BZOJ 1202: [HNOI2005]狡猾的商人 [带权并查集]

题意：

给出m个区间和，询问是否有区间和和之前给出的矛盾

---

NOIp之前做过hdu3038.....

带权并查集维护到根的权值和，向左合并

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N=1e6+;
typedef long long ll;
inline int read(){
    char c=getchar();int x=,f=;
    while(c<''||c>''){if(c=='-')f=-;c=getchar();}
    while(c>=''&&c<=''){x=x*+c-'';c=getchar();}
    return x*f;
}

int n, m, l, r, v;
int fa[N], val[N];
int find(int x) {
    if(x == fa[x]) return x;
    int root = find(fa[x]);
    val[x] += val[fa[x]];
    return fa[x] = root;
}

int main() {
    freopen("in", "r", stdin);
    int T=read();
    while(T--) {
        n=read()+; m=read();
        for(int i=; i<=n; i++) fa[i]=i, val[i]=;
        int flag=;
        for(int i=; i<=m; i++) {
            l=read(); r=read()+; v=read();
            if(!flag) continue;
            int x = find(l), y = find(r); //printf("hi %d %d  %d %d  %d %d\n", l,r,x,y,val[l],val[r]);
            if(x == y) {
                if(val[r] - val[l] != v) flag = ;
            } else fa[y] = x, val[y] = v - val[r] + val[l];
        }
        puts(flag ? "true" : "false");
    }
}