基础算法

前缀和

一维前缀和

[USACO16JAN] Subsequences Summing to Sevens S - 洛谷

这一题主要是需要结合数学知识来求解，

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>

using namespace std;

const int N = 50010;

int n;
int s[N], l[7], r[7];

int main() {
    scanf("%d", &n);

    for (int i = 1; i <= n; i ++) {
        int cur;
        scanf("%d", &cur);
        // 初始化前缀和矩阵
        s[i] = (s[i - 1] + cur) % 7;
    }
    // 从右往左，记录7的每个余数的最左边的坐标值
    for (int i = n; i ; i --) l[s[i]] = i;
    // 从左往右，记录7的每个余数的最右边的坐标值
    for (int i = 1; i <= n; i ++) r[s[i]] = i;
    l[0] = 0;

    int res = 0;
    for (int i = 0; i <= 6; i ++) {
        if (r[i]) res = max(res, r[i] - l[i]);
    }

    printf("%d\n", res);

    return 0;
}

二维前缀和

核心的原理

模板题
活动 - AcWing

例题

最大正方形 - 洛谷

这个题的N是100，比较小，可以用三重循环通过；故枚举正方形的边长，记作len
这个题的(x1,y1)等于(i,j) 这个题的(x2,y2)相当于(i+len,j+len)

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>

using namespace std;

const int N = 510;
int  a[N][N],s[N][N];
int n,m;

int main()
{
	cin>>n>>m;
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=m;j++)
		{
			cin>>a[i][j];
			s[i][j]=s[i-1][j]+s[i][j-1]-s[i-1][j-1]+a[i][j];
		}
	}
	
	//判断正方形 这个正方形里面的和是变长的平方
	// 枚举最大长度
    int res = 0;
    for (int i = 1; i <= n; i ++){
        for (int j = 1; j <= m; j ++) {
            for (int len = 0; i + len <= n && j + len <= m; len ++) {
                if (s[i + len][j + len] - s[i - 1][j + len] - s[i + len][j - 1] + s[i - 1][j - 1] == (len + 1) * (len + 1)){
                    res = max(len + 1, res);
                }
            }
        }
    }           
	cout<<res<<endl;
	
	
	return 0;
}

二维前缀和之固定边长的正方形
[HNOI2003] 激光炸弹 - 洛谷

#include<iostream>
#include<algorithm>
using namespace std;
int n, m;
int s[5050][5050];
int res = 0;



int main()
{
	cin >> n >> m;
	while (n--)
	{
		int x, y, v;
		cin >> x >> y >> v;
		s[x + 1][y + 1] += v;
	}

	for (int i = 1; i <= 5001; i++)
	{
		for (int j = 1; j <= 5001; j++)
		{
			s[i][j] += s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1];
		}
	}

	for (int i = m; i <= 5001; i++)
		for (int j = m; j <= 5001; j++)
		{
			//标准
			int value = s[i][j] - s[i - m][j] - s[i][j - m] + s[i - m][j - m];
			res = max(res, value);
		}
	printf("%d", res);

	return 0;
}

 

差分

一维差分

[Poetize6] IncDec Sequence - 洛谷

这个题里面的数学推理我现在都还没看明白

#include<bits/stdc++.h>
using namespace std;
const int N=100010;
typedef long long ll;
ll a[N];
ll b[N];
ll n;
int main()
{
	cin>>n;
	for(ll i=1;i<=n;i++)
	{
		cin>>a[i];
	}
	ll x,y;
	x=y=0;
	for(ll i=1;i<n;i++)
	{
		b[i]=a[i+1]-a[i];
		if(b[i]<0)x-=b[i];
		else y+=b[i];
	}
	cout<<max(x,y)<<endl;
	cout<<abs(x-y)+1<<endl;
	return 0;
}

二维差分

模板题

活动 - AcWing

#include<iostream>
#include<cstdio>
using namespace std;
const int N = 1e3 + 10;
int a[N][N], b[N][N];
//差分矩阵为什么是这个
void insert(int x1,int y1,int x2,int y2,int c)
{
    b[x1][y1]+=c;
    b[x2+1][y1]-=c;
    b[x1][y2+1]-=c;
    b[x2+1][y2+1]+=c;
}
// void insert(int x1, int y1, int x2, int y2, int c)
// {
//     b[x1][y1] += c;
//     b[x2 + 1][y1] -= c;
//     b[x1][y2 + 1] -= c;
//     b[x2 + 1][y2 + 1] += c;
// }
int main()
{
    int n, m, q;
    cin >> n >> m >> q;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            cin >> a[i][j];
            
    //初始化差分数组
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            insert(i, j, i, j, a[i][j]);      //构建差分数组
        }
    }
    while (q--)
    {
        int x1, y1, x2, y2, c;
        cin >> x1 >> y1 >> x2 >> y2 >> c;
        insert(x1, y1, x2, y2, c);
    }
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            b[i][j] += b[i - 1][j] + b[i][j - 1] - b[i - 1][j - 1];  //二维前缀和
        }
    }
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            printf("%d ", b[i][j]);
        }
        printf("\n");
    }
    return 0;
}

 洛谷地毯 - 洛谷

在模版的基础上稍微变一下形即可
 

#include<iostream>
#include<cstdio>
using namespace std;
const int N = 1e3 + 10;
int a[N][N], b[N][N];
//差分矩阵为什么是这个
void insert(int x1,int y1,int x2,int y2,int c)
{
    b[x1][y1]+=c;
    b[x2+1][y1]-=c;
    b[x1][y2+1]-=c;
    b[x2+1][y2+1]+=c;
}
// void insert(int x1, int y1, int x2, int y2, int c)
// {
//     b[x1][y1] += c;
//     b[x2 + 1][y1] -= c;
//     b[x1][y2 + 1] -= c;
//     b[x2 + 1][y2 + 1] += c;
// }
int main()
{
    int n, m, q;
    cin >> n >> m;
            
    //初始化差分数组
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= n; j++)
        {
            insert(i, j, i, j, 0);      //构建差分数组
        }
    }
    while (m--)
    {
        int x1, y1, x2, y2;
        cin >> x1 >> y1 >> x2 >> y2;
        insert(x1, y1, x2, y2, 1);
    }
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= n; j++)
        {
            b[i][j] += b[i - 1][j] + b[i][j - 1] - b[i - 1][j - 1];  //二维前缀和
        }
    }
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= n; j++)
        {
            printf("%d ", b[i][j]);
        }
        printf("\n");
    }
    return 0;
}

贪心 

与排序有关的贪心（注意排序的写法）

结构体排序（记忆一下）

[USACO1.3] 混合牛奶 Mixing Milk - 洛谷

bool cmp(cow farmer1,cow farmer2)
{
	return farmer1.p<farmer2.p;
}
也就是bool cmp(结构体名称 实例1，结构体名称 实例2)
{
    return 实例1的某个成员<实例2的某个成员;
}

#define _CRT_SECURE_NO_WARNINGS 1
#include<bits/stdc++.h>
using namespace std;
const int N = 2e6 + 10;
struct cow {
	int p;
	int maxnum;
}farmer[N];
bool cmp(cow farmer1,cow farmer2)
{
	return farmer1.p<farmer2.p;
}

int main()
{
	int m,n;
	int sum = 0;
	cin >> m>> n;
	for (int i = 0; i < n; i++)
	{
		cin >> farmer[i].p>>farmer[i].maxnum;
	}
	sort(farmer,farmer+n,cmp);
	
	//需要m牛奶
	int mark=0;
	while(farmer[mark].maxnum<m)
	{
		m-=farmer[mark].maxnum;
		sum+=farmer[mark].p*farmer[mark].maxnum;
		mark++;
	 } 
	 sum+=farmer[mark].p*m;
	 cout<<sum<<endl;
	
	return 0;
}