E. Selling Souvenirs 不会做

http://codeforces.com/contest/808/problem/E

不理解为什么dp = {cost, cnt1, cnt2}可以

而dp = {cost, cnt1, cnt2, cnt3}不可以

上面那个不可以的例子是：

但是这个dp是可行的，只是还有一个更新没实现起来。

dp[i] = dp[i - 2] - (花费为1的元素) + (花费为3的元素)

原来可以按1选择的数目分类，分成奇偶，然后贪心。集训队rank1的zk教我的。cf 2200+ Orz

下面评论有详解

#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>
#include <time.h>
const int maxn =  + ;
int a[][maxn];
int num[];
struct Node {
    LL val;
    int cnt;
} dp[maxn], dp2[maxn];
bool cmp(int x, int y) {
    return x > y;
}
LL sum[maxn];
LL odd[maxn], even[maxn];
LL sum3[maxn];
void work() {
    int n;
    scanf("%d", &n);
    int m;
    scanf("%d", &m);
    int lenOdd = , lenEven = ;
    for (int i = ; i <= n; ++i) {
        int id, x;
        scanf("%d", &id);
        scanf("%d", &a[id][++num[id]]);
        if (id == ) {
            odd[++lenOdd] = even[++lenEven] = a[id][num[id]];
        }
    }
    for (int i = ; i <= ; ++i) {
        sort(a[i] + , a[i] +  + num[i], cmp);
    }

    for (int i = ; i <= num[]; ++i) {
        sum3[i] = sum3[i - ] + a[][i];
    }

    for (int i = ; i <= num[]; i += ) {
        if (i +  > num[]) break;
        even[++lenEven] = a[][i] + a[][i + ];
    }
    sort(even + , even +  + lenEven, cmp);
    for (int i = ; i <= lenEven; ++i) {
        even[i] = even[i] + even[i - ];
    }
    LL ans = ;
    for (int i = ; i <= m; ++i) { //even个，暴力枚举用在另外两个的总量是多少
        LL res = even[min(lenEven, i / )];
        LL res2 = sum3[min((m - i) / , num[])];
        ans = max(res + res2, ans);
    }

    for (int i = ; i <= num[]; i += ) {
        if (i +  > num[]) break;
        odd[++lenOdd] = a[][i] + a[][i + ];
    }
    sort(odd + , odd +  + lenOdd, cmp);
    for (int i = ; i <= lenOdd; ++i) {
        odd[i] = odd[i - ] + odd[i];
    }
    LL ans2 = ;
    for (int i = ; i <= m - ; ++i) {
        LL res = odd[min(lenOdd, i / )];
        LL res2 = sum3[min(num[], (m - i - ) / )];
        ans2 = max(ans2, res + res2);
    }
    cout << max(ans, ans2 + a[][]) << endl;
}

int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    work();
    return ;
}