[Algo] 区间dp2
1. 合唱队
// 1. 合唱队
// https://www.luogu.com.cn/problem/P3205
struct CountStruct
{int cntLeft = 0;int cntRight = 0;
};
int compute(vector<int> &h)
{int n = h.size();vector<vector<CountStruct>> dp(n, vector<CountStruct>(n));for (int i = 0; i < n - 1; i++){if (h[i] < h[i + 1]){dp[i][i + 1].cntLeft = 1;dp[i][i + 1].cntRight = 1;}}for (int l = n - 3; l >= 0; l--)for (int r = l + 2; r < n; r++){CountStruct tmp = dp[l + 1][r];int a = tmp.cntLeft, b = tmp.cntRight;tmp= dp[l][r - 1];int c = tmp.cntLeft, d = tmp.cntRight;int cntL = 0, cntR = 0;if (h[l] < h[l + 1]) cntL = (cntL + a) % MOD;if (h[l] < h[r]) cntL = (cntL + b) % MOD;if (h[r] > h[l]) cntR = (cntR + c) % MOD;if (h[r] > h[r - 1]) cntR = (cntR + d) % MOD;dp[l][r].cntLeft = cntL;dp[l][r].cntRight = cntR; }return (dp[0][n - 1].cntLeft + dp[0][n - 1].cntRight) % MOD;
}
2. 移除盒子
// 2. 移除盒子
// https://leetcode.cn/problems/remove-boxes/
int removeBoxes(vector<int>& boxes) {int n = boxes.size();vector<vector<vector<int>>> dp(n, vector<vector<int>>(n, vector<int>(n, 0)));return func(boxes, 0, n - 1, 0, dp);
}
// boxes[l....r]范围上要去消除，前面跟着k个连续的和boxes[l]颜色一样的盒子
// 这种情况下，返回最大得分
int func(vector<int>& boxes, int l, int r, int k, vector<vector<vector<int>>>& dp) {if (l > r) {return 0;}if (dp[l][r][k] > 0) {return dp[l][r][k];}int s = l;while (s + 1 <= r && boxes[s + 1] == boxes[l]) {s++;}int cnt = k + (s - l + 1);// 可能性1 : 前缀先消int ans = cnt * cnt + func(boxes, s + 1, r, 0, dp);// 可能性2 : 讨论前缀跟着哪个后，一起消掉for (int m = s + 2; m <= r; m++) {if (boxes[m] == boxes[l] && boxes[m - 1] != boxes[m]) {// boxes[l] == boxes[m]是必须条件// boxes[m - 1] != boxes[m]是剪枝条件，避免不必要的调用ans = max(ans, func(boxes, s + 1, m - 1, 0, dp) + func(boxes, m, r, cnt, dp));}}dp[l][r][k] = ans;return ans;
}
3. 统计不同回文子序列
// 3. 统计不同回文子序列
// https://leetcode.cn/problems/count-different-palindromic-subsequences/description/
int countPalindromicSubsequences(string str) {const int mod = 1000000007;int n = str.size();vector<int> last(256, -1);vector<int> left(n), right(n);// left[i] : i位置的左边和s[i]字符相等且最近的位置在哪，不存在就是-1for (int i = 0; i < n; i++) {left[i] = last[str[i]];last[str[i]] = i;}// right[i] : i位置的右边和s[i]字符相等且最近的位置在哪，不存在就是nfill(last.begin(), last.end(), n);for (int i = n - 1; i >= 0; i--) {right[i] = last[str[i]];last[str[i]] = i;}// dp[i][j] : i...j范围上有多少不同的回文子序列// 如果i>j，那么认为是无效范围dp[i][j] = 0vector<vector<long>> dp(n, vector<long>(n, 0));for (int i = 0; i < n; i++) {dp[i][i] = 1;}for (int i = n - 2; i >= 0; i--) {for (int j = i + 1; j < n; j++) {if (str[i] != str[j]) {dp[i][j] = dp[i][j - 1] + dp[i + 1][j] - dp[i + 1][j - 1] + mod;} else {int l = right[i];int r = left[j];if (l > r) {// i...j的内部没有s[i]字符dp[i][j] = dp[i + 1][j - 1] * 2 + 2;} else if (l == r) {// i...j的内部有且仅有一个s[i]字符dp[i][j] = dp[i + 1][j - 1] * 2 + 1;} else {// i...j的内部至少有两个s[i]字符dp[i][j] = dp[i + 1][j - 1] * 2 - dp[l + 1][r - 1] + mod;}}dp[i][j] %= mod;}}return (int)dp[0][n - 1];
}
