P3084 [USACO13OPEN]照片Photo （dp+单调队列优化）

题目描述

Farmer John has decided to assemble a panoramic photo of a lineup of his N cows (1 <= N <= 200,000), which, as always, are conveniently numbered from 1..N. Accordingly, he snapped M (1 <= M <= 100,000) photos, each covering a contiguous range of cows: photo i contains cows a_i through b_i inclusive. The photos collectively may not necessarily cover every single cow.

After taking his photos, FJ notices a very interesting phenomenon: each photo he took contains exactly one cow with spots! FJ was aware that he had some number of spotted cows in his herd, but he had never actually counted them. Based on his photos, please determine the maximum possible number of spotted cows that could exist in his herd. Output -1 if there is no possible assignment of spots to cows consistent with FJ‘s photographic results.

输入输出格式
输入格式：

* Line 1: Two integers N and M.

* Lines 2..M+1: Line i+1 contains a_i and b_i.

输出格式：

* Line 1: The maximum possible number of spotted cows on FJ‘s farm, or -1 if there is no possible solution.

输入输出样例
输入样例#1：

5 3 
1 4 
2 5 
3 4 

输出样例#1：

1 

说明

There are 5 cows and 3 photos. The first photo contains cows 1 through 4, etc.

From the last photo, we know that either cow 3 or cow 4 must be spotted. By choosing either of these, we satisfy the first two photos as well.

#include <bits/stdc++.h>

using namespace std;
const int MAX_N = 2e5 + 5;
#define tomax(a, b) a = a>b?a:b
#define tomin(a, b) a = a<b?a:b

int N, M, l[MAX_N], r[MAX_N];
int f[MAX_N];

int main()
{
//    freopen("testdata.in", "r", stdin);
    cin >> N >> M;
    for (int i = 1; i <= N+1; i++)
        r[i] = i-1;
    for (int i = 1; i <= M; i++) {
        int x, y;
        scanf("%d%d", &x, &y);
        tomin(r[y], x-1);
        tomax(l[y+1], x);
    }
    for (int i = 2; i <= N+1; i++)
        tomax(l[i], l[i-1]);
    for (int i = N; i >= 1; i--)
        tomin(r[i], r[i+1]);
    memset(f, -1, sizeof f);
    f[0] = 0;
    for (int i = 1; i <= N+1; i++)
        for (int j = l[i]; j <= r[i]; j++) if(f[j] != -1)
            tomax(f[i], f[j] + (i!=N+1 ? 1 : 0));

    cout << f[N+1] << endl;
    return 0;
}
/*
5 3
1 4
2 4
1 1
*/

200000 2
1 100000
100001 200000

#include <bits/stdc++.h>

using namespace std;
const int MAX_N = 2e5 + 5;
#define tomax(a, b) a = a>b?a:b
#define tomin(a, b) a = a<b?a:b

int N, M, l[MAX_N], r[MAX_N];
int h, t, q[MAX_N], f[MAX_N];

int main()
{
    cin >> N >> M;
    memset(f, 0, sizeof f);
    for (int i = 1; i <= N+1; i++)
        r[i] = i-1;
    for (int i = 1; i <= M; i++) {
        int x, y;
        scanf("%d%d", &x, &y);
        tomin(r[y], x-1);
        tomax(l[y+1], x);
    }
    for (int i = 2; i <= N+1; i++)
        tomax(l[i], l[i-1]);
    for (int i = N; i >= 1; i--)
        tomin(r[i], r[i+1]);
    int j = 1;
    h = 1, t = 0, q[++t] = 0;
    for (int i = 1; i <= N+1; i++) {
        while (j <= N && j <= r[i]) {
            if (f[j] == -1) {
                ++j;
                continue;
            }
            while (h <= t && f[q[t]] <= f[j]) --t;
            q[++t] = j;
            ++j;
        }
        while (h <= t && q[h] < l[i]) ++h;
        if (h <= t) f[i] = f[q[h]] + (i!=N+1 ? 1 : 0);
        else f[i] = -1;
    }
    cout << f[N+1] << endl;
    return 0;
}