題目
發現每一行，列的差都相等
⎡⎣⎢124235346⎤⎦⎥
行
1-2=2-3=3-4
2-4=3-5=4-6
列
1-2=2-3=4-5
2-3=3-4=5-6
再發現了這個神奇的規律後，就可以用帶權並查集維護了。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
#define N 100005
using namespace std;
int T,n,m,num,fx[N],fy[N],wx[N],wy[N],min1[N],min2[N];
bool
 
 flag;
struct data{int x,y,val;}d[N];
bool cmpx(data a,data b){return a.x<b.x;}
bool cmpy(data a,data b){return a.y<b.y;}
int findx(int x){
    if(x==fx[x])return x;
    int rt=findx(fx[x]);
    wx[x]+=wx[fx[x]];
    return fx[x]=rt;
}
bool addx(int a,int b,int w){
    if(findx(a)!=findx(b)){
        int
 
 fa=findx(a),fb=findx(b);
        fx[fa]=fb;
        wx[fa]=wx[b]+w-wx[a];
        return true;
    }
    else return wx[a]==wx[b]+w;
}
int findy(int x){
    if(x==fy[x])return x;
    int rt=findy(fy[x]);
    wy[x]+=wy[fy[x]];
    return fy[x]=rt;
}
bool addy(int a,int b,int w){
    if(findy(a)!=findy(b)){
        int
 
 fa=findy(a),fb=findy(b);
        fy[fa]=fb;
        wy[fa]=wy[b]+w-wy[a];
        return true;
    }
    else return wy[a]==wy[b]+w;
}
void pno(){printf("No\n");}
int main()
{
    scanf("%d",&T);
    while(T--){
        flag=0;
        scanf("%d%d%d",&n,&m,&num);
        for(int i=1;i<=n;i++){fx[i]=i;wx[i]=0;}
        for(int i=1;i<=m;i++){fy[i]=i;wy[i]=0;}
        for(int i=1;i<=num;i++){
            scanf("%d%d%d",&d[i].x,&d[i].y,&d[i].val);
            if(d[i].val<0) flag=1;
        }
        if(flag==1){pno();continue;}
        sort(d+1,d+num+1,cmpx);
        for(int i=1;i<num;i++)
            if(d[i].x==d[i+1].x)
                if(!addy(d[i].y,d[i+1].y,d[i].val-d[i+1].val))
                    flag=1;
        if(flag==1){pno();continue;}
        sort(d+1,d+num+1,cmpy);
        for(int i=1;i<num;i++)
            if(d[i].y==d[i+1].y)
                if(!addx(d[i].x,d[i+1].x,d[i].val-d[i+1].val))
                    flag=1;
        if(flag==1){pno();continue;}
        memset(min1,0x7f,sizeof min1);
        memset(min2,0x7f,sizeof min2);
        for(int i=1;i<=num;i++){
            int rt=findx(d[i].x);
            min1[rt]=min(min1[rt],d[i].val-wx[d[i].x]);
        }
        for(int i=1;i<=n;i++){
            int rt=findx(i);
            min2[rt]=min(min2[rt],wx[i]);
        }
        for(int i=1;i<=n;i++)
            if(fx[i]==i&&min1[i]+min2[i]<0)
                flag=1;
        if(flag==1){pno();continue;}
        printf("Yes\n");
    }
    return 0;
}