P3084 [USACO13OPEN]照片Photo dp

题意：

　　有n个区间，每个区间只能有一个斑点奶牛，问最多有几个斑点奶牛。

思路：

 

首先要处理出每个点的$L[i], R[i]$。

 

$L[i]$表示$L[i]～i-1$之间一定有一个点。i也是选中的。

$R[i]$表示$R[i]～i$之间只能有i这个点是选中的。

 

通过处理出$L[i]$和$R[i]$，$dp[i]$ = 最大的$L[i]$到$R[i]$间$dp值 + 1$。这里用线段树优化，也可以用优先队列。

#include <algorithm>
#include <iterator>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <iomanip>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <stack>
#include <cmath>
#include <queue>
#include <list>
#include <map>
#include <set>
#include <cassert>

using namespace std;
#define lson (l, mid, rt << 1)
#define rson (mid + 1, r, rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue

typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
typedef pair<int, pii> p3;
typedef pair<ll, int> pli;
//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n'

#define OKC                      \
    ios::sync_with_stdio(false); \
    cin.tie(0)
#define FT(A, B, C) for (int A = B; A <= C; ++A) //用来压行
#define REP(i, j, k) for (int i = j; i < k; ++i)
#define max3(a, b, c) max(max(a, b), c);
#define min3(a, b, c) min(min(a, b), c);
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000; //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e9 + 9;
const double esp = 1e-8;
const double PI = acos(-1.0);
const double PHI = 0.61803399; //黄金分割点
const double tPHI = 0.38196601;

template <typename T>
inline T read(T &x) {
    x = 0;
    int f = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
    while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
    return x = f ? -x : x;
}
/*-----------------------showtime----------------------*/

const int maxn = 200009;
int R[maxn], L[maxn];
int mx[maxn * 4];
int dp[maxn];
int query(int L, int R, int l, int r, int rt) {
    if (l >= L && r <= R) {
        return mx[rt];
    }
    int mid = (l + r) >> 1;
    int x = -inf;
    if (mid >= L) x = max(x, query(L, R, l, mid, rt << 1));
    if (mid < R) x = max(x, query(L, R, mid + 1, r, rt << 1 | 1));
    return x;
}

void update(int p, int c, int l, int r, int rt) {
    if (l == r) {
        mx[rt] = c;
        return;
    }
    int mid = (l + r) >> 1;
    if (mid >= p)
        update(p, c, l, mid, rt << 1);
    else
        update(p, c, mid + 1, r, rt << 1 | 1);
    mx[rt] = max(mx[rt << 1], mx[rt << 1 | 1]);
}

int main() {
    int n, m;
    scanf("%d%d", &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);
        L[y + 1] = max(L[y + 1], x);
        R[y] = min(R[y], x - 1);
    }
    memset(mx, -inf, sizeof(mx));

    for (int i = 1; i <= n + 1; i++) L[i] = max(L[i - 1], L[i]);
    for (int i = n; i >= 1; i--) R[i] = min(R[i + 1], R[i]);

    update(1, 0, 1, n + 2, 1);
    for (int i = 1; i <= n + 1; i++) {
        if (L[i] <= R[i])
            dp[i] = query(L[i] + 1, R[i] + 1, 1, n + 2, 1) + 1;
        else
            dp[i] = -inf;
        update(i + 1, dp[i], 1, n + 2, 1);
    }

    // for(int i=1; i<=n+1; i++){
    //     cout<<L[i]<<" -- "<<R[i]<<endl;
    // }
    if (dp[n + 1] <= 1)
        puts("-1");
    else
        printf("%d\n", dp[n + 1] - 1);
    return 0;
}