2019百度之星复赛题解 A.B.C

1001. Diversity

题意

给你一棵n个点的树，对于节点i，你要给它标上一个 $[l_i,r_i]$[li​,ri​]之间的数，要求所有边两端节点上标的数字的差的绝对值的总和最大。

解法

一开始以为一边取大一边取小就会最优
其实不对 所以最后写了一遍树形DP

/* Algorithm: Author: anthony1314 Creat Time: Time Complexity: */

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <stack>
#include <cmath>
#include <deque>
#include <queue>
#include <list>
#include <set>
#include <map>
#define line printf("---------------------------\n")
#define mem(a, b) memset(a, b, sizeof(a))
#define pi acos(-1)
#define endl '\n'
using namespace std;
typedef long long ll;
typedef double db;
const int inf = 0x3f3f3f3f;
const int mod = 1e9+7;
const int maxn = 2e5+10;

void add(int &x, int y) {
    x = (x + y) % mod;
}
ll dp[maxn][2];
vector<int> g[maxn];
ll l[maxn], r[maxn];
void dfs(int root, int x) {
    int len = g[root].size();
    for(int i = 0; i < len; i++) {
        if(g[root][i] == x) continue;
        dfs(g[root][i], root);
    }
    for(int i = 0; i < len; i++) {
        if(g[root][i] == x) continue;
        dp[root][0] += max(dp[g[root][i]][1] + abs(l[g[root][i]] - r[root]),
                           dp[g[root][i]][0] + abs(r[g[root][i]] - r[root]));
        dp[root][1] += max(dp[g[root][i]][1] + abs(l[g[root][i]] - l[root]),
                           dp[g[root][i]][0] + abs(r[g[root][i]] - l[root]));
    }
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t, n;
    cin>>t;
    int u, v;
    while(t--) {
        for(int i = 0; i < maxn; i++) {
            g[i].clear();
        }
        memset(dp, 0, sizeof(dp));
        cin>>n;
        for(int i = 0; i < n-1; i++) {
            cin>>u>>v;
            g[u].push_back(v);
            g[v].push_back(u);
        }
        for(int i = 1; i <= n; i++) {
            cin>>l[i]>>r[i];
        }
        dfs(1, -1);
        ll ans = -1;
        for(int i = 1; i <= n; i++){
            ans = max(ans, max(dp[i][1], dp[i][0]));
        }
        cout<<ans<<endl;
    }

    return 0;
}

1002. Transformation

题意

给你一个二元组(a,b)，支持AB两种操作，分别是将其变成(a,2b−a)和(2a−b,b)。问能否通过大于等于零次操作将其变成(c,d)。

解法

答案一定满足以下规律规律：
c=((x+1)a-xb)
d=((y+1)a-yb)
那么我只要逆推回去即可
判断x和y哪个大，就用哪一种方式逆推回去

/* Algorithm: Author: anthony1314 Creat Time: Time Complexity: */

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <stack>
#include <cmath>
#include <deque>
#include <queue>
#include <list>
#include <set>
#include <map>
#define line printf("---------------------------\n")
#define mem(a, b) memset(a, b, sizeof(a))
#define pi acos(-1)
using namespace std;
typedef long long ll;
typedef double db;
const int inf = 0x3f3f3f3f;
const int mod = 1e9+7;
const int maxn = 2000+10;

void add(int &x, int y) {
	x = (x + y) % mod;
}
stack<char> ans;
bool isOdd(ll a) {
	return a % 2;
}
int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	int t;
	cin>>t;
	while(t--) {
		ll a, b, c, d;
		cin>>a>>b>>c>>d;
		if((a - b) == 0) {
			if(a == c && b == d) {
				printf("Yes\n\n");
			} else {
				printf("No\n");
			}
			continue;
		}
		ll x = (c - a) % (a - b);
		ll y = (d - b) % (b - a);
		if(x != 0 || y != 0) { //判断一下 c和d符不符合公式
			puts("No");
		} else {
			x = (c - a) / (a - b) + 1; // a的系数 
			y = (d - b) / (b - a) + 1; // b的系数 
			if(x <= 0 || y <= 0) {
				puts("No");
				continue;
			}
			bool f = 1;
			while(!ans.empty()){
				ans.pop();
			}
			while(c != a || b != d){
				if(isOdd(c + d)){
					f = 0;
					break;
				}
				if(x > y){
					c = (c + d) / 2;
					if(isOdd(x - (y - 1)))	{
						f = 0;
						break;
					}
					x = (x - (y - 1)) / 2;
					ans.push('B');
				}else {
					d = (c + d) / 2;
					if(isOdd(y - (x - 1)))	{
						f = 0;
						break;
					}
					y = (y - (x - 1)) / 2;
					ans.push('A');
				}
			}
			if(!f){
				puts("No");
			}else {
				puts("Yes");
				while(!ans.empty()){
					printf("%c", ans.top());
					ans.pop();
				}
				puts("");
			}
		}
	}
	return 0;
}

1003. Quasi Binary Search Tree

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