POJ 1873 The Fortified Forest

The Fortified Forest

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 1894		Accepted: 561

Description

Once upon a time, in a faraway land, there lived a king. This king owned a small collection of rare and valuable trees, which had been gathered by his ancestors on their travels. To protect his trees from thieves, the king ordered that a high fence be built around them. His wizard was put in charge of the operation. 
Alas, the wizard quickly noticed that the only suitable material available to build the fence was the wood from the trees themselves. In other words, it was necessary to cut down some trees in order to build a fence around the remaining trees. Of course, to prevent his head from being chopped off, the wizard wanted to minimize the value of the trees that had to be cut. The wizard went to his tower and stayed there until he had found the best possible solution to the problem. The fence was then built and everyone lived happily ever after. 

You are to write a program that solves the problem the wizard faced. 

Input

The input contains several test cases, each of which describes a hypothetical forest. Each test case begins with a line containing a single integer n, 2 <= n <= 15, the number of trees in the forest. The trees are identified by consecutive integers 1 to n. Each of the subsequent n lines contains 4 integers xi, yi, vi, li that describe a single tree. (xi, yi) is the position of the tree in the plane, vi is its value, and li is the length of fence that can be built using the wood of the tree. vi and li are between 0 and 10,000. 
The input ends with an empty test case (n = 0). 

Output

For each test case, compute a subset of the trees such that, using the wood from that subset, the remaining trees can be enclosed in a single fence. Find the subset with minimum value. If more than one such minimum-value subset exists, choose one with the smallest number of trees. For simplicity, regard the trees as having zero diameter. 
Display, as shown below, the test case numbers (1, 2, ...), the identity of each tree to be cut, and the length of the excess fencing (accurate to two fractional digits). 

Display a blank line between test cases. 

Sample Input

6
 0  0  8  3
 1  4  3  2
 2  1  7  1
 4  1  2  3
 3  5  4  6
 2  3  9  8
3
 3  0 10  2
 5  5 20 25
 7 -3 30 32
0

Sample Output

Forest 1
Cut these trees: 2 4 5 
Extra wood: 3.16

Forest 2
Cut these trees: 2 
Extra wood: 15.00

Source

World Finals 1999 

 

/*
world final的题，做起来还是挺爽的，由于tree最多只有15棵，采用比特压缩来表示的话考虑每棵树取与不取
的所有状态是 状态位是1的0 - (2 ^ 15 - 1)，比特位为1的表示砍掉，由于不可能不砍树和砍掉全部树，所以
0和2 ^ 15 - 1状态是不可能取的。遍历所有状态然后从没有砍掉的树中找到凸包，然后计算其周长，并判断砍
掉的树能否满足其周长然后更新找到的最优的状态即可
*/
#include <iostream> #include <algorithm> #include <cmath> #include <stack> #define MAX_N 20 using namespace std; int treeN; struct tree { double x, y, h; int v; }trees[MAX_N + 1]; stack<int> bstack; int expv[MAX_N + 1], minTV, minCutNum, minState; double restLen; double xProduct(const int &id1, const int &id2, const int &id3) { return (trees[id3].x - trees[id1].x) * (trees[id2].y - trees[id1].y) - (trees[id2].x - trees[id1].x) * (trees[id3].y - trees[id1].y); } bool compareXY(const int &id1, const int &id2) { return trees[id1].y == trees[id2].y ? trees[id1].x <= trees[id2].x : trees[id1].y < trees[id2].y; } double getDist(const int &id1, const int &id2) { return sqrt((trees[id1].y - trees[id2].y) * (trees[id1].y - trees[id2].y) + (trees[id1].x - trees[id2].x) * (trees[id1].x - trees[id2].x)); } int idBL; //凸包的起始参考点 bool compareBag(const int &id1, const int &id2) { double xPro = xProduct(idBL, id1, id2); if(xPro < 0) return true; else if(xPro > 0) return false; else { double dist1 = getDist(idBL, id1); double dist2 = getDist(idBL, id2); return dist1 <= dist2; } } //利用凸包计算，当前方案是否可行，返回当前木料长度减去凸包的长度值 double canGo(double curLen, int restTN, int restT[MAX_N + 1]) { int i; sort(restT, restT + restTN, compareXY); idBL = restT[0]; if(restTN >= 2) sort(&restT[1], &restT[1] + restTN - 1, compareBag); while(!bstack.empty()) bstack.pop(); bstack.push(idBL); if(restTN >= 2) bstack.push(restT[1]); for(i = 2; i < restTN; i++) { int t3 = restT[i]; while(true) //一开始居然忘掉了求凸包的这个外层while循环以至于一直WA,杯具啊 { int t2 = bstack.top(); bstack.pop(); int t1 = bstack.top(); bstack.pop(); double xPro = xProduct(t1, t2, t3); bstack.push(t1); if(xPro < 0) { bstack.push(t2); break; } if(bstack.size() == 1) break; } bstack.push(t3); } double lenNeed = 0; int first = bstack.top(); bstack.pop(); int last = first, cur; while(!bstack.empty()) { cur = bstack.top(); bstack.pop(); lenNeed += getDist(cur, last); last = cur; } lenNeed += getDist(first, last); return curLen - lenNeed; } void solve(int caseNum) { int endState = pow(2.0, treeN) - 1; int curState, curTV, i, cutNum; int restT[MAX_N + 1], restN; double curTL; //遍历所有状态，除了最小值和最大值之外 for(curState = 1; curState < endState; curState++) { curTL = curTV = restN = cutNum = 0; for(i = 0; i < treeN; i++) { if(curState & expv[i]) //cut { curTL += trees[i].h; curTV += trees[i].v; cutNum++; } else //keep restT[restN++] = i; } if(curTV <= minTV) { if(curTV == minTV && cutNum >= minCutNum) continue; double curRestLen; if((curRestLen = canGo(curTL, restN, restT)) >= 0) { minTV = curTV; minCutNum = cutNum; restLen = curRestLen; minState = curState; } } } printf("Forest %d/n", caseNum); printf("Cut these trees:"); for(i = 0; i < treeN; i++) if(minState & expv[i]) printf(" %d", i + 1); printf("/nExtra wood: %.2lf/n/n", restLen); } int main() { int i, c = 0; expv[0] = 1; for(i = 1; i < MAX_N; i++) expv[i] = expv[i - 1] * 2; while(scanf("%d", &treeN) && treeN != 0) { //输入 c++; for(i = 0; i < treeN; i++) scanf("%lf%lf%d%lf", &trees[i].x, &trees[i].y, &trees[i].v, &trees[i].h); minTV = minCutNum = INT_MAX; solve(c); } return 0; }