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; }