HDU 4041 Eliminate Witches! （ACM ICPC 2011 北京賽區網絡賽題目） Eliminate Witches! Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 863 Accepted Submission(s): 342 Problem Description Kaname Madoka is a Magical Girl(Mahou Shoujo/Puella Magi). The duty of a Magical Girl is to eliminate Witches(Majo). Though sounds horrific, it is not a hard job for her as a powerful magical girl. One day Madoka is eliminating Witches as usual. This time she is facing a maze full of Witches. The maze consists of rooms, each lives exactly one Witch. And there is exactly one path from one room to another. So you see, the maze can be represented as a tree, with rooms regarded as nodes on the tree. Madoka eliminates Witches according to the following rules: 1. At first, Madoka enters the root node room of the maze. 2. If the room Madoka enters lives a Witch, Madoka will eliminate it at once, and the Witch disappear. 3. If the room has child node rooms with Witches, Madoka will choose the leftmost one and enter it. 4. Madoka won't go back to the parent node room of a room X until Witches living in child node rooms of X are all eliminated. See the figure below for details about a sample maze. The numbers inside nodes indicate the order of elimination of the corresponding Witches, the strings below nodes are names of Witches, and the arrow shows the tracks Madoka travels: After finishes her task, Madoka just make a brief log like this: "walpurgis(charlotte(patricia,gertrud),elly,gisela)" which represents the tree-like maze identifying rooms by the names of Witches living in them. Akemi Homura, a classmate of Madoka, also a Magical Girl, is a mad fan of her. She wants to take detailed notes of everything Madoka do! Apparently the log Madoka made is hard to read, so Homura decide to make a new one of her own. The new log should contain the following information: 1. The number of rooms of the maze 2. Names of Witches in all rooms. 3. The tracks Madoka travels. (represented by the number identifying the node) So the new log should be like this: 6 walpurgis charlotte patricia gertrud elly gisela 1 2 2 3 3 2 2 4 4 2 2 1 1 5 5 1 1 6 6 1 However, the maze may be very large, so Homura nees a program to help her. Input The first line contains an integer T(T<=20), indicating the number of test cases. For each case there is only one string on a line, Madoka's log. It is guaranteed that the maze consists of at most 50000 rooms, and the names of Witches is a string consists of at most 10 lowercase characters, while the string of Madoka's log consists of at most 1000000 characters, which are lowercase characters, '(', ')' or ','. Output For each case, you should output the detailed log. The first line an integer N, the number of rooms of the maze. The following N lines, each line a string, the name of the Witches, in the order of elimination. The following 2(N-1) lines, each line two integers, the number of two nodes indicating the path Madoka passes. Output a blank line after each case. Sample Input 3 walpurgis(charlotte(patricia,gertrud),elly,gisela) wuzetian nanoha(fate(hayate)) Sample Output 6 walpurgis charlotte patricia gertrud elly gisela 1 2 2 3 3 2 2 4 4 2 2 1 1 5 5 1 1 6 6 1 1 wuzetian 3 nanoha fate hayate 1 2 2 3 3 2 2 1 Source The 36th ACM/ICPC Asia Regional Beijing Site —— Online Contest Recommend lcy 題目大意：T組測試數據，接下來T組字符串，表示一棵樹，從字符串看樹的結構顯而易見，然後輸出這棵樹訪問的過程。 解題思路：模擬題，這題就是一種先序的方式，但是不是二叉樹，所以自己寫程序模擬一下過程就可以了
#include <iostream>
#include <cstring>
#include <cstdio>
#include <string>
#include <map>
#include <cstring>
#include <vector>
using namespace std;
const int maxlen=10000010;
vector <string> ans;
vector <int> v;
map <string,int> mp;
char str[maxlen];
int len,num,pos;
void initial(){
ans.clear();
mp.clear();
v.clear();
num=1;
pos=0;
}
void input(){
scanf("%s",str);
len=strlen(str);
}
void dfs(){
if(pos>=len) return;
string st;
while(str[pos]>='a' && str[pos]<= 'z'){
st.push_back(str[pos]);
pos++;
}
ans.push_back(st);
int now=num++;
v.push_back(now);
if(str[pos]=='('){
pos++;
dfs();
v.push_back(now);
while(str[pos]==','){
pos++;
dfs();
v.push_back(now);
}
pos++;
}else return;
}
void computing(){
dfs();
}
void output(){
printf("%d\n",num-1);
for(int i=0;i<ans.size();i++){
printf("%s\n",ans[i].c_str());
}
for(int i=0;i<v.size()-1;i++){
printf("%d %d\n",v[i],v[i+1]);
}
printf("\n");
}
int main(){
int t;
cin>>t;
while(t-- >0){
initial();
input();
computing();
output();
}
return 0;
}
拓展：樹的遍歷方式（轉載） 1)前序遍歷 a)遞歸方式: void preorder_recursive(Bitree T) /* 先序遍歷二叉樹的遞歸算法 */ { if (T) { visit(T); /* 訪問當前結點 */ preorder_recursive(T->lchild); /* 訪問左子樹 */ preorder_recursive(T->rchild); /* 訪問右子樹 */ } } b)非遞歸方式 void preorder_nonrecursive(Bitree T) /* 先序遍歷二叉樹的非遞歸算法 */ { initstack(S); push(S,T); /* 根指針進棧 */ while(!stackempty(S)) { while(gettop(S,p)&&p) { /* 向左走到盡頭 */ visit(p); /* 每向前走一步都訪問當前結點 */ push(S,p->lchild); } pop(S,p); if(!stackempty(S)) { /* 向右走一步 */ pop(S,p); push(S,p->rchild); } } } 2)中序遍歷 a)遞歸方式 void inorder_recursive(Bitree T) /* 中序遍歷二叉樹的遞歸算法 */ { if (T) { inorder_recursive(T->lchild); /* 訪問左子樹 */ visit(T); /* 訪問當前結點 */ inorder_recursive(T->rchild); /* 訪問右子樹 */ } } b)非遞歸方式 void inorder_nonrecursive(Bitree T) { initstack(S); /* 初始化棧 */ push(S, T); /* 根指針入棧 */ while (!stackempty(S)) { while (gettop(S, p) && p) /* 向左走到盡頭 */ push(S, p->lchild); pop(S, p); /* 空指針退棧 */ if (!stackempty(S)) { pop(S, p); visit(p); /* 訪問當前結點 */ push(S, p->rchild); /* 向右走一步 */ } } } 3)後序遍歷 a)遞歸方式 void postorder_recursive(Bitree T) /* 中序遍歷二叉樹的遞歸算法 */ { if (T) { postorder_recursive(T->lchild); /* 訪問左子樹 */ postorder_recursive(T->rchild); /* 訪問右子樹 */ visit(T); /* 訪問當前結點 */ } } b)非遞歸方式 typedef struct { BTNode* ptr; enum {0,1,2} mark; } PMType; /* 有mark域的結點指針類型 */ void postorder_nonrecursive(BiTree T) /* 後續遍歷二叉樹的非遞歸算法 */ { PMType a; initstack(S); /* S的元素為PMType類型 */ push (S,{T,0}); /* 根結點入棧 */ while(!stackempty(S)) { pop(S,a); switch(a.mark) { case 0: push(S,{a.ptr,1}); /* 修改mark域 */ if(a.ptr->lchild) push(S,{a.ptr->lchild,0}); /* 訪問左子樹 */ break; case 1: push(S,{a.ptr,2}); /* 修改mark域 */ if(a.ptr->rchild) push(S,{a.ptr->rchild,0}); /* 訪問右子樹 */ break; case 2: visit(a.ptr); /* 訪問結點 */ } } }
