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Uva5211/POJ1873 The Fortified Forest 凸包

預處理 scan min bit mail 壓縮 spa init next

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題意:給出點集,每個點有個價值v和長度l,問把其中幾個點取掉,用這幾個點的長度能把剩下的點圍住,要求剩下的點價值和最大,拿掉的點最少且剩余長度最長。

思路:1999WF中的水題。考慮到其點的數量最多只有15個,那麽可以使用暴力枚舉所有取點情況,二進制壓縮狀態,預處理出該狀態下的價值,同時記錄該狀態擁有的點,並按價值排序。按價值枚舉狀態,並對擁有的這些點求凸包,check是否合法,找到一組跳出即可。然而POJ似乎沒有SPJ,同樣的代碼POJ會超時,UVA60ms,可以在常數上優化,不預處理,實時判價值最大值,不使用結構體等

/** @Date    : 2017-07-16 14:48:08
  * @FileName: POJ 1873 凸包.cpp
  * @Platform: Windows
  * @Author  : Lweleth ([email protected]
/* */) * @Link : https://github.com/ * @Version : $Id$ */ #include <stdio.h> #include <iostream> #include <string.h> #include <algorithm> #include <utility> #include <vector> #include <map> #include <set> #include <string> #include <stack> #include <queue> #include <math.h> //#include <bits/stdc++.h> #define LL long long #define PII pair<int ,int> #define MP(x, y) make_pair((x),(y)) #define fi first #define se second #define PB(x) push_back((x)) #define MMG(x) memset((x), -1,sizeof(x)) #define MMF(x) memset((x),0,sizeof(x)) #define MMI(x) memset((x), INF, sizeof(x)) using namespace std; const int INF = 0x3f3f3f3f; const int N = 1e5+20; const double eps = 1e-8; struct point { double x, y; point(){} point(double _x, double _y){x = _x, y = _y;} point operator -(const point &b) const { return point(x - b.x, y - b.y); } double operator *(const point &b) const { return x * b.x + y * b.y; } double operator ^(const point &b) const { return x * b.y - y * b.x; } }; double xmult(point p1, point p2, point p0) { return (p1 - p0) ^ (p2 - p0); } double distc(point a, point b) { return sqrt((double)((b - a) * (b - a))); } int sign(double x) { if(fabs(x) < eps) return 0; if(x < 0) return -1; else return 1; } //////////// point pt[50]; int v[50], l[50]; struct yuu { point p[50]; int val; int cnt; int len; int sta; double cst; void init(){cnt = len = val = cst = 0;} }tt[1 << 16]; point stk[50]; /*int cmpA(point a, point b)//以p[0]基準 極角序排序 { int t = xmult(a, b, p[0]); if(t > 0) return 1; if(t == 0) return distc(a, p[0]) < distc(b, p[0]); if(t < 0) return 0; }*/ int cmpB(yuu a, yuu b)//預處理用 { if(a.val == b.val) return a.cnt < b.cnt; return a.val < b.val; } int cmpC(point a, point b)//水平序排序 { return sign(a.x - b.x) < 0 || (sign(a.x - b.x) == 0 && sign(a.y - b.y) < 0); } /*void grahamS() { while(!s.empty()) s.pop(); for(int i = 0; i < min(n, 2); i++) s.push(i); int t = 1; for(int i = 2; i < n; i++) { while(s.size() > 1) { int p2 = s.top(); s.pop(); int p1 = s.top(); if(xmult(p[p1], p[p2], p[i]) > 0) { s.push(p2); break; } } s.push(i); } }*/ int graham(int st)//水平序 { sort(tt[st].p, tt[st].p + tt[st].cnt, cmpC); int top = 0; for(int i = 0; i < tt[st].cnt; i++) { while(top >= 2 && sign(xmult(stk[top - 2], stk[top - 1], tt[st].p[i])) <= 0) top--; stk[top++] = tt[st].p[i]; } int tmp = top; for(int i = tt[st].cnt - 2; i >= 0; i--) { while(top > tmp && sign(xmult(stk[top - 2],stk[top - 1] ,tt[st].p[i] )) <= 0) top--; stk[top++] = tt[st].p[i]; } if(tt[st].cnt > 1) top--; return top; } int check(int st) { int ma = graham(st); double ans = 0; stk[ma] = stk[0]; for(int i = 0; i < ma; i++) ans += distc(stk[i + 1], stk[i]); tt[st].cst = ans; if(sign(tt[st].len - ans) < 0) return 0; return 1; } int main() { int n; int cc = 0; while(~scanf("%d", &n) && n) { if(cc) printf("\n"); cc++; double x, y; for(int i = 0; i < n; i++) { scanf("%lf%lf%d%d", &x, &y, v + i, l + i); pt[i] = point(x, y); } int c = 0; for(int st = 0; st < (1 << n); st++)//預處理 { tt[st].sta = st; tt[st].init(); c = 0; for(int i = 0; i < n; i++) { if(st & (1 << i)) tt[st].len += l[i], tt[st].val += v[i]; else tt[st].p[c++] = pt[i]; } tt[st].cnt = c; } sort(tt, tt + (1 << n), cmpB); ////// for(int st = 0; st < (1 << n); st++)//遍歷 { if(check(st)) { printf("Forest %d\n", cc); printf("Cut these trees:"); for(int i = 0; i < n; i++) { if(tt[st].sta & (1 << i)) printf(" %d", i + 1); } printf("\nExtra wood: %.2lf\n", ((double)tt[st].len - tt[st].cst + eps)); break; } } } return 0; }

Uva5211/POJ1873 The Fortified Forest 凸包