BZOJ 4195 && NOI 2015 并查集

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假设x1,x2,x3,代表程序中出现的变量,给定n个形如xi=xjxixj的变量相等/不等的约束条件,请判定是否可以分别为每一个变量赋予恰当的值,使得上述所有约束条件同时被满足。

并查集真心水啊QAQ。
不过毕竟编译提交啥的都一次就通过了还是很愉悦的(本来就该这样?)
水一发QAQ

#include <cstdio>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std;inline char get(void) {    static char buf[1000000], *p1 = buf, *p2 = buf;    if (p1 == p2) {        p2 = (p1 = buf) + fread(buf, 1, 1000000, stdin);        if (p1 == p2) return EOF;    }    return *p1++;}inline void read(int &x) {    x = 0; char c = get(); int sign = 1;    for (; c < '0' || c > '9'; c = get()) if(c == '-') sign = 0;    for (; c >= '0' && c <= '9'; x = (x << 1) + (x << 3) + c - '0', c = get());    x = sign ? x : -x;}const int N = 1001010;struct event {    int x, y;    bool eq;    event(int _x = 0, int _y = 0, bool e = 0):x(_x), y(_y), eq(e) {}    friend bool operator <(const event &a, const event &b) {        return a.eq > b.eq;    }};event a[N];int mp[N << 1], fa[N << 1], f[N << 1];int n, m, opt, test, cnt, x, y, l, r, mid, f1, f2, p;inline int find(int x) {    l = 0; r = cnt - 1;    while (l < r) {        mid = (l + r) >> 1;        if (x <= mp[mid]) r = mid;        else l = mid + 1;    }    return r + 1;}inline int getfa(int x) {    return x == f[x] ? x : f[x] = getfa(f[x]);}int main(void) {    freopen("1.in", "r", stdin);    read(test);    while (test--) {        read(n); cnt = 0;        for (int i = 0; i < n; i++) {            read(x); read(y); read(opt);            mp[cnt++] = x; mp[cnt++] = y;            a[i] = event(x, y, opt);        }        sort(mp, mp + cnt);        cnt = unique(mp, mp + cnt) - mp;        sort(a, a + n);        for (int i = 1; i <= cnt; i++) f[i] = i;        for (p = 0; a[p].eq && p < n; p++) {            f1 = getfa(find(a[p].x));            f2 = getfa(find(a[p].y));            if (f1 != f2) f[f1] = f2;        }        for (; p < n; p++) {            if (getfa(find(a[p].x)) == getfa(find(a[p].y))) {                puts("NO"); goto W;            }        }        puts("YES"); W:;    }    return 0;}
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