NOIP 模拟 $19; m u$

题解 (by;zjvarphi)

二维差分的题目

维护两个标记，一个向下传，一个向右下传；

对于每次更新，我们可以直接更新 ((r,c)+s,(r+l,c)-s) ; ((r,c+1)-s,(r+l,c+l+1)+s)，每组的后一个都是为了消除影响

Code

#include<bits/stdc++.h>
#define ri register signed
#define p(i) ++i
using namespace std;
namespace IO{
    char buf[1<<21],*p1=buf,*p2=buf;
    #define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++
    template<typename T>inline void read(T &x) {
        ri f=1;x=0;register char ch=gc();
        while(ch<'0'||ch>'9') {if (ch=='-') f=0;ch=gc();}
        while(ch>='0'&&ch<='9') {x=(x<<1)+(x<<3)+(ch^48);ch=gc();}
        x=f?x:-x;
    }
}
using IO::read;
namespace nanfeng{
    #define FI FILE *IN
    #define FO FILE *OUT
    template<typename T>inline T cmax(T x,T y) {return x>y?x:y;}
    template<typename T>inline T cmin(T x,T y) {return x>y?y:x;}
    typedef long long ll;
    static const int N=2e3+7;
    int n,q,r,c,l,s;
    ll fg1[N][N],fg2[N][N],res;
    inline int main() {
        // FI=freopen("nanfeng.in","r",stdin);
        // FO=freopen("nanfeng.out","w",stdout);
        read(n),read(q);
        for (ri i(1);i<=q;p(i)) {
            read(r),read(c),read(l),read(s);
            fg1[r][c]+=s,fg2[r][c+1]-=s;
            fg1[r+l][c]-=s,fg2[r+l][c+l+1]+=s;
        }
        for (ri i(1);i<=n;p(i)) {
            register ll fg=0;
            for (ri j(1);j<=n;p(j)) {
                fg+=fg1[i][j]+fg2[i][j];
                res^=fg;
                fg1[i+1][j]+=fg1[i][j];
                fg2[i+1][j+1]+=fg2[i][j];
            }
        }
        printf("%lld
",res);
        return 0;
    }  
}
int main() {return nanfeng::main();}