cf1136D. Nastya Is Buying Lunch(贪心)
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2022-03-27 12:04:28
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 ......
题意
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 << '\n';} 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; }