2019/10/24 CSP-S 模拟

T1 tom

题意：

考虑一定是属于\(a\)的在一坨，属于\(b\)的在一坨，找到这条连接\(a\)和\(b\)的边，然后分别直接按\(dfs\)序染色即可
注意属于\(a\)的连通块或属于\(b\)的连通块可能在\(dfs\)树上不都体现为一棵完整的子树，所以需要都判断一下

#include<bits/stdc++.h>
#define N (200000 + 10)
using namespace std;
inline int read() {
	int cnt = 0, f = 1; char c = getchar();
	while (!isdigit(c)) {if (c == '-') f = -f; c = getchar();}
	while (isdigit(c)) {cnt = (cnt << 3) + (cnt << 1) + (c ^ 48); c = getchar();}
	return cnt * f;
}
int n, a, b, x, y, fa[N], siz[N], id[N], val;
int first[N], to[N], nxt[N], tot;
void add (int x, int y) {nxt[++tot] = first[x], first[x] = tot, to[tot] = y;}
void get_siz(int x, int father) {
	siz[x] = 1; fa[x] = father;
	for (register int i = first[x]; i; i = nxt[i]) {
		int v = to[i];
		if (v == father) continue;
		get_siz(v, x), siz[x] += siz[v];
	}
}
void print(int x, int fa, int d) {
	for (register int i = first[x]; i; i = nxt[i]) {
		int v = to[i];
		if (v == fa) continue;
		print(v, x, d);
	}
	val += d;
	id[x] = val; 
}
void work () {
	bool ok = 0;
	get_siz(1, 0);
	for (register int i = 1; i <= n; ++i) 
		if (siz[i] == a) {
			val = 0;
			print(i, fa[i], 1);
			val = 0;
			print(fa[i], i, -1);
			ok = 1;
		} else if (siz[i] == b) {
			val = 0;
			print(i, fa[i], -1);
			val = 0;
			print(fa[i], i, 1);
			ok = 1;
		}
	if (!ok) puts("-1");
	else for (register int i = 1; i <= n; ++i) printf("%d ", id[i]);
	putchar('\n');
}
int main() {
	n = read(), a = read(), b = read();
	for (register int i = 1; i <= n - 1; ++i) {
		x = read(), y = read();
		add(x, y), add(y, x);
	}
	work();
	return 0;
}

T2 Jerry

首先能发现一个性质，括号结构最多嵌套两层
如果有三层的嵌套可以这样化简
( ( ( ) ) )
↓
( ( ) ( ) )
符号反三层和反一层等价
设\(f[i][0~2]\)表示当前处理到第\(i\)位，前面有\(0/1/2\)个未配对的左括号
在当前数\(>0\)时状态转移考虑补右括号，\(<0\)时考虑先强行在“-”后补一个左括号（如果对答案不优，自然会在后面的更新中去掉：-(a) + b），再进行转移
具体方程看代码

#include<bits/stdc++.h>
#define int long long
using namespace std;
inline int read() {
	int cnt = 0, f= 1;char c = getchar();
	while (!isdigit(c)) {if (c == '-') f = -f; c = getchar();}
	while (isdigit(c)) {cnt = (cnt << 3) + (cnt << 1) + (c ^ 48); c = getchar();}
	return cnt * f;
}
int T, n, a[500010], dp[500010][3];
signed main() {
	T = read();
	while (T--) {
		n = read();
		for (register int i = 1; i <= n; ++i) a[i] = read();
		dp[0][0] = 0, dp[0][1] = dp[0][2] = -1e18;
		for (register int i = 1; i <= n; ++i) 
			if (a[i] > 0) {
				dp[i][0] = max(max(dp[i - 1][0] + a[i], dp[i - 1][1] + a[i]), dp[i - 1][2] + a[i]);
				dp[i][1] = max(dp[i - 1][1] - a[i], dp[i - 1][2] - a[i]);
				dp[i][2] = dp[i - 1][2] + a[i];
			} else {
				dp[i][0] = -1e18;
				dp[i][1] = max(max(dp[i - 1][0] + a[i], dp[i - 1][1] + a[i]), dp[i - 1][2] + a[i]);
				dp[i][2] = max(dp[i - 1][1] - a[i], dp[i - 1][2] - a[i]);
			}
		printf("%lld\n", max(max(dp[n][0], dp[n][1]), dp[n][2]));
	}
	return 0;
}

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T3太麻烦了不想写，先咕了