牛客练习赛19 托米的简单表示法

托米的简单表示法

思路：网上说dfs建树然后树形dp，我不是很懂。。。 

我的写法是，扫一遍字符串找到每个括号的匹配括号，然后给每个括号一个标，每个匹配括号的

宽度为两者在字符串中的下标差值，高度为它们两个之间的括号的标号最大值-自身的标号 + 1，然后算就好啦。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define pii pair<int, int>

using namespace std;

const int N = 4e5 + 7;
const int M = 1e6 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 +7;

int n, a[N], id, mx[N << 2], l[N], mp[N];
char s[N];

void update(int pos, int v, int l, int r, int rt) {
    if(l == r) {
        mx[rt] = v;
        return;
    }

    int mid = l + r >> 1;
    if(pos <= mid) update(pos, v, l, mid, rt << 1);
    if(pos > mid) update(pos, v, mid + 1, r, rt << 1 | 1);
    mx[rt] = max(mx[rt << 1], mx[rt << 1 | 1]);
}

int getMx(int L, int R, int l, int r, int rt) {
    if(l >= L && r <= R) return mx[rt];
    int mid = l + r >> 1;
    int ans = 0;
    if(L <= mid) ans = max(ans, getMx(L, R, l, mid, rt << 1));
    if(R > mid) ans = max(ans, getMx(L, R, mid + 1, r, rt << 1 | 1));
    return ans;
}

int main(){
    int T; scanf("%d", &T);
    while(T--) {
        scanf("%s", s + 1);
        n = strlen(s + 1);
        for(int i = 1; i <= n; i++) {
            if(s[i] == '(') a[i] = ++id, mp[id] = i;
            else a[i] = id, l[i] = mp[id--];
            update(i, a[i], 1, n, 1);
        }

        LL ans = 0;
        for(int i = 1; i <= n; i++) {
            if(s[i] == ')') {
                int op = (a[i] & 1) ? 1 : -1;
                int wide = i - l[i];
                int hight = getMx(l[i], i, 1, n, 1) - a[i] + 1;

                ans += 1ll * op * wide * hight;
            }
        }

        printf("%lld
", ans);
    }
    return 0;
}

/*
*/