hdu-4451 思维题

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这个题一开始想的就是只有有一套衣服就加2然后有重复的就需要减一，就用这个方法，写了三遍，wa了三遍，最后发现那个数组没有遍历最后一个元素

所以那个题一直错，后来队友用了另外的方法搞定了，自己比赛结束前也没有看出来这个错误，下面贴上两个代码，

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>#include <vector>#include <string>#include <map>#include <math.h>#include <stack>#define LL long longusing namespace std;const int INF = 0x3f3f3f3f;int dir[4][2] = {{1,0},{0,1},{-1,0},{0,-1}};const int maxn = 1000 + 10;int m,n,k;int a[maxn];int b[maxn];int N;int a1,b1;string c1,c2;int main(){    while(scanf("%d%d%d",&n,&m,&k)&&(n||m||k))    {        memset(a,0,sizeof(a));        memset(b,0,sizeof(b));        int sum = 0;        scanf("%d",&N);        for(int i = 0; i < N; i++)        {            cin>>c1>>a1>>c2>>b1;            if(c1 == "clothes")            {                a[b1]++;            }            else            {                b[a1]++;            }        }        //cout<<a[1]<<b[1]<<endl;        for(int i = 1; i <= m; i++)        {            sum+=((n - a[i]) * (k - b[i]));        }        printf("%d\n",sum);    }    return 0;}

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>#include <vector>#include <string>#include <map>#include <math.h>#include <stack>#define LL long longusing namespace std;const int INF = 0x3f3f3f3f;int dir[4][2] = {{1,0},{0,1},{-1,0},{0,-1}};const int maxn = 1000 + 10;int m,n,k;vector<int> a[maxn];int b[maxn];int a1,b1;int N;string c1,c2;int main(){    while(scanf("%d%d%d",&n,&m,&k)&&(n||m||k))    {        for(int i=1;i<=n;i++)            a[i].clear();        memset(b,0,sizeof(b));        int ans = 0,num=0,sum=0;        scanf("%d",&N);        if(N==0)        {            printf("%d\n",m*n*k);            continue;        }        for(int i = 0; i < N; i++)        {            cin>>c1>>a1>>c2>>b1;            if(c1 == "clothes")            {                a[a1].push_back(b1);            }            else            {                  b[a1]++;            }        }       for(int i=1;i<=n;i++)           if(a[i].size())           {               for(int j=0;j<a[i].size();j++)               {                   int zz=a[i][j];                   ans=ans+b[zz];                   num++;               }           }           sum+=num*k;           num=0;         for(int i=1;i<=m;i++)             if(b[i])             {                num+=b[i];             }          sum+=num*n;          int hh=n*m*k-sum+ans;        printf("%d\n",hh);    }    return 0;}

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