BZOJ 1202 狡猾的商人(带权并查集)

给出了l,r,w。我们就得知了s[r]-s[l-1]=w.也就是说，点l-1和点r的距离为w。

于是可以使用带权并查集，定义dis[i]表示点i到根节点的距离。查询和合并的时候维护一下就OK了。

如果账本有错误，那么这两点的距离一定不等于在并查集上面的距离。

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-3
# define MOD 100000007
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int res=0, flag=0;
    char ch;
    if((ch=getchar())=='-') flag=1;
    else if(ch>='0'&&ch<='9') res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')  res=res*10+(ch-'0');
    return flag?-res:res;
}
void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
const int N=10005;
//Code begin...

int fa[105], dis[105];
int find(int x)
{
    int tmp;
    if (fa[x]!=x) {
        tmp=find(fa[x]);
        dis[x]+=dis[fa[x]];
        fa[x]=tmp;
    }
    return fa[x];
}
int main ()
{
    int n, m, T, l, r, w;
    scanf("%d",&T);
    while (T--) {
        mem(dis,0);
        int flag=1;
        scanf("%d%d",&n,&m);
        FOR(i,0,n) fa[i]=i;
        while (m--) {
            scanf("%d%d%d",&l,&r,&w);
            if (!flag) continue;
            int u=find(l-1), v=find(r);
            if (u!=v) {
                dis[u]=dis[r]+w-dis[l-1];
                fa[u]=v;
            }
            else {
                if (dis[l-1]-dis[r]!=w) flag=0;
            }
        }
        puts(flag?"true":"false");
    }
    return 0;
}