SGU - 321 - The Spy Network

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先上題目：

321. The Spy NetworkTime limit per test: 0.5 second(s)
Memory limit: 65536 kilobytesinput: standard
output: standard

The network of spies consists of N intelligence officers. They are numbered with the code numbers from 1 to N so that nobody could discover them. The number 1 belongs to the radiowoman Kat. There is exactly N - 1 communication channels between the spies. It is known that a message from any spy to Kat can reach her. All channels are unidirectional.

A channel can have one of two types: protected and almost protected. It is known that a message will not be intercepted almost surely by the hostile security service if at least half of the channels along the path to radiowoman Kat are protected. What is the minimum number of channels to be made protected from almost protected, so that any message from any spy will not be intercepted almost surely ? What are those channels?

Input

The first line of the input contains the integer number N (1 ≤ N ≤ 200000). The following N - 1 lines contain the description of the communication channels. Each channel is described by a pair of the code numbers of spies (the direction of the channel is from the first spy to the second one) and the parameter pi. If pi = 

protected

, the channel is protected and if pi = 

almost protected

, the channel is almost protected.

Output

Write the number of channels to be converted to protected to the first line of the output. To the next line write numbers of channels to be made protected. If there are several solutions, choose any of them.

Example(s)

sample input	sample output
55 1 almost protected3 1 almost protected2 3 protected4 3 almost protected	21 2

 

 

　　　　題意：給你一棵樹，樹的邊是有向的，每條邊有一個狀態0或者1，現在告訴你根是1，問你最少需要改變多少多少條邊的狀態才能使每個點到達跟的路徑上有一半以上的邊是1，並且輸出數目和這些邊的編號。

　　　　做法：先從根節點dfs一次，對於記錄下每個節點還需要多少個1以及這個節點以及它的子樹中的節點最多需要1的節點需要1的數目。然後在從根節點再dfs一次，這次的目的是對於目前狀態為0的邊我們可以考慮是否將它轉化為1，因為對於這種轉化來說，轉移深度小的邊對樹的影響比轉移深度大的邊對樹的影響更大，所以我們如果一棵子樹裡面的節點還需要1的話，就轉換深度小的邊。也就是說這樣做符合貪心的思想。

 

上代碼：

 

 1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #define MAX 200002 5 using namespace std; 6  7 typedef struct { 8     int id,u,v,w,next; 9 } Edge;10 11 Edge e[MAX];12 char ch[100];13 int p[MAX],need[MAX],am[MAX];14 int tot,top;15 16 inline void add(int id,int u,int v,int w) {17     e[tot].id=id;18     e[tot].u=u;19     e[tot].v=v;20     e[tot].w=w;21     e[tot].next=p[u];22     p[u]=tot++;23 }24 25 void dfs1(int r,int dep,int a) {26     am[r]=(dep+1)/2 - a;27     for(int i=p[r]; i!=-1; i=e[i].next) {28         dfs1(e[i].v,dep+1,a+e[i].w);29         am[r]=max(am[e[i].v],am[r]);30     }31 }32 33 void dfs2(int r,int num) {34     for(int i=p[r]; i!=-1; i=e[i].next) {35         if(num<am[e[i].v]) {36             if(e[i].w==0) {37                 need[top++]=e[i].id;38                 dfs2(e[i].v,num+1);39             } else dfs2(e[i].v,num);40         }41     }42 }43 44 int main() {45     int n,u,v;46     //freopen("data.txt","r",stdin);47     while(~scanf("%d",&n)) {48         memset(p,-1,sizeof(p));49         tot=top=0;50         for(int i=1; i<n; i++) {51             scanf("%d %d %s",&v,&u,ch);52             if(ch[0]==‘a‘) {53                 scanf("%s",ch);54                 add(i,u,v,0);55             } else {56                 add(i,u,v,1);57             }58         }59         dfs1(1,0,0);60         dfs2(1,0);61         printf("%d\n",top);62         for(int i=0; i<top; i++) {63             if(i) printf(" ");64             printf("%d",need[i]);65         }66         printf("\n");67     }68     return 0;69 }

/*321*/

 

SGU - 321 - The Spy Network