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;

}