题目链接

今天的题比较简单 人均40分钟ak

最小生成树

水题

#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
#define inf 1e9
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;

inline ll read()
{
	ll x=0,w=1; char c=getchar();
	while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
	while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
	return w==1?x:-x;
}
const int N=1e5+10;
int a[N],n;
int main()
{
    n=read();
    
    ll ans=0;
    rep(i,1,n) a[i]=read(),ans+=a[i];
    if(n==1){
        puts("0");return 0;
    }
    
    sort(a+1,a+1+n);
    ans+=1ll*(n-2)*a[1];
    cout<<ans<<endl;
}

B-病毒感染

做法：树形dp+换根 dp[u] 代表u这个子树 子节点到u节点的距离之和，dp[u]+=dp[v]+sz[v]  然后随便换根一下就好了。

C-Shopping

做法：水题，记录 1 的个数num，然后前 min(num,m) 大的取一半 剩余的全部取就好了。

#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
#define inf 1e9
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;

inline ll read()
{
	ll x=0,w=1; char c=getchar();
	while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
	while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
	return w==1?x:-x;
}
const int N=1e3+10;
int a[N],b[N],n,m;
bool cmp(int x,int y)
{
    return x>y;
}
int main()
{
    int _=read();while(_--)
    {
        n=read(),m=read();
        int num=0;
        rep(i,1,n) {
            a[i]=read(),b[i]=read();
            num+=b[i];
        }
        num=min(num,m);
        double ans=0;
        sort(a+1,a+1+n,cmp );
        for(int i=1;i<=num&&i<=n;++i){
            ans+=1.0*a[i]/2;
        }
        for(int i=num+1;i<=n;++i) ans+=a[i];
        printf("%.1f\n",ans);
    }
}

D-铺地毯

水题。

#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
#define inf 1e9
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;

inline ll read()
{
	ll x=0,w=1; char c=getchar();
	while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
	while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
	return w==1?x:-x;
}
const int N=1e4+10;
struct node
{
    int x1,y1,x2,y2;
}a[N];
int n,x,y;
int main()
{
    n=read();
    rep(i,1,n) {
        a[i].x1=read(),a[i].y1=read(),a[i].x2=read(),a[i].y2=read();
        a[i].x2+=a[i].x1;
        a[i].y2+=a[i].y1;
    }
    x=read(),y=read();
    int ans=-1;
    rep(i,1,n)
    {
        if(a[i].x1<=x&&x<=a[i].x2&&a[i].y1<=y&&y<=a[i].y2) ans=i;
    }
    cout<<ans<<endl;
}

E-金币馅饼

做法：简单dp  考虑到走的方式比较奇特  先枚举列 再枚举行就好了。

#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
#define inf 1e9
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;

inline ll read()
{
	ll x=0,w=1; char c=getchar();
	while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
	while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
	return w==1?x:-x;
}
const int N=5e2;
ll dp[N][N],a[N][N],n,m;
int vis[N][N];
int main()
{
    n=read(),m=read();
    rep(i,1,n) rep(j,1,m) a[i][j]=read();
    dp[1][1]=a[1][1];
    vis[1][1]=1;

    for(int j=1;j<=m;++j){
        for(int i=1;i<=n;++i){
            if(vis[i][j]==0) continue;
            vis[i-1][j+1]=1;
            vis[i][j+1]=1;
            vis[i+1][j+1]=1;

            dp[i-1][j+1]=max(dp[i-1][j+1],dp[i][j]+a[i-1][j+1]);
            dp[i][j+1]=max(dp[i][j+1],dp[i][j]+a[i][j+1]);
            dp[i+1][j+1]=max(dp[i+1][j+1],dp[i][j]+a[i+1][j+1]);
        }
    }
    printf("%lld\n",dp[n][m]);
}