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题目出自 Codeforces Round #126 (Div. 2) 的E。题意大致如下：给定a，b，c，s，求三个非负整数x，y，z，满足0<=x<=y<=z，ax+by+cz=s，使得f(x,y,z)=|ax-by|+|by-cz|最小
思路：枚举z，得到一个方程ax+by=s-cz，用扩展欧几里得求出这个方程的一个解，然后三分通解的整系数，求出最小f值。至于为什么可以三分画画图就清楚了，两个绝对值函数叠加在一起最多只有三种状态(第一维表示临界点较小的那个绝对值函数)：(降,降)，(升,降)，(升,升)，无论两个函数哪个变化快，最终趋势都是：降然后升(由于临界点的情况不同，可能变成了单调的，但并不影响我们用三分求解)。
思来想去，决定搞一个三分的框架来避免头疼的临界问题（这是求最小值，求最大值时只需把<=换成>=即可，另外函数值在一段范围内不发生变化可能导致结果出错）：
1 2 3 4 5 6 7 | int L = ..., R = ...;while (L < R) { int M1 = L + (R - L) / 3, M2 = R - (R - L) / 3; if (F(M1) <= F(M2)) R = M2 - 1; else L = M1 + 1; } solve(L); |
出while循环后 L=R=目标解
下面是题目的源码：
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 | /* ******************************************************************************** */#include <iostream> //#include <cstdio> //#include <cmath> //#include <cstdlib> //#include <cstring> //#include <vector> //#include <ctime> //#include <deque> //#include <queue> //#include <algorithm> //#include <map> //#include <cmath> //using namespace std; // //#define pb push_back //#define mp make_pair //#define X first //#define Y second //#define all(a) (a).begin(), (a).end() //#define fillchar(a, x) memset(a, x, sizeof(a)) // //typedef pair<int, int> pii; //typedef long long ll; //typedef unsigned long long ull; // //#ifndef ONLINE_JUDGE //void RI(vector<int>&a,int n){a.resize(n);for(int i=0;i<n;i++)scanf("%d",&a[i]);} //void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R> //void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?1:-1; //while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T> //void print(const T t){cout<<t<<endl;}template<typename F,typename...R> //void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T> //void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;} //#endif // ONLINE_JUDGE //template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);} //template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);} //template<typename T> //void V2A(T a[],const vector<T>&b){for(int i=0;i<b.size();i++)a[i]=b[i];} //template<typename T> //void A2V(vector<T>&a,const T b[]){for(int i=0;i<a.size();i++)a[i]=b[i];} // //const double PI = acos(-1.0); //const int INF = 1e9 + 7; // ///* -------------------------------------------------------------------------------- */ll x, y, z, a, b, c, a0, b0;ll gcd(ll a, ll b) { return b? gcd(b, a % b) : a;}void gcd(ll a, ll b, ll &d, ll &x, ll &y) { if (!b) { d = a; x = 1; y = 0; } else { gcd(b, a % b, d, y, x); y -= x * (a / b); }}ll f(ll k) { ll xx = x + k * b0, yy = y - k * a0; return abs(xx * a - yy * b) + abs(yy * b - z * c);}bool chk(ll k1, ll k2) { if (x + k1 * b0 < 0) return false; if (x + k1 * b0 > z) return true; if (y - k1 * a0 < 0) return true; if (y - k1 * a0 > z) return false; if (x + k2 * b0 < 0) return false; if (x + k2 * b0 > z) return true; if (y - k2 * a0 < 0) return true; if (y - k2 * a0 > z) return false; if (x + k1 * b0 > y - k1 * a0) return true; if (x + k2 * b0 > y - k2 * a0) return true; return f(k1) <= f(k2);}int main() {#ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout);#endif // ONLINE_JUDGE int n, s; cin >> n >> s; int cnt[3] = {}; for (int i = 0; i < n; i ++) { int x; scanf("%d", &x); cnt[x - 3] ++; } a = cnt[0], b = cnt[1], c = cnt[2]; ll ans = INF, ix, iy, iz; for (z = 1; z * c <= s; z ++) { ll g; gcd(a, b, g, x, y); if ((s - z * c) % g) continue; ll K = (s - z * c) / g; x *= K; y *= K; a0 = a / g; b0 = b / g; ll L = -INF, R = INF; while (L < R) { ll M1 = L + (R - L) / 3, M2 = R - (R - L) / 3; if (chk(M1, M2)) R = M2 - 1; else L = M1 + 1; } ll xx = x + L * b0, yy = y - L * a0; if (0 <= xx && xx <= yy && yy <= z) { if (umin(ans, f(L))) { ix = xx; iy = yy; iz = z; } } } if (ans < INF) cout << ix << " " << iy << " " << iz << endl; else puts("-1"); return 0;}/* ******************************************************************************** */ |
[codeforces 200 E Tractor College]枚举，扩展欧几里得，三分
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原文地址：http://www.cnblogs.com/jklongint/p/4700892.html
