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

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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+5;
typedef long long ll;
inline int read(){
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';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()+1; m=read();
        for(int i=1; i<=n; i++) fa[i]=i, val[i]=0;
        int flag=1;
        for(int i=1; i<=m; i++) {
            l=read(); r=read()+1; v=read();
            if(!flag) continue;
            int x = find(l), y = find(r); //printf("hi %d %d  %d %d  %d %d
", l,r,x,y,val[l],val[r]);
            if(x == y) {
                if(val[r] - val[l] != v) flag = 0; 
            } else fa[y] = x, val[y] = v - val[r] + val[l];
        }
        puts(flag ? "true" : "false");
    }
}