洛谷 [P1995] 程序自动分析

并查集+ 离散化

首先本题的数据范围很大,需要离散化,
STL离散化代码:

	//dat是原数据,id是编号,sub是数据的副本
		sort(sub + 1, sub + tot + 1);
		size = unique(sub + 1, sub + tot + 1) - sub - 1;
		for(int i = 1; i <= tot; i++) {
			id[i] = lower_bound(sub + 1, sub + size + 1, dat[i]) - sub;
		}

并查集所能维护的是具有传递性的关系,比如本题中 等于 就是这样的关系,而不等就不是.
所以本题的思路非常简单,首先处理出来等于的关系,对于每一个不等的关系找矛盾即可

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
using namespace std;
const int MAXN = 100005;
int init() {
	int rv = 0, fh = 1;
	char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-') fh = -1;
		c = getchar();
	}
	while(c >= '0' && c <= '9') {
		rv = (rv<<1) + (rv<<3) + c - '0';
		c = getchar();
	}
	return fh *rv;
}
struct opt{
	int x, y;
	bool f;
}e[MAXN];
int T, n, id[MAXN<<1], dat[MAXN<<1], tot, sub[MAXN<<1], size, fa[MAXN<<2];
int find(int x) {
	if(fa[x] != x) fa[x] = find(fa[x]);
	return fa[x];
}
void merge(int x, int y) {
	if(x == y) return;
	int r1 = find(x), r2 = find(y);
	if(r1 != r2) fa[r1] = r2;
}
int main() {
	T = init();
	while(T--){
		memset(e, 0, sizeof(e));
		n = init();
		tot = 0; size = 0;
		for(int i = 1; i <= n; i++) {
			e[i].x = init(); e[i].y = init(); e[i].f = init();
			tot++; dat[tot] = sub[tot] = e[i].x;
			tot++; dat[tot] = sub[tot] = e[i].y;
		}
		sort(sub + 1, sub + tot + 1);
		size = unique(sub + 1, sub + tot + 1) - sub - 1;
		for(int i = 1; i <= tot; i++) {
			id[i] = lower_bound(sub + 1, sub + size + 1, dat[i]) - sub;
		}
		for(int i = 1; i <= size; i++) {
			fa[i] = i;
		}
		/*for(int i = 1; i <= tot ; i++) printf("%d %d
", id[i], dat[i]);
		printf("
");*/
		bool fff = 0;
		for(int i = 1; i <= n; i++) {
			e[i].x = id[i * 2 - 1]; e[i].y = id[i * 2];
			if(e[i].f) {
				merge(e[i].x, e[i].y);
			}
		}
		for(int i = 1; i <= n; i++) {
			if(!e[i].f) {
				if(find(e[i].x) == find(e[i].y)) {fff = 1;break;}
			}
		}
		if(!fff) {
			printf("YES
");
		}else printf("NO
");
	}
	return 0;
}