P3067 [USACO12OPEN] Balanced Cow Subsets G

合集 - 洛谷(99)

1.P8755 [蓝桥杯 2021 省 AB2] 负载均衡2023-11-222.P1102 A-B 数对的三种解法2023-11-263.P1090 [NOIP2004 提高组] 合并果子2023-11-264.P3379 【模板】最近公共祖先（LCA）2023-11-285.P8599 [蓝桥杯 2013 省 B] 带分数2023-11-286.P1433 吃奶酪2023-11-297.P3385 【模板】负环2023-12-068.P1576 最小花费2023-12-069.Zuma2023-12-0610.P3205 [HNOI2010] 合唱队2023-12-0611.P1220 关路灯2023-12-0712.P4170 [CQOI2007] 涂色（天赋哥不要点进来）2023-12-1113.P8805 [蓝桥杯 2022 国 B] 机房2023-12-1214.P3386 【模板】二分图最大匹配2023-12-2015.P2197 【模板】Nim 游戏2023-12-2016.P1036 [NOIP2002 普及组] 选数2023-12-2117.T397291 【模板】拓扑排序（加强版）2023-12-2618.P2865 [USACO06NOV] Roadblocks G2023-12-2619.P1168 中位数2024-01-0220.P1631 序列合并2024-01-0221.P3372 【模板】线段树 12024-01-0222.P1352 没有上司的舞会2024-01-0223.P2015 二叉苹果树2024-01-0324.P2014 [CTSC1997] 选课2024-01-0325.P1613 跑路2024-01-0326.P2911 [USACO08OCT] Bovine Bones G2024-01-0427.P4316 绿豆蛙的归宿2024-01-0428.P3870 [TJOI2009] 开关2024-01-1029.P3373 【模板】线段树 22024-01-1430.P1558 色板游戏2024-01-1531.P1638 逛画展2024-01-1632.P1043 [NOIP2003 普及组] 数字游戏2024-01-2633.P1147 连续自然数和2024-01-2634.P1063 [NOIP2006 提高组] 能量项链2024-01-2735.P4342 [IOI1998] Polygon2024-01-2736.P1122 最大子树和2024-02-0537.P2016 战略游戏2024-02-0538.P2585 [ZJOI2006] 三色二叉树2024-02-0739.P1273 有线电视网2024-02-0740.P2986 [USACO10MAR] Great Cow Gathering G2024-02-0741.P1114 “非常男女”计划2024-02-0842.P5026 Lycanthropy2024-02-0843.P4231 三步必杀2024-02-0944.P2985 [USACO10FEB] Chocolate Eating S2024-02-0945.P4090 [USACO17DEC] Greedy Gift Takers P2024-02-1046.P4933 大师2024-02-1047.P1082 [NOIP2012 提高组] 同余方程2024-02-1148.P3811 【模板】模意义下的乘法逆元2024-02-1249.P9725 [EC Final 2022] Chase Game 22024-02-1250.P1330 封锁阳光大学2024-02-1251.P8674 [蓝桥杯 2018 国 B] 调手表2024-02-1252.P10111 [GESP202312 七级] 纸牌游戏2024-02-1453.P9975 [USACO23DEC] Cowntact Tracing 2 B2024-02-1554.P8783 [蓝桥杯 2022 省 B] 统计子矩阵2024-02-1555.P8786 [蓝桥杯 2022 省 B] 李白打酒加强版2024-02-1556.P8807 [蓝桥杯 2022 国 C] 取模2024-02-1557.P8732 [蓝桥杯 2020 国 ABC] 答疑2024-02-1558.P8725 [蓝桥杯 2020 省 AB3] 画中漂流2024-02-1559.P3958 [NOIP2017 提高组] 奶酪2024-02-1560.P4084 [USACO17DEC] Barn Painting G2024-02-1561.P1892 [BOI2003] 团伙2024-02-1662.P1198 [JSOI2008] 最大数2024-02-1663.P1204 [USACO1.2] 挤牛奶Milking Cows2024-02-1664.P5490 【模板】扫描线2024-02-1765.P8784 [蓝桥杯 2022 省 B] 积木画2024-02-1766.P9325 [CCC 2023 S2] Symmetric Mountains2024-02-1767.Iva & Pav2024-02-1768.P9889 [ICPC2018 Qingdao R] Plants vs. Zombies2024-02-1769.P9847 [ICPC2021 Nanjing R] Crystalfly2024-02-1870.E. Anna and the Valentine's Day Gift2024-02-1971.F. Chat Screenshots2024-02-1972.G. One-Dimensional Puzzle2024-02-2073.P2899 [USACO08JAN] Cell Phone Network G2024-02-2074.P2023 [AHOI2009] 维护序列2024-02-2075.P1531 I Hate It2024-02-2076.P5057 [CQOI2006] 简单题2024-02-2077.P2357 守墓人2024-02-2078.P8775 [蓝桥杯 2022 省 A] 青蛙过河2024-02-2179.P1656 炸铁路2024-02-2280.Watering an Array2024-02-2381.Dora and Search2024-02-2482.P1137 旅行计划2024-02-2483.P2835 刻录光盘2024-02-2484.P1197 [JSOI2008] 星球大战2024-02-2485.P3388 【模板】割点（割顶）2024-02-2486.P8435 【模板】点双连通分量2024-02-2587.P8436 【模板】边双连通分量2024-02-2588.P2860 [USACO06JAN] Redundant Paths G2024-02-25

89.P3067 [USACO12OPEN] Balanced Cow Subsets G2024-02-26

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收起

原题链接

题解

设前半部分对两个集合贡献的差为a，后半部分贡献为b

若 $a==b$ 则
差为a的组合数（被选上，和在哪个集合无关）sa 与b的组合数的sb
此时对答案的贡献为 $sa·sb$

所以穷举所有差的组合，然后累加

设差为集合A-集合B
每个元素对差的贡献有三种可能，要么加要么减要么不加也不减
由于只要用过就算进组合里（无论在哪个集合），所以差的正负值不重要（这是代码特性决定的）
而且同一组合可以诞生不同的差

code1

#define ll long long
#include<bits/stdc++.h>
using namespace std;

// Custom input and output functions
inline void read(ll &x) {
    x = 0;
    ll flag = 1;
    char c = getchar();
    while(c < '0' || c > '9'){
        if(c == '-')flag = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        x = (x << 3) + (x << 1) + (c ^ 48);
        c = getchar();
    }
    x *= flag;
}

inline void write(ll x)
{
    if(x < 0){
        putchar('-');
        x = -x;
    }
    if(x > 9)
        write(x / 10);
    putchar(x % 10 + '0');
}

ll n, mid;
ll a[28] = {0};
map<int,int> id;
ll judge[(1<<22)+5] = {0};
ll len = 0;
vector<ll> use[(1<<22)+5];

void ss1(ll x, ll sum, ll now)
{
    if(x > mid)
    {
        if(!id[sum]) id[sum] = ++len;
        use[id[sum]].push_back(now);
        return;
    }

    ss1(x + 1, sum + a[x], now | (1LL << (x - 1)));
    ss1(x + 1, sum - a[x], now | (1LL << (x - 1)));
    ss1(x + 1, sum, now);
}

void ss2(ll x, ll sum, ll now)
{
    if(x > n)
    {
        ll who = id[sum];
        if(who)
            for(ll i = 0; i < use[who].size(); i++)
                judge[now | use[who][i]] = 1;
        return;
    }

    ss2(x + 1, sum + a[x], now | (1LL << (x - 1)));
    ss2(x + 1, sum - a[x], now | (1LL << (x - 1)));
    ss2(x + 1, sum, now);
}

int main()
{
    read(n);
    for(ll i = 1; i <= n; i++) read(a[i]);

    mid = n >> 1; // Optimized division by 2
    ss1(1, 0, 0);
    ss2(mid + 1, 0, 0);

    ll ans = 0;
    for(ll i = 1; i <= (1LL << n)  ; i++) ans += judge[i];

    write(ans);
    putchar('\n');
    return 0;
}

code2

#define ll long long
#include<bits/stdc++.h>
using namespace std;

inline void read(ll &x) {
    x = 0;
    ll flag = 1;
    char c = getchar();
    while(c < '0' || c > '9'){
        if(c == '-')flag = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        x = (x << 3) + (x << 1) + (c ^ 48);
        c = getchar();
    }
    x *= flag;
}

inline void write(ll x)
{
    if(x < 0){
        putchar('-');
        x = -x;
    }
    if(x > 9)
        write(x / 10);
    putchar(x % 10 + '0');
}

ll n, mid,ans=0;
ll a[13] = {0} ,b[13]={0};
unordered_map<ll, bitset<1024> >haxi;
bitset<1024> use1[1024];
ll len = 0;

void ss1(ll x, ll sum, ll now)
{
    if(x > mid)
    {
        haxi[sum].set(now);//把sum下的这种可能点亮
        return;
    }

    ss1(x + 1, sum + a[x], now | (1LL << (x - 1)));
    ss1(x + 1, sum - a[x], now | (1LL << (x - 1)));
    ss1(x + 1, sum, now);
}

void ss2(ll x, ll sum, ll now)
{
    if(x > n-mid)
    {
        bitset<1024> use (haxi[sum]);//代表前半部分达到sum的有哪些可能的组合
        use&=~use1[now];//去重，use1代表当后半部分的组合为now时，它（指代后半部分组合）在此之前已经和哪些前半部分组合过了
        ans+=use.count();
        use1[now]|=use;
        return;
    }

    ss2(x + 1, sum + b[x], now | (1LL << (x - 1)));
    ss2(x + 1, sum - b[x], now | (1LL << (x - 1)));
    ss2(x + 1, sum, now);
}

int main()
{
    read(n);mid = n >> 1;
    for(ll i = 1; i <= mid; i++) read(a[i]);

    for(ll i = 1; i <= n-mid; i++) read(b[i]);

    ss1(1, 0, 0);
    ss2(1, 0, 0);

    write(ans-1);
    putchar('\n');
    return 0;
}