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Time Limit: 20 Sec
Memory Limit: 256 MB
http://codeforces.com/gym/100513/problem/M
In computer programming, variable shadowing occurs when a variable declared within a certain scope has the same name as a variable declared in an outer scope. The outer variable is said to be shadowed by the inner variable, and this can lead to a confusion. If multiple outer scopes contain variables with the same name, the variable in the nearest scope will be shadowed.
Formally, a declared variable shadows another declared variable if the following conditions are met simultaneously:
Here is an example containing exactly one variable shadowing:
/* Prints a+max(b,c) */
int main() {
int a, b, c;
cin » a » b » c;
if (b > c) {
int a = b; // <– variable ‘a‘ shadows outer ‘a‘
int x = c;
b = x;
c = a;
}
int x = a + c; // <– no shadowing here
cout « x « endl;
}
Variable shadowing is permitted in many modern programming languages including C++, but compilers can warn a programmer about variable shadowing to avoid possible mistakes in a code.
Consider a trivial programming language that consists only of scopes and variable declarations. The program consists of lines, each line contains only characters ‘{‘, ‘}‘, ‘a‘ ... ‘z‘ separated by one or more spaces.
Given a syntactically correct program (i.e. curly brackets form a regular bracket sequence), write an analyzer to warn about each fact of variable shadowing. Warnings should include exact positions of shadowing and shadowed variables. Your output should follow the format shown in the examples below.
Input
The first line contains integer n (1 ≤ n ≤ 50) — the number of lines in the program. The following n lines contain the program. Each program line consists of tokens ‘{‘, ‘}‘, ‘a‘ ... ‘z‘ separated by one or more spaces. The length of each line is between 1 and 50 characters. Each program line contains at least one non-space character.
The curly brackets in the program form a regular bracket sequence, so each opening bracket ‘{‘ has uniquely defined matching closing bracket ‘}‘ and vice versa. A variable is declared in a scope at most once. Any scope (including global) can be empty, i.e. can contain no variable declarations.
Output
For each fact of shadowing write a line in form "r1:c1: warning: shadowed declaration of ?, the shadowed position is r2:c2", where "r1:c1" is the number of line and position in line of shadowing declaration and "r2:c2" is the number of line and position in line of shadowed declaration. Replace ‘?‘ with the letter ‘a‘ ... ‘z‘ — the name of shadowing/shadowed variable. If multiple outer scopes have variables named as the shadowing variable, the variable in the nearest outer scope is shadowed.
Print warnings in increasing order of r1, or in increasing order of c1 if values r1 are equal. Leave the output empty if there are no variable shadowings.
Sample Input
1
{ a { b { a } } } b
Sample Output
1:11: warning: shadowed declaration of a, the shadowed position is 1:3
题意
给一段代码，要求找到每一处外部变量被内部变量覆盖的地方
题解：
乍一看有点难，甚至想到堆栈什么的，但是由于规模很小，只要借助{、}，开一个flag变量，然后暴力往回扫描找到被覆盖的变量即可。
//debug：在把s改成tmp的过程中忘记吧s.length改成tmp.length
代码
#include <cstdio> #include <cmath> #include <cstring> #include <ctime> #include <iostream> #include <algorithm> #include <set> #include <vector> #include <sstream> #include <queue> #include <typeinfo> #include <fstream> #include <map> #include <stack> typedef long long ll; using namespace std; #define test freopen("1.txt","r",stdin) #define maxn 2000001 #define mod 10007 #define eps 1e-9 const int inf=0x3f3f3f3f; const ll infll = 0x3f3f3f3f3f3f3f3fLL; inline int read() { ll x=0,f=1;char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} return x*f; } inline void out(int x) { if(x>9) out(x/10); putchar(x%10+‘0‘); } //************************************************************************************** int numr[3000]; int numc[3000]; int main() { int n=read(); string s; for(int i=0,p=0;i<n;i++){ string tmp; getline(cin,tmp); s+=tmp; for(int j=0;j<tmp.length();j++){ numr[p]=i; numc[p++]=j; } } for(int i=0;i<s.length();i++){ if(s[i]>=‘a‘&&s[i]<=‘z‘){ int flag=0,j=i-1; for(;j>=0;j--){ if(s[j]==‘}‘) flag--; else if(s[j]==‘{‘&&flag!=1) flag++; if(flag==1&&s[i]==s[j]){ //printf("%d %c %c\n",flag,s[i],s[j]); printf("%d:%d: warning: shadowed declaration of %c, the shadowed position is %d:%d\n",numr[i]+1,numc[i]+1,s[i],numr[j]+1,numc[j]+1); break; } } } } return 0; }
Codeforces Gym 100513M M. Variable Shadowing 暴力
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原文地址：http://www.cnblogs.com/diang/p/4741493.html
