BZOJ 2880: Ipsc2009 Easy representation

一个合法的括号序列能够映射成一棵树
用栈处理括号序列之后将树建出来，可以发现对应矩形高度为该节点子树内的最大深度，矩形宽度为该节点之间的括号个数+1，dfs统计答案即可

#include <bits/stdc++.h>
#define pb push_back
#define fi first
#define se second
#define pii pair<ll, int>
#define lp p << 1
#define rp p << 1 | 1
#define mid ((l + r) >> 1)
#define lowbit(i) ((i) & (-i))
#define ll long long
#define ull unsigned long long
#define db double
#define rep(i,a,b) for(int i=a;i<b;i++)
#define per(i,a,b) for(int i=b-1;i>=a;i--)
#define Edg int ccnt=1,head[N],to[E],ne[E];void addd(int u,int v){to[++ccnt]=v;ne[ccnt]=head[u];head[u]=ccnt;}void add(int u,int v){addd(u,v);addd(v,u);}
#define Edgc int ccnt=1,head[N],to[E],ne[E],c[E];void addd(int u,int v,int w){to[++ccnt]=v;ne[ccnt]=head[u];c[ccnt]=w;head[u]=ccnt;}void add(int u,int v,int w){addd(u,v,w);addd(v,u,w);}
#define es(u,i,v) for(int i=head[u],v=to[i];i;i=ne[i],v=to[i])
const int MOD = 1e9 + 7;
void M(int &x) {if (x >= MOD)x -= MOD; if (x < 0)x += MOD;}
int qp(int a, int b = MOD - 2) {int ans = 1; for (; b; a = 1LL * a * a % MOD, b >>= 1)if (b & 1)ans = 1LL * ans * a % MOD; return ans % MOD;}
template<class T>T gcd(T a, T b) { while (b) { a %= b; std::swap(a, b); } return a; }
template<class T>bool chkmin(T &a, T b) { return a > b ? a = b, 1 : 0; }
template<class T>bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
char buf[1 << 21], *p1 = buf, *p2 = buf;
inline char getc() {
	return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;
}
inline int _() {
	int x = 0, f = 1; char ch = getc();
	while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getc(); }
	while (ch >= '0' && ch <= '9') { x = x * 10ll + ch - 48; ch = getc(); }
	return x * f;
}

const int N = 5e5 + 7, E = 5e5 + 7;
Edg
char s[N];
int st[N];
int n, top, id[N], tol, len[N], dep[N];
bool vis[N];
ll ans;

void dfs(int u, int opt) {
	vis[u] = 1;
	dep[u] = 1;
	es (u, i, v) {
		assert(vis[v] == 0);
		dfs(v, -1 * opt);
		chkmax(dep[u], dep[v] + 1);
	}
	ans += (ll)len[u] * dep[u] * opt;
}

void solve() {
	scanf("%s", s + 1);
	n = strlen(s + 1);
	rep (i, 1, n + 1) {
		if (s[i] == '(') {
			id[i] = ++tol;
			if (top) addd(id[st[top]], id[i]);
			st[++top] = i;
		} else {
			id[i] = id[st[top]];
			len[id[i]] = i - st[top--];
		}
	}
	rep (i, 1, tol + 1) {
		if (!vis[i]) dfs(i, 1);
	}
	printf("%lld\n", ans);
}
void init() {
	ccnt = 1;
	rep (i, 1, tol + 1) head[i] = vis[i] = 0;
	tol = ans = 0;
}

int main() {
#ifdef LOCAL
	freopen("ans.out", "w", stdout);
#endif
	int T;
	scanf("%d", &T);
	while(T--) solve(), init();
	return 0;
}