A 找到冠军 I
枚举求强于其他所有队的队
cpp
class Solution {
public:
int findChampion(vector<vector<int>> &grid) {
int n = grid.size();
int res = 0;
for (int i = 0; i < n; i++) {
int t = 0;
for (int j = 0; j < n; j++)
if (j != i)
t += grid[i][j];
if (t == n - 1) {
res = i;
break;
}
}
return res;
}
};
B 找到冠军 II
计数:若图中入度为 0 0 0 的点只有一个则该点为冠军,否则返回 − 1 -1 −1
cpp
class Solution {
public:
int findChampion(int n, vector<vector<int>> &edges) {
vector<int> indeg(n);
for (auto &ei: edges)
indeg[ei[1]]++;
vector<int> li;
for (int i = 0; i < n; i++)
if (indeg[i] == 0)
li.push_back(i);
if (li.size() != 1)
return -1;
return li[0];
}
};
C 在树上执行操作以后得到的最大分数
动态规划:设 p [ c u r ] [ 0 ] p[cur][0] p[cur][0] 为在以 c u r cur cur 为根的子树上执行若干操作使得该子树是健康的 能得到的最大分数,设 p [ c u r ] [ 1 ] p[cur][1] p[cur][1] 为以 c u r cur cur 为根的子树各节点 v a l u e s values values 之和,有状态转移方程: p [ c u r ] [ 0 ] = { 0 , c u r 是叶子节点 m a x { v a l u e s [ c u r ] + ∑ j 是 c u r 的子节点 p [ j ] [ 0 ] , ∑ j 是 c u r 的子节点 p [ j ] [ 1 ] } , c u r 不是叶子节点 p[cur][0]=\left\{\begin{matrix} 0 & ,cur 是叶子节点\\ max\{ values[cur]+\sum_{j 是 cur的子节点} p[j][0],\;\sum_{j 是 cur的子节点} p[j][1] \} &,cur 不是叶子节点 \end{matrix}\right. p[cur][0]={0max{values[cur]+∑j是cur的子节点p[j][0],∑j是cur的子节点p[j][1]},cur是叶子节点,cur不是叶子节点
cpp
class Solution {
public:
using ll = long long;
long long maximumScoreAfterOperations(vector<vector<int>> &edges, vector<int> &values) {
int n = edges.size() + 1;
vector<int> e[n];
for (auto &ei: edges) {
e[ei[0]].push_back(ei[1]);
e[ei[1]].push_back(ei[0]);
}
ll p[n][2];
memset(p, -1, sizeof(p));//初始化状态
function<ll(int, int, int)> get = [&](int cur, int type, int par) {//记忆化搜索
if (p[cur][type] != -1)
return p[cur][type];
if (type == 0) {
if (cur != 0 && e[cur].size() == 1)
return p[cur][type] = 0;
ll r1 = 0, r2 = values[cur];
for (auto j: e[cur])
if (j != par) {
r1 += get(j, 1, cur);
r2 += get(j, 0, cur);
}
return p[cur][type] = max(r1, r2);
} else {
ll r2 = values[cur];
for (auto j: e[cur])
if (j != par)
r2 += get(j, 1, cur);
return p[cur][type] = r2;
}
};
return get(0, 0, 0);
}
};
D 平衡子序列的最大和
动态规划 + 离散化 + 树状数组:设数组 c [ i ] = n u m s [ i ] − i c[i]=nums[i]-i c[i]=nums[i]−i ,则 c c c 数组中非降序子序列的下标序列即为平衡子序列,所有可以对 c c c 数组进行离散化,然后定义状态 p [ i ] p[i] p[i] 为: c c c 数组中末尾元素为 i i i 的非降序子序列对应的平衡子序列在 n u m s nums nums 中的最大和,有状态转移方程: p [ i ] = m a x { n u m s [ i ] m a x { p [ j ] ∣ j ≤ i } + n u m s [ i ] p[i]=max\left\{\begin{matrix} nums[i]\\ max\{p[j] \;|\; j\le i \}+nums[i] \end{matrix}\right. p[i]=max{nums[i]max{p[j]∣j≤i}+nums[i] ,通过树状数组实现其中的前缀极值查询和单点更新
cpp
class Solution {
public:
using ll = long long;
int N;
vector<ll> a;
inline int lowbit(int x) {
return x & -x;
}
void update(int loc, ll val) {// li[loc]=max(li[loc], val);
for (; loc < N; loc += lowbit(loc))
a[loc] = max(a[loc], val);
}
ll query(int loc) {// max{li[k] | 1<=k<=loc}
ll res = INT64_MIN;
for (; loc > 0; loc -= lowbit(loc))
res = max(res, a[loc]);
return res;
}
long long maxBalancedSubsequenceSum(vector<int> &nums) {
int n = nums.size();
vector<int> c(n);
for (int i = 0; i < n; i++)
c[i] = nums[i] - i;
vector<int> t = c;
sort(t.begin(), t.end());
t.erase(unique(t.begin(), t.end()), t.end());
for (int i = 0; i < n; i++)//离散化c
c[i] = lower_bound(t.begin(), t.end(), c[i]) - t.begin();
N = n + 1;
a = vector<ll>(N, INT64_MIN);
for (int i = 0; i < n; i++) {
ll t1 = query(c[i] + 1);
update(c[i] + 1, max(t1, 0LL) + nums[i]);
}
return query(n);
}
};