前缀和--差分总结

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一维前缀和

一维前缀和就是用另一个数组记录前 i 项的和，然后可以算区间的和，还可以判断是否两个位置之间为0；
例如：
a[ i ]=1 ，2 ，3，4，5
sum[i] = 1，3，6,10,15
则任意区间[L,R]的和为sum[R]-sum[L-1];
前缀和

#include<bits/stdc++.h>
#define ll long long
#define PI 3.141592653589793
#define E 2.718281828459045
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
#define mem(a,b) memset(a,b,sizeof(a))
#define FOR( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define FO( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define lowbit(a) ((a)&-(a))
typedef unsigned long long ull;
const ll mod=10007;
const ll INF=0x3f3f3f3f;
const ll Max=1e7;
using namespace std;
ll t,n,m,l;
ll st[Max];//记录每个位置的状态
ll sum[Max];
/*queue<ll> q;
stack<ll> s;
//升序队列
priority_queue <int,vector<int>,greater<int> > q;
//降序队列
priority_queue <int,vector<int>,less<int> >q;*/
int main()
{
    ios::sync_with_stdio(false);
    cin>>n>>m;
    for(ll i=1;i<=n;i++)
    {
        cin>>t;
        sum[i]=sum[i-1]+t;
    }
    while(m--)
    {
        ll a,b;
        cin>>a>>b;
        cout<<sum[b]-sum[a-1]<<endl;
    }
    return 0;
}

二维前缀和

二维前缀和记录的是一个矩阵的和

这是一个二维数组，根据这个二维数组构造的前缀和矩阵就是将该点（x,y）到（1,1）所构成的矩阵的和，如下图

则对于这个前缀和矩阵的公式为:s[i][j]=s[i-1][j]+s[i][j-1]-s[i-1][j-1]+a[i][j];
由图可知黄色区域被加了两次，所以要减去一次，故最后的值为蓝色区域的和，再减去黄色区域，再加上6本身；

例如：对于i=2;j=3;a[1][1]和a[1][2]会被加两次，所以减去是s[1][2]
子矩阵的和
代码：

#include<bits/stdc++.h>
#define ll long long
#define PI 3.141592653589793
#define E 2.718281828459045
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
#define mem(a,b) memset(a,b,sizeof(a))
#define FOR( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define FO( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define lowbit(a) ((a)&-(a))
#define PII pair<ll ,ll >
#define ft first
#define sd second
typedef unsigned long long ull;
const ll mod=10007;
const ll INF=0x3f3f3f3f;
const ll Max=1e3+10;
using namespace std;
ll t,n,m,l,r;
ll ans;
ll st[Max],copst[Max];//记录每个位置的状态
ll a[Max][Max];
ll sum[Max][Max];
/*queue<ll> q;
stack<ll> s;
//升序队列
priority_queue <int,vector<int>,greater<int> > q;
//降序队列
priority_queue <int,vector<int>,less<int> >q;*/
int main()
{
    ios::sync_with_stdio(false);
    cin>>n>>m>>t;
    for(ll i=1;i<=n;i++)
        for(ll j=1;j<=m;j++)
    {
        cin>>a[i][j];
        sum[i][j]=sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1]+a[i][j];
    }
    while(t--)
    {
        ll x1,x2,y1,y2;
        cin>>x1>>y1>>x2>>y2;
        cout<<sum[x2][y2]-sum[x2][y1-1]-sum[x1-1][y2]+sum[x1-1][y1-1]<<endl;
    }
    return 0;
}

一维差分

差分是前缀和的逆用

利用两个数组：a[Max]储存原来的数据，b[Max]用来逆用前缀和
a[1]=b[1]
a[2]=b[1]+b[2]
a[3]=b[1]+b[2]+b[3];
…
a[i]=b[1]+…+b[i];
对于区间L-R修改C
只需要b[L]+=C;可以保证a[L]之后的都加上C（包括a[L]）
只需要b[R+1]-=C;可以保证a[R+1]后面都减去C(包括a[R+1]);
这两部可以保证区间[L,R]都修改C;
下面是代码

#include<bits/stdc++.h>
#define ll long long
#define PI 3.141592653589793
#define E 2.718281828459045
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
#define mem(a,b) memset(a,b,sizeof(a))
#define FOR( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define FO( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define lowbit(a) ((a)&-(a))
#define PII pair<ll ,ll >
#define ft first
#define sd second
typedef unsigned long long ull;
const ll mod=10007;
const ll INF=0x3f3f3f3f;
const ll Max=1e5+10;
using namespace std;
ll t,n,m,l,r,k;
ll ans;
ll a[Max];
ll b[Max];
/*queue<ll> q;
stack<ll> s;
//升序队列
priority_queue <int,vector<int>,greater<int> > q;
//降序队列
priority_queue <int,vector<int>,less<int> >q;*/
void insert(ll l,ll r,ll x)
{
    b[l]+=x;
    b[r+1]-=x;
}
int main()
{
    ios::sync_with_stdio(false);
    cin>>n>>m;
    for(ll i=1;i<=n;i++){cin>>a[i];insert(i,i,a[i]);}
    while(m--)
    {
        cin>>l>>r>>k;
        insert(l,r,k);
    }
    for(ll i=1;i<=l;i++){a[i]=a[i-1]+b[i];cout<<a[i]<<' ';}
    cout<<endl;

    return 0;
}

模板题
差分

二维差分

对于原数组构建差分数组
对于x1，y1，x2,y2区域进行修改C
（1）b[x1][y1]+=C;
则后面包含b[x1][y1]的矩阵都加上C

（2）b[x1][y2+1]-=C;

(3)b[x2+1][y1]-=C;(黑色区域修改了-C)

(4)b[x2+1][y2+1]+=C;

总结：经过上述操作就得到原矩阵的差分矩阵，再利用二维前缀和的逆用
二维差分

#include<bits/stdc++.h>
#define ll long long
#define PI 3.141592653589793
#define E 2.718281828459045
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
#define mem(a,b) memset(a,b,sizeof(a))
#define FOR( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define FO( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define lowbit(a) ((a)&-(a))
#define PII pair<ll ,ll >
#define ft first
#define sd second
typedef unsigned long long ull;
const ll mod=10007;
const ll INF=0x3f3f3f3f;
const ll Max=1e3+10;
using namespace std;
ll t,n,m,l,k,q;
ll ans;
ll a[Max][Max];
ll b[Max][Max];
/*queue<ll> q;
stack<ll> s;
//升序队列
priority_queue <int,vector<int>,greater<int> > q;
//降序队列
priority_queue <int,vector<int>,less<int> >q;*/
void insert(ll x1,ll y1,ll x2,ll y2,ll c)
{
    b[x1][y1]+=c;
    b[x1][y2+1]-=c;
    b[x2+1][y1]-=c;
    b[x2+1][y2+1]+=c;
}
int main()
{
    ios::sync_with_stdio(false);
    scanf("%lld%lld%lld",&n,&m,&q);
    for(ll i=1;i<=n;i++)
        for(ll j=1;j<=m;j++)
        scanf("%lld",&a[i][j]);
    for(ll i=1;i<=n;i++)
        for(ll j=1;j<=m;j++)
    {
        insert(i,j,i,j,a[i][j]);
    }
    while(q--)
    {   ll x1,x2,y1,y2,c;
        scanf("%lld%lld%lld%lld%lld",&x1,&y1,&x2,&y2,&c);
        insert(x1,y1,x2,y2,c);
    }
    for(ll i=1;i<=n;i++)
        {for(ll j=1;j<=m;j++)
    {
        b[i][j]=b[i-1][j]+b[i][j-1]-b[i-1][j-1]+b[i][j];//二维数组前缀和
        printf("%lld ",b[i][j]);
    }   puts("");
        }
    return 0;
}