HDU 5596 GTW likes gt（逆向思维）

最新推荐文章于 2016-01-10 19:45:29 发布

原创最新推荐文章于 2016-01-10 19:45:29 发布 · 474 阅读

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本文介绍了一种利用两个Set维护动态数据结构的方法，并通过实际代码展示了如何优化算法至O(n)的时间复杂度，适用于算法竞赛中的特定问题。

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题意:

$有中文题面$

分析:

$我们发现b_1,b_2,...,b_x都加1就相当于b_{x+1},b_{x+2},...,b_n都减1$
$我们只要用两个set来维护两组人, 同时维护一个减去的delta, 每次被消灭的从对应的set中删去就好了$
$由于元素最多被删去n次, 复杂度为O(nlogn)$

坑:

$- - 元素可能重复,应该用multiset！$

代码:

//
//  Created by TaoSama on 2015-12-12
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 5e4 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, m;
int a[N], b[N], c[N];

int main() {
#ifdef LOCAL
    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
    ios_base::sync_with_stdio(0);

    int t; scanf("%d", &t);
    while(t--) {
        scanf("%d%d", &n, &m);
        for(int i = 1; i <= n; ++i) scanf("%d%d", a + i, b + i);
        memset(c, 0, sizeof c);
        for(int i = 1; i <= m; ++i) {
            int x; scanf("%d", &x);
            ++c[x];
        }
        multiset<int> x, y;
        multiset<int>::iterator iter;
        int delta = 0;
        for(int i = 1; i <= n; ++i) {
            b[i] -= delta;
            if(a[i]) {
                iter = y.upper_bound(b[i]);
                y.erase(y.begin(), iter);
                x.insert(b[i]);
            } else {
                iter = x.upper_bound(b[i]);
                x.erase(x.begin(), iter);
                y.insert(b[i]);
            }
            delta += c[i];
        }
        printf("%d\n", x.size() + y.size());
    }
    return 0;
}

分析:

$与上面的思想类似, 我们可以不用set, 我们考虑倒过来做, 这个时候要预处理c_i前缀和$
$我们只要维护从后往前两组对应的最大值, 就知道当前这个人会不会被消灭了, 因为他被消灭只和他后面的人有关系$
$这样就优化到了O(n)$

代码:

//
//  Created by TaoSama on 2015-12-12
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, m;
int a[N], b[N], c[N];

int main() {
#ifdef LOCAL
    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
    ios_base::sync_with_stdio(0);

    int t; scanf("%d", &t);
    while(t--) {
        scanf("%d%d", &n, &m);
        for(int i = 1; i <= n; ++i) scanf("%d%d", a + i, b + i);
        memset(c, 0, sizeof c);
        for(int i = 1; i <= m; ++i) {
            int x; scanf("%d", &x);
            ++c[x];
        }
        for(int i = 1; i <= n; ++i)
            c[i] += c[i - 1];

        int maxv[] = { -INF, -INF}, ans = n;
        for(int i = n; i; --i) {
            int t = a[i];
            b[i] -= c[i - 1];
            if(b[i] < maxv[t ^ 1]) --ans;
            maxv[t] = max(maxv[t], b[i]);
        }
        printf("%d\n", ans);
    }
    return 0;
}