D - Simple String CSU - 1550

http://acm.csu.edu.cn/csuoj/problemset/problem?pid=1550

很久都没补这题，最近想学网络流，就看看，队友以前用网络流过的，Orz，

但是这题只需要简单的判断，可能想起来有点麻烦。

考虑一定要从A串取出n个，B串也一定要取出n个，那么A和C的交集一定要大于等于n，同理B。

同时为了得到C串，还需要A + B和C的交集一定要大于等于n * 2

怎么说呢。可以画个图表示下，反正找不到反例。就先码一波。

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cmath>

#include <algorithm>

#include <assert.h>

#define IOS ios::sync_with_stdio(false)

using namespace std;

#define inf (0x3f3f3f3f)

typedef long long int LL;

#include <iostream>

#include <sstream>

#include <vector>

#include <set>

#include <map>

#include <queue>

#include <string>

#include <bitset>

const int maxn = ;

char str[maxn], sub1[maxn], sub2[maxn];

int hasStr[], hasSub1[], hasSub2[];

void work() {

    int lenstr = strlen(str + );

    memset(hasStr, , sizeof hasStr);

    memset(hasSub1, , sizeof hasSub1);

    memset(hasSub2, , sizeof hasSub2);

    for (int i = ; i <= lenstr; ++i) hasStr[str[i]]++;

    for (int i = ; i <= lenstr; ++i) hasSub1[sub1[i]]++;

    for (int i = ; i <= lenstr; ++i) hasSub2[sub2[i]]++;

    int ans1 = , ans2 = , ans3 = ;

    for (int i = 'A'; i <= 'Z'; ++i) {

        ans1 += min(hasStr[i], hasSub1[i]);

        ans2 += min(hasStr[i], hasSub2[i]);

        ans3 += min(hasStr[i], hasSub1[i] + hasSub2[i]);

    }

    if (ans1 < lenstr /  || ans2 < lenstr /  || ans3 < lenstr) {

        printf("NO\n");

    } else printf("YES\n");

}

int main() {

#ifdef local

    freopen("data.txt", "r", stdin);

//    freopen("data.txt", "w", stdout);

#endif

    while (scanf("%s%s%s", sub1 + , sub2 + , str + ) > ) work();

    return ;

}