HDU1055 POJ2054 Color a Tree

Color a Tree

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1068 Accepted Submission(s): 349

Problem Description Bob is very interested in the data structure of a tree. A tree is a directed graph in which a special node is singled out, called the "root" of the tree, and there is a unique path from the root to each of the other nodes.

Bob intends to color all the nodes of a tree with a pen. A tree has N nodes, these nodes are numbered 1, 2, ..., N. Suppose coloring a node takes 1 unit of time, and after finishing coloring one node, he is allowed to color another. Additionally, he is allowed to color a node only when its father node has been colored. Obviously, Bob is only allowed to color the root in the first try.

Each node has a “coloring cost factor”, Ci. The coloring cost of each node depends both on Ci and the time at which Bob finishes the coloring of this node. At the beginning, the time is set to 0. If the finishing time of coloring node i is Fi, then the coloring cost of node i is Ci * Fi.

For example, a tree with five nodes is shown in Figure-1. The coloring cost factors of each node are 1, 2, 1, 2 and 4. Bob can color the tree in the order 1, 3, 5, 2, 4, with the minimum total coloring cost of 33.



Given a tree and the coloring cZ喎?http://www.Bkjia.com/kf/ware/vc/" target="_blank" class="keylink">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"brush:java;">5 1 1 2 1 2 4 1 2 1 3 2 4 3 5 0 0
Sample Output

33

Source Asia 2004, Beijing (Mainland China)

參考

#include 
#include 
#define maxn 1002

struct Node{
	int pre, c, t;
	double w;
} V[maxn];

int findMax(int n, int root)
{
	int i, u;
	double maxv = -1;
	for(i = 1; i <= n; ++i){
		if(V[i].w > maxv && i != root){
			maxv = V[i].w; u = i;
		}
	}
	return u;
}

int main()
{
	int n, root, i, ans, a, b, m, pre, j;
	while(scanf("%d%d", &n, &root), n || root){
		for(i = 1, ans = 0; i <= n; ++i){
			scanf("%d", &V[i].c);
			ans += V[i].c; V[i].t = 1;
			V[i].w = V[i].c;
		}
		for(i = 1; i < n; ++i){
			scanf("%d%d", &a, &b);
			V[b].pre = a;
		}
		for(i = 1; i < n; ++i){
			m = findMax(n, root);
			V[m].w = 0; pre = V[m].pre;
			ans += V[m].c * V[pre].t;
			for(j = 1; j <= n; ++j)
				if(V[j].pre == m) V[j].pre = pre;
			V[pre].c += V[m].c;
			V[pre].t += V[m].t;
			V[pre].w = (double)V[pre].c / V[pre].t;
		}
		printf("%d\n", ans);
	}
	return 0;
}