hdu 4451水题

挺水的题。我的做法是用ans = N*M*K减去不和谐的数目。首先，对每一个paints shoes对，ans -= N，这是没有问题的。这个操作全执行完以后就能记录每个paints跟多少shoes和谐(用paintspair[i]表示)。然后对于每一个clothes paints对，ans -= paintspair[i]，这样就避免了重复计数。

/*
 * hdu4451/win.cpp
 * Created on: 2012-10-29
 * Author    : ben
 */
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
#include <stack>
#include <string>
#include <vector>
#include <deque>
#include <list>
#include <functional>
#include <numeric>
#include <cctype>
using namespace std;
const int MAXM = 1005;
int N, M, K;
int paintspair[MAXM];
int main() {
#ifndef ONLINE_JUDGE
    freopen("data.in", "r", stdin);
#endif
    int p, ans, a, b;
    char str1[100], str2[100];
    vector<int> cp;
    while(scanf("%d%d%d", &N, &M, &K) == 3) {
        if(N == 0 && M == 0 && K == 0) {
            break;
        }
        ans = N * M * K;
        cp.clear();
        fill(paintspair, paintspair + MAXM, K);
        scanf("%d", &p);
        while(p--) {
            scanf(" %s %d %s %d", str1, &a, str2, &b);
            if(strcmp(str1, "clothes") == 0) {
                cp.push_back(b);
            }else {
                paintspair[a]--;
                ans -= N;
            }
        }
        for(int i = 0, len = cp.size(); i < len; i++) {
            ans -= paintspair[cp[i]];
        }
        printf("%d\n", ans);
    }
    return 0;
}