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  • HDU 1242 Rescue

    Rescue

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 27106    Accepted Submission(s): 9600

    Problem Description
    Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the prison.

    Angel's friends want to save Angel. Their task is: approach Angel. We assume that "approach Angel" is to get to the position where Angel stays. When there's a guard in the grid, we must kill him (or her?) to move into the grid. We assume that we moving up, down, right, left takes us 1 unit time, and killing a guard takes 1 unit time, too. And we are strong enough to kill all the guards.

    You have to calculate the minimal time to approach Angel. (We can move only UP, DOWN, LEFT and RIGHT, to the neighbor grid within bound, of course.)
    Input
    First line contains two integers stand for N and M.
    Then N lines follows, every line has M characters. "." stands for road, "a" stands for Angel, and "r" stands for each of Angel's friend.
    Process to the end of the file.
     
    Output
    For each test case, your program should output a single integer, standing for the minimal time needed. If such a number does no exist, you should output a line containing "Poor ANGEL has to stay in the prison all his life." 
     
    Sample Input
    7 8 #.#####. #.a#..r. #..#x... ..#..#.# #...##.. .#...... ........
     
    Sample Output
    13
     
    Author
    CHEN, Xue
     
    Source
     
    Recommend
    Eddy
     1 #include <iostream>
     2 #include <map>
     3 #include <algorithm>
     4 #include <string>
     5 using namespace std;
     6 map<string,int> m1,m2;
     7 int main()
     8 {
     9     int n,ans;
    10     string str;
    11     while(cin>>n)
    12     {
    13         ans=0;
    14         m1.clear(); m2.clear();
    15         for(int i=0;i<n;i++)
    16         {
    17             cin>>str;
    18             if(m1.count(str))
    19                 m1[str]++;
    20             else
    21                 m1[str]=1;
    22         }
    23         for(int i=0;i<n;i++)
    24         {
    25             cin>>str;
    26             if(m2.count(str))
    27                 m2[str]++;
    28             else
    29                 m2[str]=1;
    30         }
    31         map<string,int>::iterator it;
    32         for(it=m1.begin();it!=m1.end();it++)
    33         {
    34             str=it->first;
    35             ans+=min(m1[str],m2[str]);
    36         }
    37         cout<<ans<<endl;
    38     }
    39 }
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  • 原文地址:https://www.cnblogs.com/cumulonimbus/p/5874526.html
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