HDU 4679 Terrorist’s destroy
Terrorist’s destroy
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 621 Accepted Submission(s): 184
Total Submission(s): 621 Accepted Submission(s): 184
Problem Description
There is a city which is built like a tree.A terrorist wants to destroy the city's roads. But now he is alone, he can only destroy one road, then the city will be divided into two cities. Impression of the city is a number defined as the distance between the farthest two houses (As it relates to the fare).When the terrorist destroyed a road, he needs to spend some energy, assuming that the number is a.At the same time,he will get a number b which is maximum of the Impression of two cities. The terrorist wants to know which road to destroy so that the product of a and b will be minimized.You should find the road's id.
Note that the length of each road is one.
Note that the length of each road is one.
Input
The first line contains integer T(1<=T<=20), denote the number of the test cases.
For each test cases,the first line contains a integer n(1 < n <= 100000);denote the number of the houses;
Each of the following (n-1) lines contains third integers u,v,w, indicating there is a road between house u and houses v,and will cost terrorist w energy to destroy it.The id of these road is number from 1 to n-1.(1<=u<=n , 1<=v<=n , 1<=w<=10000)
For each test cases,the first line contains a integer n(1 < n <= 100000);denote the number of the houses;
Each of the following (n-1) lines contains third integers u,v,w, indicating there is a road between house u and houses v,and will cost terrorist w energy to destroy it.The id of these road is number from 1 to n-1.(1<=u<=n , 1<=v<=n , 1<=w<=10000)
Output
For each test case, output the case number first,and then output the id of the road which the terrorist should destroy.If the answer is not unique,output the smallest id.
Sample Input
2 5 4 5 1 1 5 1 2 1 1 3 5 1 5 1 4 1 1 3 1 5 1 1 2 5 1
Sample Output
Case #1: 2 Case #2: 3
Source
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zhuyuanchen520
各种bfs都是为了最后的枚举边，在枚举边的时候如果边不在直径上，就w*直径,在直径上借助先前预处理出来的当断裂时两城市的最大距离（dp加速）
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#define N 100100
#define INF 0x7ffffff
using namespace std;
struct infor
{
int u,v,w,next,pt;
}a[2*N],Map[2*N];
int b[N],level[N],dep[N],pre[N],temp[N];
int dp1[N],dp2[N],sum[N],Top,n;
bool check[N],bir[N],inque[N];
int main()
{
//freopen("data.in","r",stdin);
void addeage(int x,int y,int pos);
void ra_bfs(int &s,int &e);
void DP(int *p);
int t,tem=1;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
memset(b,-1,sizeof(b));
Top=1;
for(int i=1;i<=n-1;i++)
{
int x,y,w;
scanf("%d %d %d",&x,&y,&w);
Map[i].u = x;
Map[i].v = y;
Map[i].w = w;
addeage(x,y,i);
addeage(y,x,i);
}
//求直径
int s=1,e;
ra_bfs(s,e); //s->e;
ra_bfs(e,s);//e->s
Top=0;
int point=s;
memset(check,false,sizeof(check));
while(true)
{
dep[Top++] = point;
check[point] = true;
int k = pre[point];
bir[a[k].pt] = false;
if(!pre[point])
{
break;
}
point=a[k].u;
}
//倒求
DP(dp2);
//正求
for(int i=0;i<=Top-1;i++)
{
temp[i] = dep[Top-1-i];
}
for(int i=0;i<=Top-1;i++)
{
dep[i] = temp[i];
}
DP(dp1);
//枚举结果
int Min=INF,key;
for(int i=1;i<=n-1;i++)
{
int com;
if(bir[i])
{
com = (Top-1)*(Map[i].w);
}else
{
int x = Map[i].u;
int y = Map[i].v;
if(level[x]<level[y])
{
com = max(dp1[x],dp2[y])*(Map[i].w);
}else
{
com = max(dp1[y],dp2[x])*(Map[i].w);
}
}
if(com<Min)
{
Min = com;
key = i;
}
}
printf("Case #%d: %d\n",tem++,key);
}
return 0;
}
void addeage(int x,int y,int pos)
{
a[Top].u = x;
a[Top].v = y;
a[Top].next = b[x];
a[Top].pt = pos;
b[x] = Top++;
bir[pos] = true;
}
void ra_bfs(int &s,int &e)
{
level[s] = 1;
queue<int>que;
memset(inque,false,sizeof(inque));
que.push(s);
inque[s] = true;
pre[s] = 0;
while(!que.empty())
{
int x = que.front();
que.pop();
for(int i=b[x];i!=-1;i=a[i].next)
{
int y = a[i].v;
if(!inque[y])
{
pre[y] = i;
level[y] = level[x]+1;
que.push(y);
inque[y] = true;
}
}
}
int Max=0;
for(int i=1;i<=n;i++)
{
if(level[i]>Max)
{
Max = level[i];
e = i;
}
}
}
void DP(int *p)
{
p[dep[0]] = 0;
memset(sum,0,sizeof(sum));
for(int i=1;i<=Top-1;i++)
{
int x = dep[i];
queue<int>que;
que.push(x);
sum[x] = 1;
int Max = 1;
while(!que.empty())
{
int x = que.front();
que.pop();
for(int i=b[x];i!=-1;i=a[i].next)
{
int y = a[i].v;
if(sum[y]==0&&!check[y])
{
sum[y] = sum[x] + 1;
Max = max(sum[y],Max);
que.push(y);
}
}
}
p[x] = max(p[dep[i-1]],i+Max-1);
}
}