t=%d
",last,top.s,top.t);
            ans=max(ans,last-top.s);
            last=last+top.t;
            q.pop();
        }
        //printf("last=%d ans=%d
",last,ans);
    }
    printf("%lld
",ans);
    return 0;
}

H.Mooyo Mooyo

dfs求连通块+模拟

#include <iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

int dire[4][2]={0,-1,-1,0,0,1,1,0};
int n,k;
const int m=10;
char Map[110][15],t[110];
int vis[110][15];

inline bool in(int x,int y) {return x>=1&&x<=n&&y>=1&&y<=m;}

int dfs(int x,int y){
  //printf("(%d,%d)  %c
",x,y,Map[x][y]);
  vis[x][y]=1;
  int ret=1;
  for(int i=0;i<4;i++){
    int nx=x+dire[i][0];
    int ny=y+dire[i][1];
    if(in(nx,ny)&&Map[nx][ny]==Map[x][y]&&!vis[nx][ny]) ret+=dfs(nx,ny);
  }
  return ret;
}

void del(int x,int y){
  char c=Map[x][y];
  Map[x][y]='0';
  for(int i=0;i<4;i++){
    int nx=x+dire[i][0];
    int ny=y+dire[i][1];
    if(in(nx,ny)&&Map[nx][ny]==c) del(nx,ny);
  }
}

int main()
{
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++){
        scanf("%s",Map[i]+1);
    }
    while(1){
      bool flag=0;
      for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            memset(vis,0,sizeof(vis));
            if(Map[i][j]!='0'&&dfs(i,j)>=k){
              flag=1,del(i,j);
            }
        }
      }
      if(!flag) break;
      for(int j=1;j<=m;j++){
        int cnt=0;
        for(int i=1;i<=n;i++) if(Map[i][j]!='0') t[++cnt]=Map[i][j];
        for(int i=n-cnt+1,p=1;i<=n;i++,p++) Map[i][j]=t[p];
        for(int i=1;i<=n-cnt;i++) Map[i][j]='0';
      }
      //for(int i=1;i<=n;i++) puts(Map[i]+1);
    }
    for(int i=1;i<=n;i++) puts(Map[i]+1);
    return 0;
}

E.排队

 bitset求所有组合情况

#include <iostream>
#include<cstdio>
#include<bitset>
#include<vector>
#include<algorithm>
using namespace std;

int pre[1005],vis[1005];
vector<int> a;
bitset<1005> bs;

int main()
{
    int n,p;scanf("%d%d",&n,&p);
    for(int i=1;i<=n;i++){
        scanf("%d",&pre[i]);
        if(pre[i]) vis[pre[i]]=1;
    }
    for(int i=1;i<=n;i++){
        if(vis[i]) continue;
        int x=i,t=0,flag=0;
        while(x){
            t++;
            if(x==p) flag=1;
            x=pre[x];
        }
        if(!flag) a.push_back(t);
    }
    bs[0]=1;
    for(auto t:a){
        bs|=(bs<<t);
    }
    int x=p,k=0;
    while(x){
        k++;
        x=pre[x];
    }
    for(int i=0;i<=n;i++) if(bs[i]) printf("%d
",i+k);
    return 0;
}