loj2305 NOI2017 游戏

题目链接

思路

既然(x)的数量那么小，我们就可以先把每个(x)搜索一遍。

枚举x的时候不需要把(a,b,c)全枚举一遍，只要枚举其中的两个就可以枚举到当前位置选任何车的情况。

然后就变成了只有('a','b','c')的序列。寻找满足题目要求的方案。

(2-sat)模型。

连边的时候注意一些技巧，否则(if)写到自闭。。

在(UOJ)上会被卡掉(3)分。实在懒得去卡常了233

代码

/*
* @Author: wxyww
* @Date:   2019-04-29 09:08:01
* @Last Modified time: 2019-04-30 14:32:38
*/
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<ctime>
using namespace std;
typedef long long ll;
const int N = 100010; 
ll read() {
	ll x=0,f=1;char c=getchar();
	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;
}
vector<int>e[N];
char a[N],tmp[N];
int n,D,m;
int opp[N];
#define add(x,y) e[x].push_back(y)
int val[N];

int tot,dfn[N],low[N],vis[N],sta[N],coljs,col[N],top;

void tarjan(int u) {
	int k = e[u].size();
	dfn[u] = low[u] = ++tot;
	sta[++top] = u;vis[u] = 1;
	for(int i = 0;i < k;++i) {
		int v = e[u][i];
		if(!dfn[v]) {
			tarjan(v);low[u] = min(low[u],low[v]);
		}
		else if(vis[v]) low[u] = min(low[u],low[v]);
	}
	if(dfn[u] == low[u]) {
		++coljs;
		do {
			int x = sta[top--];
			col[x] = coljs;
			vis[x] = 0;
		}while(sta[top + 1] != u);
	}
}


int t1[N],t2[N];
char h1[N],h2[N];

int get_u(int x,char y) {
	if(a[x] == 'a') return y == 'B' ? x : x + n;
	return y == 'A' ? x : x + n;
}

#define ote(x) x > n ? x - n : x + n

void solve() {
	memset(dfn,0,sizeof(dfn));memset(low,0,sizeof(low));
	memset(col,0,sizeof(col));
	coljs = 0,tot = 0;top = 0;
	for(int i = 1;i <= n + n;++i) e[i].clear();
	for(int i = 1;i <= m;++i) {
		if(a[t1[i]] == h1[i] - 'A' + 'a') continue;
		int u = get_u(t1[i],h1[i]);
		if(a[t2[i]] == h2[i] - 'A' + 'a') {
			add(u,ote(u));
			continue;
		}
		int v = get_u(t2[i],h2[i]);
		add(u,v);add(ote(v),ote(u));
	}
	for(int i = 1;i <= n + n;++i) if(!dfn[i]) tarjan(i);

	for(int i = 1;i <= n;++i) if(col[i] == col[i + n]) return;

	for(int i = 1;i <= n;++i) {
		if(col[i] < col[i + n]) {
			if(a[i] == 'a') putchar('B');
			else putchar('A');
		}
		else {
			if(a[i] == 'c') putchar('B');
			else putchar('C');
		}
	}
	exit(0);
}

void dfs(int pos) {
	if(pos == n + 1) {
		solve();
		return;
	}
	if(tmp[pos] != 'x') dfs(pos + 1);
	else {
		a[pos] = 'a';
		dfs(pos + 1);
		a[pos] = 'b';
		dfs(pos + 1);
	}
}

int main() {
	// freopen("2305/game3.in","r",stdin);
	n = read(),D = read();
	scanf("%s",tmp + 1);
	memcpy(a + 1,tmp + 1,n);
	m = read();
	for(int i = 1;i <= m;++i) {
		t1[i] = read();h1[i] = getchar();while(h1[i] != 'A' && h1[i] != 'B' && h1[i] != 'C') h1[i] = getchar();
		t2[i] = read();h2[i] = getchar();while(h2[i] != 'A' && h2[i] != 'B' && h2[i] != 'C') h2[i] = getchar();
	}

	dfs(1);
	puts("-1");
	return 0;
}