AGC027 C - ABland Yard 拓扑排序

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目录

• [TOC]

题目链接

AGC027 C - ABland Yard

题解

发现有解的充要条件是有一个形为AABBAABBAABB的环

此时每一个点至少与两个不同颜色的点相连

对于初始不合法的点直接删掉，判断删掉后与其相连的点是否变为不合法

类似拓扑排序

代码

#include<bits/stdc++.h>
#define gc getchar()
#define pc putchar
inline int read() {
	int x = 0,f = 1;
	char c = gc;
	while(c < '0' || c > '9') {if(c == '-')f = -1; c = gc;}
	while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = gc;
	return x * f;
}
void print(long long  x) {
	if(x < 0) {
		pc('-');
		x = -x;
	}
	if(x >= 10) print(x / 10);
	pc(x % 10 + '0');
}
const int maxn = 1000007;
int val[maxn];
int n,m;
struct node {
	int v,next;
} edge[maxn];
int head[maxn],num = 0;
inline void add_edge(int u,int v) {
	edge[++ num].v = v; edge[num].next = head[u];head[u] = num;
}
int d[maxn][2];
std::queue<int>q;
char s[maxn];
bool vis[maxn];
int main() {
	n = read(),m = read();
	scanf("%s",s + 1);
	for(int i = 1;i <= n;++ i) {
		char c = s[i];
		if(c == 'A') val[i] = 0;
		else val[i] = 1;
	}
	for(int u,v,i = 1;i <= m;++ i) {
		u = read(),v = read();
		add_edge(u,v); add_edge(v,u);
		d[u][val[v]] ++,d[v][val[u]]++;
	}
	int count = 0;
	for(int i = 1;i <= n;++ i)
		if(!d[i][1] || !d[i][0])
			q.push(i),count ++ ,vis[i] = 1;
	while(!q.empty()) {
		int u = q.front(); q.pop();
		for(int i = head[u];i;i = edge[i].next) {
			int v = edge[i].v;
			d[v][val[u]] --;
		 	if((!d[v][1] || !d[v][0] )&& !vis[v]) {
		 		q.push(v); count++; vis[v] = 1;
			 }
		}
	}
	puts(count == n ? "No" : "Yes");
	return 0;
}