《算法分析》瑞格部分实验题

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一、递归与分治

1.排序

这里用归并排序：

#include <bits/stdc++.h>
using namespace std;

int n,a[1010],tmp[1010];

void merge_sort(int a[],int l,int r)
{
    if(l>=r) return ;

    int mid=l+r>>1;
    merge_sort(a,l,mid),merge_sort(a,mid+1,r);

    int k=0,i=l,j=mid+1;
    while(i<=mid && j<=r)
        if(a[i]<=a[j]) tmp[k++]=a[i++];
        else tmp[k++]=a[j++];

    while(i<=mid) tmp[k++]=a[i++];
    while(j<=r) tmp[k++]=a[j++];

    for(int i=l,j=0;i<=r;i++,j++)
        a[i]=tmp[j];
}

int main()
{
    cin>>n;
    for(int i=1;i<=n;i++) cin>>a[i];

    merge_sort(a,1,n);

    for(int i=1;i<=n;i++)
        cout<<a[i]<<" ";
    cout<<endl;

    return 0;
}

2.L型骨牌棋盘问题

#include <iostream>

using namespace std;

static int tile=1;
int board[100][100];
void chessboard(int tr,int tc,int dr,int dc,int size)
{
    if(size==1) return;
	int t=tile++,s=size/2;
	if(dr<tr+s&&dc<tc+s)
	   chessboard(tr,tc,dr,dc,s);
	else {
		board[tr+s-1][tc+s-1]=t;
		chessboard(tr,tc,tr+s-1,tc+s-1,s);
	}
	if(dr<tr+s&&dc>=tc+s)
	   chessboard(tr,tc+s,dr,dc,s);
	else {
		board[tr+s-1][tc+s]=t;
		chessboard(tr,tc+s,tr+s-1,tc+s,s);
	}
	if(dr>=tr+s&&dc<tc+s)
	   chessboard(tr+s,tc,dr,dc,s);
	else {
		board[tr+s][tc+s-1]=t;
		chessboard(tr+s,tc,tr+s,tc+s-1,s);
	}
	if(dr>=tr+s&&dc>=tc+s)
	    chessboard(tr+s,tc+s,dr,dc,s);
	else {
		board[tr+s][tc+s]=t;
		chessboard(tr+s,tc+s,tr+s,tc+s,s);
	}
}

int main()
{
	int n,dr,dc;
	cin >> n >> dr >> dc;
	dr--,dc--;
	chessboard(0,0,dr,dc,n);
	for(int i=0;i<n;i++){
		for(int j=0;j<n;j++){
			printf("%2d ",board[i][j]);
		}
		printf("\n");
	}
	return 0;
}

---

二、DP

1.最长公共子序列

#include <bits/stdc++.h>
using namespace std;

const int N=110;
int n,m;
char a[N],b[N],ans[N];
int f[N][N];
bool st[N];//标记数组

int main()
{
    cin>>n>>m;
    cin>>a+1>>b+1;

    int k=1;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
        {
            f[i][j]=max(f[i-1][j],f[i][j-1]);
            if(a[i]==b[j])
            {
                f[i][j]=max(f[i][j],f[i-1][j-1]+1);
                if(!st[i])//没用过 记录
                    st[i]=true,ans[k++]=a[i];
            }
        }

    for(int i=1;i<k;i++)
        cout<<ans[i];
    cout<<endl;

    return 0;
}

2.数字三角形（记录答案）

#include <bits/stdc++.h>
using namespace std;

const int N=110;
int n,a[N][N],backup[N][N];
int cnt_x[N][N],cnt_y[N][N];
stack<int>q;

int main()
{
    cin>>n;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=i;j++)
            cin>>a[i][j],backup[i][j]=a[i][j];

    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=i;j++)
        {
            if(a[i-1][j-1]>=a[i-1][j])
            {
                a[i][j]+=a[i-1][j-1];
                cnt_x[i][j]=i-1;
                cnt_y[i][j]=j-1;
            }
            else
            {
                a[i][j]+=a[i-1][j];
                cnt_x[i][j]=i-1;
                cnt_y[i][j]=j;
            }
        }
    }
    int maxx=-1,f_x=n,f_y=1;
    for(int i=1;i<=n;i++)
        if(maxx<a[n][i])
            maxx=a[n][i],f_y=i;

    while(f_x&&f_y)
    {
        int t=backup[f_x][f_y];
        q.push(t);
        int tx=f_x,ty=f_y;
        f_x=cnt_x[tx][ty],f_y=cnt_y[tx][ty];
    }

    cout<<maxx<<endl;
    while(!q.empty())
    {
        cout<<q.top()<<" ";
        q.pop();
    }
    cout<<endl;

    return 0;
}

---

三、贪心

1.水题

#include <bits/stdc++.h>
using namespace std;

int n,k,x,ans;

int main()
{
    cin>>n>>k>>x;

    int now=n;
    while(k--)
    {
        int nx;
        cin>>nx;
        if(now-x>=x)
        {
            now-=x;
            x=nx;
            continue;
        }
        else
        {
            ans++,now=n;
            //cout<<
        }
    }
    cout<<ans<<endl;

    return 0;
}

2.结束时间优先策略 👈

终于到这里了，写这篇博客主要就是为了ma这题的出题人，太**傻*了 整个题面说求 去掉的最少闭区间数 ，结果评测的是去掉后剩的 最多闭区间个数 ，这题思路也就是这样求的。（搞得我debug输出看来看去不知道错哪了）

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

int n;
struct node
{
    int l, r;
} poi[110];

bool cmp(node a, node b)
{
    return a.r < b.r;
}

int main()
{
    cin >> n;
    for (int i = 0; i < n; i++)
    {
        cin >> poi[i].l >> poi[i].r;
        if(poi[i].l>poi[i].r)
        {
            int t=poi[i].l;
            poi[i].l=poi[i].r;
            poi[i].r=t;
        }
        //cout << poi[i].l <<" "<< poi[i].r<<endl;
    }
    sort(poi, poi + n, cmp);

    poi[109].r=-1;
    int cnt = 0, tmp = 109;
    for (int i = 0; i < n; i++)
    {
        if (poi[i].l > poi[tmp].r)
        {
            cnt++;
            tmp = i;
            //cout<<" :"<<poi[i].l<<" "<<poi[i].r<<endl;
        }
    }
    //cout << n-cnt << endl;
    cout << cnt << endl;

    //system("pause");
    return 0;
}