2020-2021 ICPC - Gran Premio de Mexico - Repechaje

A. Atsa’s Checkers Board

• 给定一个$n\ast m$大小的矩阵，在该矩阵内放置黑色白色的石头，要保证每个$2\ast 2$的矩阵都有两个黑石头和两个白石头，问有多少种摆放方式
• 可以发现当有两块颜色相同的石头摆放在一起的时候，整个矩阵的摆放方式就确定了
• 可以发现第一行至少有一对颜色相同石子相连的方案数是$1<<m-2$
• 而剩下两种没有颜色相同的石子相连的情况，黑白…和白黑…
• 取其中一种情况作计算。可以发现当第一行确定为黑白…的时候，第一列的排列方式数就为$1<<(n-1)$。同理可得第一行确定为白黑…的时候也为$1<<(n-1)$，两者总和即为$1<<n$
• 因此总情况数就是$(1<<m)+(1<<n)-2$

#include <bits/stdc++.h>
#include <unordered_map>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int N = 3200;
const int INF = 0x3f3f3f3f;
const double pi = acos(-1.0);
const double eps = 1e-8;
inline ll rd() {
    
    
    ll x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
    
    
        if (ch == '-')
            f = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
    
    
        x = (x << 1) + (x << 3) + (ch ^ 48);
        ch = getchar();
    }
    return x * f;
}
int main() {
    
    
    ll n, m;
    n = rd();
    m = rd();
    if (n < 2 || m < 2) {
    
    
        puts("0");
        return 0;
    }
    printf("%lld\n", (1LL << n) + (1LL << m) - 2);
    return 0;
} 

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E. Enterprise Recognition Program

• 给定一个树形图代表一个公司的主从结构，并且公司中有$n$个员工，老板节点编号为$0$，老板不属于员工
• 操作$m$次，并且有两种操作
• 第一种操作是给员工$e$发$v$元的奖金
• 第二种操作是给员工$e$及其所有上司发$v$元的奖金
• 询问$q$次，并且有两种询问
• 第一种询问是询问员工$x$有多少奖金
• 第二种询问是询问员工$x$的下属和员工$x$共有多少奖金
• 设置一个$earn$数组，$earn[i][0]$表示公司用操作一发给员工$i$的奖金，$earn[i][1]$表示公司用操作二发给员工$i$的奖金，$earn[i][2]$表示员工$i$的下属们的总奖金
• 从老板节点开始往下$dfs$，从叶子节点往上累加奖金即可
• 以下为两个递推公式，$u$为当前员工，$v$为$u$的下属
• $earn[u][1]+=earn[v][1];$
• $earn[u][2]+=earn[v][0]+earn[v][1]+earn[v][2];$
• 具体逻辑在代码中体现

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <functional>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <unordered_map>
#include <vector>
#define lowbit(x) ((x) & -(x))
#define endl "\n"
using namespace std;
typedef pair<int, int> pii;
typedef long long ll;
typedef unsigned long long ull;
const int N = 1e6 + 10, NN = 2e3 + 10, INF = 0x3f3f3f3f, LEN = 20;
const ll MOD = 1e9 + 7;
const ull seed = 31;
const double pi = acos(-1.0), eps = 1e-8;
inline int read() {
    
    
    int x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
    
    
        if (ch == '-')
            f = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
    
    
        x = (x << 1) + (x << 3) + (ch ^ 48);
        ch = getchar();
    }
    return x * f;
}
struct Edge {
    
    
    int next, to;
} edge[N];
int num_edge, n, m, q, boss;
int head[N], cha[N];
ll earn[N][3];
void add_edge(int from, int to) {
    
    
    edge[num_edge].next = head[from];
    edge[num_edge].to = to;
    head[from] = num_edge++;
}
void init() {
    
    
    memset(head, -1, sizeof head);
}
void dfs(int u) {
    
    
    for (int i = head[u]; i != -1; i = edge[i].next) {
    
    
        int v = edge[i].to;
        dfs(v);
        earn[u][1] += earn[v][1];
        earn[u][2] += earn[v][0] + earn[v][1] + earn[v][2];
    }
}
int main() {
    
    
    // ios::sync_with_stdio(false);
    // cin.tie(0);
    // cout.tie(0);
    // freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);
    n = read(), m = read(), q = read();
    init();
    for (int i = 1; i <= n; ++i) {
    
    
        int x = read();
        add_edge(x, i);
        if (x == 0)
            boss = i;
    }
    for (int i = 1; i <= m; ++i) {
    
    
        int m = read(), e = read(), v = read();
        earn[e][m - 1] += v;
    }
    dfs(boss);
    while (q--) {
    
    
        int op = read(), x = read();
        if (op == 1)
            printf("%lld\n", earn[x][0] + earn[x][1]);
        else
            printf("%lld\n", earn[x][2] + earn[x][1] + earn[x][0]);
    }
    return 0;
}

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K. Kitchen Waste

• 有$n$个面包，$m$口锅，每口锅中都装了一定体积的汤，要将汤淋在面包上，面包也有体积，一体积汤能淋一体积面包
• 一次淋面包的操作必须要将整个面包都淋上汤，若锅中剩余汤的体积不足以将整个面包都淋上，则将这个锅中的汤都倒掉
• 签到题，根据题意直接模拟即可

#include <bits/stdc++.h>
#include <unordered_map>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int N = 1e7 + 10;
const int INF = 0x3f3f3f3f;
inline ll read() {
    
    
    ll x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
    
    
        if (ch == '-')
            f = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
    
    
        x = (x << 1) + (x << 3) + (ch ^ 48);
        ch = getchar();
    }
    return x * f;
}
int n, m;
int a[N], b[N];
int main() {
    
    
    n = read();
    m = read();
    for (int i = 1; i <= n; ++i) 
        a[i] = read();
    for (int i = 1; i <= m; ++i)
        b[i] = read();
    int pos = 1;
    for (int i = 1; i <= m; ++i) {
    
    
        while (b[i] >= a[pos]) {
    
    
            b[i] -= a[pos];
            ++pos;
            if (pos > n)
                break;
        }
        if (pos > n)
            break;
    }
    int ans = 0;
    for (int i = 1; i <= m; ++i)
        ans += b[i];
    printf("%d\n", ans);
    return 0;
}