Codeforces Round #277.5 (Div. 2) A B C D-优快云博客

本文链接：https://blog.youkuaiyun.com/u014204835/article/details/41257893

本文探讨了排序算法中的A-SwapSort方法，通过贪心策略实现最小元素的前移；并介绍了BerSUBall算法，利用贪心原则在数组间进行元素匹配，同时提供了关于给定长度和数字总和的贪心解决方案。文章还讨论了基于贪心原则的其他应用，如不可忍受的争议性问题和数据结构与算法的实践。

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A - SwapSort:

每次找打最小值放在前面。

#include<stdio.h>
#include<iostream>
#include<math.h>
#include<algorithm>
using namespace std;
int n,a[3100];
int main()
{
    cin >> n;
    for(int i = 0; i < n; i++)
        cin >> a[i];
    cout << n << endl;
    for(int i = 0; i < n; i++)
    {
        int k = min_element(a + i, a + n) - a;//找到[a+i,a+n)里面的最小元素的下标
        //cout << k << endl;
        swap(a[i], a[k]);
        cout << i << " " << k << endl;
    }
    return 0;
}

B - BerSU Ball：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

#define N 105

int a[N], b[N];
int hash1[N], hash2[N];

int main()
{
    int n, m;
    while(~scanf("%d", &n))
    {
        memset(hash1, 0, sizeof(hash1));
        memset(hash2, 0, sizeof(hash2));
        for( int i = 1; i <= n; i++ )
        {
            scanf("%d", &a[i]);
            hash1[a[i]]++;
        }
        scanf("%d", &m);
        for( int i = 1; i <= m; i++ )
        {
            scanf("%d", &b[i]);
            hash2[b[i]]++;
        }
        
        int cnt = 0;
        sort(a+1, a+1+n);
        sort(b+1, b+1+m);
        
        if ( n > m )
        {
            for( int i = 1; i <= n; i++ )
            {
                if( hash2[a[i]-1] )
                {
                    hash2[a[i]-1]--;
                    cnt++;
                    continue;
                }
                if( hash2[a[i]] )
                {
                    hash2[a[i]]--;
                    cnt++;
                    continue;
                }
                if( hash2[a[i]+1] )
                {
                    hash2[a[i]+1]--;
                    cnt++;
                    continue;
                }
            }
        }
        else
        {
            for( int i = 1; i <= m; i++ )
            {
                if( hash1[b[i]-1] )
                {
                    hash1[b[i]-1]--;
                    cnt++;
                    continue;
                }
                if( hash1[b[i]] )
                {
                    hash1[b[i]]--;
                    cnt++;
                    continue;
                }
                if( hash1[b[i]+1] )
                {
                    hash1[b[i]+1]--;
                    cnt++;
                    continue;
                }
            }
        }
        printf("%d\n", cnt);
    }
    return 0;
}

C - Given Length and Sum of Digits...

贪心。

#include <iostream>

using namespace std;

void lesss(int m, int s)
{
    if (s == 0)
        cout<<"0"<<endl;
    else
    {
        if ((m-1)*9+1 >= s)
        {
            cout<<1;
            s--;
            m--;
        }
        while(m)
        {
            if ((m-1)*9 >= s)
            {
                cout<<0;
            }
            else
            {
                cout<<s-(9*(m-1));
                s -= s-(9*(m-1));
            }
            m--;
        }
    }
}

void best(int m, int s){

    while(m){
        if (s > 9){
            cout<<9;
            s -= 9;
        }
        else{
            cout<<s;
            s = 0;
        }
        m--;
    }
}

int main(){
    int m, s;
    cin>>m>>s;

    if (s == 0 && m != 1)
    {
        cout<<"-1 -1"<<endl;
    }
    else
    {
        if (s > m*9)
        {
            cout<<"-1 -1"<<endl;
        }
        else
        {
            lesss(m, s);
            cout<<" ";
            best(m, s);
            cout<<endl;
        }
    }
    return 0;
}

D - Unbearable Controversy of Being

贪心：
每次边连出去的边到达的那个点计数一下，最后统计就行了

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>

#define N 3005

using namespace std;

struct node{
	int to, nxt;
}edge[N*10];
int head[N];
int cnt;
int n, m;
int ans;
int mat[N][N];

void init()
{
	memset(head, -1, sizeof(head));
	memset(mat, 0, sizeof(mat));
	cnt = 1;
	ans = 0;
}

void addedge( int u, int v )
{
	edge[cnt].to = v;
	edge[cnt].nxt = head[u];
	head[u] = cnt++;
}

int f( int x )
{
	return x * (x-1) / 2;
}

int main()
{
	while(~scanf("%d%d", &n, &m))
	{
		init();
		int a, b;
		while(m--)
		{
			scanf("%d%d", &a, &b);
			addedge(a, b);
		}
		for( int i = 1; i <= n; i++ )
		{
			for( int j = head[i]; ~j; j = edge[j].nxt )
			{
				int v = edge[j].to;
				for( int k = head[v]; ~k; k = edge[k].nxt)
				{
					int o = edge[k].to;
					mat[i][o]++;
				}
			}
		}
		for( int i = 1; i <= n; i++)
		{
			for( int j = 1; j <= n; j++ )
			{
				if(i == j)
					continue;
				ans += f(mat[i][j]);
			}
		}
		printf("%d\n", ans);
	}
	return 0;
}