ACM预备队-week11(综合)

1.汤姆斯的天堂梦

题目链接：P1796 汤姆斯的天堂梦 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

此题类似于数字三角形，要求从顶点到三角形最后一行的最大距离和，dp

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int N=105;
 4 int n,k,f[N][N];
 5 signed main()
 6 {
 7     ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
 8     cin>>n;
 9     memset(f,0x3f,sizeof f);
10     f[0][1]=0;
11     for(int i=1;i<=n;i++)
12     {
13         cin>>k;
14         for(int j=1;j<=k;j++)
15         {
16             int a,b;
17             cin>>a;
18             while(a!=0)
19             {
20                 cin>>b;
21                 f[i][j]=min(f[i][j],f[i-1][a]+b);
22                 cin>>a;
23             }
24         }
25     }
26     int ans=0x3f3f3f3f;
27     for(int i=1;i<=k;i++)
28     {
29         ans=min(ans,f[n][i]);
30     }
31     cout<<ans<<endl;
32     return 0;
33 }

2.跑步

题目链接：P1806 跑步 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

本题类似于01背包，可以从将数组从二维优化成一维

 1 #include <bits/stdc++.h>
 2 #define int long long
 3 using namespace std;
 4 const int N=510;
 5 int ans;
 6 int f[N][N];//f[i][j]表示总共跑i圈,最后一圈跑了j圈的方案数 
 7 //感觉就是01背包了
 8 signed main()
 9 {
10     ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
11     int n;
12     cin>>n;
13     for(int i=1;i<=n;i++)f[i][i]=1;
14     for(int i=1;i<=n;i++)
15          for(int j=1;j<i;j++)
16              for(int k=1;k<j&&k+j<=i;k++)
17                 f[i][j]=f[i][j]+f[i-j][k];
18     for(int i=1;i<n;i++) ans+=f[n][i];
19     cout<<ans;
20     return 0;
21 }

3.砝码称重

题目链接：P8742 [蓝桥杯 2021 省 AB] 砝码称重 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

f[i][j]表示从前i个砝码选，总重量是j的bool值

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 const int N=110,M=2e5+10,B=M/2;//f[i][j]表示从前i个砝码选，总重量是j的bool值 
 6 
 7 int n,m;//m表示所有方法的总重量 
 8 int w[N];
 9 bool f[N][M];
10 
11 signed main()
12 {
13     ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
14     cin>>n;
15     for(int i=1;i<=n;i++)cin>>w[i],m+=w[i];
16     f[0][B]=true;
17     for(int i=1;i<=n;i++)
18         for(int j=-m;j<=m;j++)
19         {
20             f[i][j+B]=f[i-1][j+B];
21             if(j-w[i]>=-m)f[i][j+B]|=f[i-1][j-w[i]+B];
22             if(j+w[i]<=m)f[i][j+B]|=f[i-1][j+w[i]+B];
23         }
24     int res=0;
25     for(int j=1;j<=m;j++)
26         if(f[n][j+B])res++;
27     
28     cout<<res<<endl;
29     return 0;
30 }

4.遗址

题目链接：P1959 遗址 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

如果枚举四个点，三个点，都会TLE，可以降到N^2的复杂度，我们可以从其中的两个点，推测中其中另外两个点

#include<bits/stdc++.h>
using namespace std;
int vis[5005][5005];
int x[3005];
int y[3005];
int main()
{
	int n;
	cin >> n;
	for (int i = 1; i <= n; i++)
	{
		cin >> x[i] >> y[i];
		vis[x[i]][y[i]] = 1;
	}
	int ans = 0;
	for (int i = 1; i <= n; i++)
	{
		for (int j = i+1; j <= n; j++)
		{
			int x1 = x[i] - x[j];
			int y1 = y[i] - y[j];
			int x2 = x[i] + y1;
			int y2 = y[i] - x1;
			int x3 = x[j] + y1;
			int y3 = y[j] - x1;
			if (x2 < 1 || x2>5000 || y2 < 1 || y2>5000 || x3 < 1 || x3>5000 || y3 < 1 || y3>5000) continue;
			if (vis[x2][y2] == 1 && vis[x3][y3] == 1)
			{
				ans = max(ans, x1 * x1 + y1 * y1);
			}
		}
	}
	cout << ans;
	return 0;
}

5.环境治理

题目链接：P8794 [蓝桥杯 2022 国 A] 环境治理 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

Floyd和二分的综合题目

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define LL long long
 4 LL  temp,n, Q;
 5 LL f[105][105],min_f[105][105],cut[105],dis[105][105];//cut为减少多少，dis表示减少后的
 6 bool check(LL day)
 7 {
 8     temp = 0;//赋初值
 9     LL times = day / n;
10     LL now = day%n;
11     for (int i = 1; i <= n; i++)
12     {
13         if (i <= now)cut[i] = times + 1;
14         else cut[i] = times;
15     }
16     /*for (int i = 1; i <= n; i++)
17         dis[i][i] = 0;*/
18     for (int i = 1; i <= n; i++)
19     {
20         for (int j = 1; j <= n; j++)
21         {
22             dis[i][j] = max(f[i][j] - cut[i] - cut[j], min_f[i][j]);
23         }
24     }
25     for (int k = 1; k <= n; k++)//Floyd模板
26         for (int i = 1; i <= n; i++)
27             for (int j = 1; j <= n; j++)
28                 dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]);
29  
30     for (int i = 1; i <= n; i++)
31     {
32         for (int j = 1; j <= n; j++)
33         {
34             temp += dis[i][j];
35         }
36     }
37     if (temp <= Q)return true;
38     else return false;
39 }
40 int main()
41 {
42     cin >> n>>Q;
43     for (int i = 1; i <= n; i++)
44     {
45         for (int j = 1; j <= n; j++)
46         {
47             cin >> f[i][j];
48         }
49     }
50     for (int i = 1; i <= n; i++)
51     {
52         for (int j = 1; j <= n; j++)
53         {
54             cin >> min_f[i][j];
55         }
56     }
57     LL left = 0, right = 1e9,mid/*,ans=1e9*/;
58     while (left<right)
59     {
60         mid = (left + right) / 2;
61         if (check(mid))/*ans=min(ans,mid),*/ right = mid;
62         else left = mid + 1;
63     }
64     if (left == 1e9)cout << "-1" << endl;
65     else cout << left << endl;
66     return 0;
67 }

 

posted @ 2023-01-16 20:50  Zac-saodiseng  阅读(545)  评论(0)    收藏  举报