HDU 4679 Terrorist’s destroy (2013多校8 1004题 树形DP)
Terrorist’s destroy
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 24 Accepted Submission(s): 6
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
Recommend
zhuyuanchen520
各种dfs,导致爆栈了,加个栈挂,C++交果断AC了。
我是先求树的直径。
如果去掉的不是树的直径上的边,那么去掉后乘积就是w*直径长度
如果去掉直径上的边,那么直径两端各dfs一次就可以了
1 /* *********************************************** 2 Author :kuangbin 3 Created Time :2013/8/15 14:48:54 4 File Name :F:\2013ACM练习\2013多校8\1004.cpp 5 ************************************************ */ 6 #pragma comment(linker, "/STACK:1024000000,1024000000") 7 8 #include <stdio.h> 9 #include <string.h> 10 #include <iostream> 11 #include <algorithm> 12 #include <vector> 13 #include <queue> 14 #include <set> 15 #include <map> 16 #include <string> 17 #include <math.h> 18 #include <stdlib.h> 19 #include <time.h> 20 using namespace std; 21 22 const int MAXN = 100010; 23 struct Edge 24 { 25 int to,next; 26 int id; 27 int w; 28 }edge[MAXN*2]; 29 int mm[MAXN]; 30 int maxn[MAXN]; 31 int smaxn[MAXN]; 32 int head[MAXN],tot; 33 void init() 34 { 35 memset(head,-1,sizeof(head)); 36 tot = 0; 37 } 38 void addedge(int u,int v,int w,int id) 39 { 40 edge[tot].to = v; 41 edge[tot].w = w; 42 edge[tot].id = id; 43 edge[tot].next = head[u]; 44 head[u] = tot++; 45 edge[tot].to = u; 46 edge[tot].w = w; 47 edge[tot].id = id; 48 edge[tot].next = head[v]; 49 head[v] = tot++; 50 } 51 void dfs(int u,int pre) 52 { 53 mm[u] = 0; 54 maxn[u] = 0; 55 smaxn[u] = 0; 56 for(int i = head[u];i != -1;i = edge[i].next) 57 { 58 int v = edge[i].to; 59 if(v == pre)continue; 60 dfs(v,u); 61 if(maxn[v]+1 > smaxn[u]) 62 { 63 smaxn[u] = maxn[v] + 1; 64 if(smaxn[u] > maxn[u]) 65 { 66 swap(smaxn[u],maxn[u]); 67 } 68 } 69 if(mm[v] > mm[u]) 70 mm[u] = mm[v]; 71 } 72 mm[u] = max(mm[u],maxn[u]+smaxn[u]); 73 } 74 int ans; 75 int dep[MAXN]; 76 int p[MAXN]; 77 bool used[MAXN]; 78 int cnt; 79 int index; 80 int a[MAXN]; 81 void solve(int u,int pre) 82 { 83 for(int i = head[u];i != -1;i = edge[i].next) 84 { 85 int v = edge[i].to; 86 int w = edge[i].w; 87 if(v == pre)continue; 88 solve(v,u); 89 if(used[v]) 90 { 91 a[edge[i].id] = max(a[edge[i].id],w*mm[v]); 92 } 93 else 94 { 95 a[edge[i].id] = max(a[edge[i].id],w*cnt); 96 } 97 } 98 } 99 ; 100 void dfs1(int u,int pre) 101 { 102 p[u] = pre; 103 dep[u] = dep[pre] + 1; 104 for(int i = head[u]; i != -1;i = edge[i].next) 105 { 106 int v = edge[i].to; 107 if(v==pre)continue; 108 dfs1(v,u); 109 } 110 } 111 112 int main() 113 { 114 //freopen("in.txt","r",stdin); 115 //freopen("out.txt","w",stdout); 116 int T; 117 int n; 118 scanf("%d",&T); 119 int u,v,w; 120 int iCase = 0; 121 while(T--) 122 { 123 iCase ++; 124 init(); 125 scanf("%d",&n); 126 for(int i = 1;i < n;i++) 127 { 128 scanf("%d%d%d",&u,&v,&w); 129 addedge(u,v,w,i); 130 } 131 dep[0] = 0; 132 dfs1(1,0); 133 u = 1; 134 for(int i = 1;i <= n;i++) 135 if(dep[u] < dep[i]) 136 u = i; 137 dfs1(u,0); 138 v = 1; 139 for(int i =1;i <= n;i++) 140 if(dep[v] < dep[i]) 141 v = i; 142 cnt = dep[v]-1; 143 memset(used,false,sizeof(used)); 144 int tmp = v; 145 while(tmp) 146 { 147 used[tmp] = true; 148 tmp = p[tmp]; 149 } 150 for(int i = 1;i <= n;i++) 151 a[i] = 0; 152 ans = 1000000000; 153 dfs(u,0); 154 solve(u,-1); 155 dfs(v,0); 156 solve(v,-1); 157 for(int i = 1;i < n;i++) 158 if(a[i]<ans) 159 { 160 ans = a[i]; 161 index = i; 162 } 163 printf("Case #%d: %d\n",iCase,index); 164 } 165 166 return 0; 167 }
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