loj#2483. 「CEOI2017」Building Bridges(dp cdq 凸包)
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2022-06-10 17:18:30
题意 "题目链接" Sol $$f[i], f[j] + (h[i] h[j])^2 + (w[i 1] w[j]))$$ 然后直接套路斜率优化,发现$k, x$都不单调 写个cdq就过了 ~~辣鸡noi.ac居然出裸题&&原题~~ cpp include define Pair pair defi ......
题意
sol
\[f[i], f[j] + (h[i] - h[j])^2 + (w[i - 1] - w[j]))\]
然后直接套路斜率优化,发现\(k, x\)都不单调
写个cdq就过了
辣鸡noi.ac居然出裸题&&原题
#include<bits/stdc++.h> #define pair pair<double, double> #define mp(x, y) make_pair(x, y) #define fi first #define se second #define int long long #define ll long long #define db double #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 = 1e18 + 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;} 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; struct sta { int id; db h, w, x, y, f, ad; void get() { x = 2 * h; y = f + h * h - w; } }a[maxn], st[maxn]; vector<pair> v; int comp(const sta &a, const sta &b) { return a.id < b.id; } double getk(pair a, pair b) { if((b.fi - a.fi) < eps) return inf; return (b.se - a.se) / (b.fi - a.fi); } int sid[maxn], cur; void getconvexhull(int l, int r) { while(cur) sid[cur--] = 0; v.clear(); for(int i = l; i <= r; i++) { double x = a[i].x, y = a[i].y; while((v.size() > 1 && ((getk(v[v.size() - 1], mp(x, y)) < getk(v[v.size() - 2], v[v.size() - 1])))) || ((v.size() > 0) && (v[v.size() - 1].fi == x) && (v[v.size() - 1].se >= y))) v.pop_back(), sid[cur--] = 0; v.push_back(mp(x, y)); sid[++cur] = a[i].id; } } int cnt = 0; db find(int id, db k) { int tmp = v.size(); int l = 0, r = v.size() - 1, ans = 0; while(l <= r) { int mid = l + r >> 1; if((mid == v.size() - 1) || (getk(v[mid], v[mid + 1]) > k)) r = mid - 1, ans = mid; else l = mid + 1; } return v[ans].se - k * v[ans].fi; } void cdq(int l, int r) { if(l == r) { a[l].get(); return ; } int mid = l + r >> 1; cdq(l, mid); getconvexhull(l, mid); for(int i = mid + 1; i <= r; i++) { chmin(a[i].f, find(i, a[i].h) + a[i].ad); } cdq(mid + 1, r); int tl = l, tr = mid + 1, tot = tl - 1; while(tl <= mid || tr <= r) { if((tr > r) || (tl <= mid && a[tl].x < a[tr].x)) st[++tot] = a[tl++];//?????tl <= mid else st[++tot] = a[tr++]; } for(int i = l; i <= r; i++) a[i] = st[i]; } signed main() { n = read(); for(int i = 1; i <= n; i++) a[i].h = read(); for(int i = 1; i <= n; i++) { a[i].w = read() + a[i - 1].w; a[i].id = i; a[i].f = a[i - 1].f + sqr(a[i].h - a[i - 1].h); a[i].ad = sqr(a[i].h) + a[i - 1].w; if(i == 1) a[1].f = 0; } cdq(1, n); sort(a + 1, a + n + 1, comp); // for(int i = 1; i <= n; i++) cout << i << ' ' << (ll)a[i].f << '\n'; cout << (ll)a[n].f; return 0; }
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