cf1136D. Nastya Is Buying Lunch(贪心)

题意

题目链接

给出一个排列，以及(m)个形如((x, y))的限制，表示若(x)在(y)之前则可以交换(x, y)。

问(n)位置上的数最多能前进几步

(n leqslant 3* 10^5, m leqslant 5 * 10^5)

Sol

每次遇到这种动来动去的题基本都做不出来qwq

我最开始想到的是图论模型，然后发现不管怎么建都有反例。结果标算是个神仙贪心？。。

考虑这样一种贪心：从前往后处理每一个数，记一个(num)数组表示该位置的数最多能往后移动几次，每次把当前数的限制条件加到(num)里。如果当前的(num)大于等于和开始时最后一个数的距离，那么(ans++)。(好像看代码会更明白一些)

#include<bits/stdc++.h> 
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
//#define int long long 
#define LL long long 
#define ull unsigned long long 
#define Fin(x) {freopen(#x".in","r",stdin);}
#define Fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int MAXN = 1e6 + 10, mod = 1e9 + 7, INF = 1e9 + 10;
const double eps = 1e-9;
template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline LL add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
template <typename A, typename B> inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;}
template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod + mod) % mod;}
template <typename A> inline void debug(A a){cout << a << '
';}
template <typename A> inline LL sqr(A x){return 1ll * x * x;}
template <typename A, typename B> inline LL fp(A a, B p, int md = mod) {int b = 1;while(p) {if(p & 1) b = mul(b, a);a = mul(a, a); p >>= 1;}return b;}
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, M, p[MAXN], num[MAXN];
vector<int> v[MAXN];
signed main() {
    N = read(); M = read();
    for(int i = 1; i <= N; i++) p[i] = read();
    for(int i = 1; i <= M; i++) {
    	int x = read(), y = read();
    	v[y].push_back(x);
    }
    for(auto &x : v[p[N]]) num[x]++;
    int ans = 0;
    for(int i = N - 1; i >= 1; i--) {
    	if(num[p[i]] >= N - ans - i) ans++;
    	else for(auto &x : v[p[i]]) num[x]++;
	}
    cout << ans;
    return 0;
}