ZOJ3195 Design the city [2017年6月计划 树上问题04]

Design the city

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Time Limit: 1 Second      Memory Limit: 32768 KB

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Cerror is the mayor of city HangZhou. As you may know, the traffic system of this city is so terrible, that there are traffic jams everywhere. Now, Cerror finds out that the main reason of them is the poor design of the roads distribution, and he want to change this situation.

In order to achieve this project, he divide the city up to N regions which can be viewed as separate points. He thinks that the best design is the one that connect all region with shortest road, and he is asking you to check some of his designs.

Now, he gives you an acyclic graph representing his road design, you need to find out the shortest path to connect some group of three regions.

Input

The input contains multiple test cases! In each case, the first line contian a interger N (1 < N < 50000), indicating the number of regions, which are indexed from 0 to N-1. In each of the following N-1 lines, there are three interger Ai, Bi, Li (1 < Li < 100) indicating there's a road with length Li between region Ai and region Bi. Then an interger Q (1 < Q < 70000), the number of group of regions you need to check. Then in each of the following Q lines, there are three interger Xi, Yi, Zi, indicating the indices of the three regions to be checked.

Process to the end of file.

Output

Q lines for each test case. In each line output an interger indicating the minimum length of path to connect the three regions.

Output a blank line between each test cases.

Sample Input

4
0 1 1
0 2 1
0 3 1
2
1 2 3
0 1 2
5
0 1 1
0 2 1
1 3 1
1 4 1
2
0 1 2
1 0 3

Sample Output

3
2

2
2

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Author: HE, Zhuobin
Source: ZOJ Monthly, May 2009

 

三个点的距离等于任意两点间距离加和除以2

 

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cstdlib>
inline void read(int &x)
{
	char ch = getchar();char c = ch;x = 0;
	while(ch < '0' || ch > '9')c = ch, ch = getchar();
	while(ch <= '9' && ch >= '0')x = x * 10 + ch - '0', ch = getchar();
	if(c == '-')x = -x;
}
inline void swap(int& a, int& b){int tmp = a;a = b;b = tmp;}
const int MAXN = 50000 + 10;
struct Edge{int u,v,w,next;}edge[MAXN << 1];
int head[MAXN], cnt, n, m;
inline void insert(int a,int b, int c){edge[++cnt] = Edge{a,b,c,head[a]};head[a] = cnt;}
int log2[MAXN], pow2[30];
int p[30][MAXN], deep[MAXN], len[MAXN];int b[MAXN];

void dfs(int u)
{
	for(int pos = head[u];pos;pos = edge[pos].next)
	{
		int v = edge[pos].v;
		if(b[v])continue;
		b[v] = true;
		len[v] = len[u] + edge[pos].w;
		deep[v] = deep[u] + 1;
		p[0][v] = u;
		dfs(v);
	}
}

inline void yuchuli()
{
	b[1] = true;
	deep[1] = 0;
	dfs(1);
	for(register int i = 1;i <= log2[n];i ++)
		for(register int j = 1;j <= n;j ++)
			p[i][j] = p[i - 1][p[i - 1][j]];
}

inline int lca(int va, int vb)
{
	if(deep[va] < deep[vb])swap(va,vb);
	for(register int i = log2[n];i >= 0;i --)
		if(deep[va] - pow2[i] >= deep[vb])
			va= p[i][va];
	if(va == vb)return va;
	for(register int i = log2[n];i >= 0;i --)
	{
		if(p[i][va] != p[i][vb])
		{
			va = p[i][va];
			vb = p[i][vb];
		}
	}
	return p[0][va];
}

inline int l(int va, int vb)
{
	int k = lca(va, vb);
	return len[va] + len[vb] - (len[lca(va, vb)] << 1);
}

int main()
{
	register int tmp1,tmp2,tmp3;
	log2[0] = -1;
	for(register int i = 1;i <= MAXN;++ i)log2[i] = log2[i >> 1] + 1;
	pow2[0] = 1;
	for(register int i = 1;i <= 30;++ i)pow2[i] = pow2[i - 1] << 1;
	bool ok = false;
	while(scanf("%d", &n) != EOF)
	{
		if(ok)putchar('\n'),putchar('\n');
		cnt = 0;memset(head, 0, sizeof(head));
		memset(edge, 0, sizeof(edge));
		memset(deep, 0, sizeof(deep));
		memset(p, 0, sizeof(p));m = 0;
		memset(b, 0, sizeof(b));
		memset(len, 0, sizeof(len));
		for(register int i = 1;i < n;++ i)
		{
			read(tmp1);read(tmp2);read(tmp3);
			insert(tmp1 + 1, tmp2 + 1, tmp3);
			insert(tmp1 + 1, tmp1 + 1, tmp3);
		}
		yuchuli();
		read(m);
		read(tmp1);read(tmp2);read(tmp3);
		++ tmp1;++ tmp2;++ tmp3;
		printf("%d", (l(tmp1, tmp2) + l(tmp2, tmp3) + l(tmp1, tmp3))>> 1);
		for(register int i = 2;i <= m;++ i)
		{
			read(tmp1);read(tmp2);read(tmp3);
			tmp1 ++;tmp2 ++;tmp3 ++;
			int a = l(tmp1, tmp2);int b = l(tmp2, tmp3);int c = l(tmp1, tmp3);
			printf("\n%d", (l(tmp1, tmp2) + l(tmp2, tmp3) + l(tmp1, tmp3))>> 1);
		}
		ok = true;
	}
	return 0;
} 

 

转载于:https://www.cnblogs.com/huibixiaoxing/p/7076239.html