【算法笔记自学】第 11 章 提高篇（5）——动态规划专题
创始人
2025-01-09 09:08:15
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11.1动态规划的递归写法和递推写法
#include const int MOD = 10007; const int MAXN = 10000 + 1; int fib[MAXN]; int main() { int n; scanf("%d", &n); fib[1] = fib[2] = 1; for (int i = 3; i <= n; i++) { fib[i] = (fib[i - 1] + fib[i - 2]) % MOD; } printf("%d", fib[n]); return 0; } 
#include #include using namespace std; const int MAXN = 100 + 1; int a[MAXN][MAXN]; int dp[MAXN][MAXN]; int main() { int n; scanf("%d", &n); for (int i = 1; i <= n; i++) { for (int j = 1; j <= i; j++) { scanf("%d", &a[i][j]); } } for (int i = 1; i <= n; i++) { dp[n][i] = a[n][i]; } for (int i = n - 1; i >= 1; i--) { for (int j = 1; j <= i; j++) { dp[i][j] = max(dp[i + 1][j], dp[i + 1][j + 1]) + a[i][j]; } } printf("%d", dp[1][1]); return 0; } 11.2最大连续子序列和
#include #include using namespace std; const int MAXN = 10000; const int INF = 0x3fffffff; int a[MAXN]; int dp[MAXN]; int main() { int n; scanf("%d", &n); for (int i = 0; i < n; i++) { scanf("%d", &a[i]); } dp[0] = a[0]; for (int i = 1; i < n; i++) { dp[i] = max(a[i], dp[i - 1] + a[i]); } int maxResult = -INF; for (int i = 0; i < n; i++) { maxResult = max(maxResult, dp[i]); } printf("%d", maxResult); return 0; } 11.3最长不下降子序列（LIS）
#include #include using namespace std; const int MAXN = 100; int a[MAXN]; int dp[MAXN]; int main() { int n; scanf("%d", &n); for (int i = 0; i < n; i++) { scanf("%d", &a[i]); } int maxLen = 0; for (int i = 0; i < n; i++) { dp[i] = 1; for (int j = 0; j < i; j++) { if (a[j] <= a[i] && dp[j] + 1 > dp[i]) { dp[i] = dp[j] + 1; } } maxLen = max(maxLen, dp[i]); } printf("%d", maxLen); return 0; } 11.4最长公共子序列（LCS）
#include #include #include using namespace std; const int MAXN = 100 + 1; int dp[MAXN][MAXN]; int main() { string s1, s2; cin >> s1 >> s2; for (int i = 1; i <= s1.length(); i++) { for (int j = 1; j <= s2.length(); j++) { if (s1[i - 1] == s2[j - 1]) { dp[i][j] = dp[i - 1][j - 1] + 1; } else { dp[i][j] = max(dp[i][j - 1], dp[i - 1][j]); } } } printf("%d", dp[s1.length()][s2.length()]); return 0; } 11.5最长回文子串
#include #include #include using namespace std; const int MAXN = 100; bool dp[MAXN][MAXN]; int main() { string s; cin >> s; int maxLength = 1; memset(dp, false, sizeof(dp)); for (int i = 0; i < s.length(); i++) { dp[i][i] = true; } for (int i = 0; i < (int)s.length() - 1; i++) { if (s[i] == s[i + 1]) { dp[i][i + 1] = true; maxLength = 2; } } for (int len = 3; len <= s.length(); len++) { for (int i = 0; i + len - 1 < s.length(); i++) { int j = i + len - 1; if (s[i] == s[j] && dp[i + 1][j - 1]) { dp[i][j] = true; maxLength = len; } } } printf("%d", maxLength); return 0; } 11.6DAG最长路
#include #include #include using namespace std; const int MAXN = 100; const int INF = 1e9; int G[MAXN][MAXN]; int dp[MAXN]; int getDAGMaxLength(int i, int n) { if (dp[i]) { return dp[i]; } for (int j = 0; j < n; j++) { if (G[i][j] != INF) { dp[i] = max(dp[i], getDAGMaxLength(j, n) + G[i][j]); } } return dp[i]; } int main() { memset(dp, 0, sizeof(dp)); fill(G[0], G[0] + MAXN * MAXN, INF); int n, m; scanf("%d%d", &n, &m); int u, v, w; for (int i = 0; i < m; i++) { scanf("%d%d%d", &u, &v, &w); G[u][v] = w; } int maxLength = 0; for (int i = 0; i < n; i++) { maxLength = max(maxLength, getDAGMaxLength(i, n)); } printf("%d", maxLength); return 0; } 
#include #include #include using namespace std; const int MAXN = 100; const int INF = 1e9; int G[MAXN][MAXN]; int dp[MAXN]; int getDAGMaxLength(int i, int n) { if (dp[i] >= 0) { return dp[i]; } for (int j = 0; j < n; j++) { if (G[i][j] != INF) { dp[i] = max(dp[i], getDAGMaxLength(j, n) + G[i][j]); } } return dp[i]; } int main() { fill(dp, dp + MAXN, -INF); fill(G[0], G[0] + MAXN * MAXN, INF); int n, m, t; scanf("%d%d%d", &n, &m, &t); dp[t] = 0; int u, v, w; for (int i = 0; i < m; i++) { scanf("%d%d%d", &u, &v, &w); G[u][v] = w; } int maxLength = 0; for (int i = 0; i < n; i++) { maxLength = max(maxLength, getDAGMaxLength(i, n)); } printf("%d", maxLength); return 0; } 11.7背包问题
#include #include #include using namespace std; const int MAXN = 100 + 1; const int MAXV = 1000 + 1; int w[MAXN], c[MAXN]; int dp[MAXV]; int main() { int n, maxW; scanf("%d%d", &n, &maxW); for (int i = 1; i <= n; i++) { scanf("%d", &w[i]); } for (int i = 1; i <= n; i++) { scanf("%d", &c[i]); } memset(dp, 0, sizeof(dp)); for (int i = 1; i <= n; i++) { for (int v = maxW; v >= w[i]; v--) { dp[v] = max(dp[v], dp[v - w[i]] + c[i]); } } printf("%d", dp[maxW]); return 0; } 
#include #include #include using namespace std; const int MAXN = 100 + 1; const int MAXV = 1000 + 1; int w[MAXN], c[MAXN]; int dp[MAXV]; int main() { int n, maxW; scanf("%d%d", &n, &maxW); for (int i = 1; i <= n; i++) { scanf("%d", &w[i]); } for (int i = 1; i <= n; i++) { scanf("%d", &c[i]); } memset(dp, 0, sizeof(dp)); for (int i = 1; i <= n; i++) { for (int v = w[i]; v <= maxW; v++) { dp[v] = max(dp[v], dp[v - w[i]] + c[i]); } } printf("%d", dp[maxW]); return 0; } 11.8总结
#include #include using namespace std; const int MAXN = 10000; int h[MAXN]; int dp[MAXN]; int main() { int n; scanf("%d", &n); for (int i = 0; i < n; i++) { scanf("%d", &h[i]); } dp[0] = 0; for (int i = 1; i < n; i++) { dp[i] = dp[i - 1] + abs(h[i] - h[i - 1]); if (i - 2 >= 0) { dp[i] = min(dp[i], dp[i - 2] + abs(h[i] - h[i - 2])); } } printf("%d", dp[n - 1]); return 0; } 
#include #include using namespace std; const int MAXN = 100; int matrix[MAXN][MAXN]; int dp[MAXN][MAXN] = {0}; int main() { int n, m; scanf("%d%d", &n, &m); for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { scanf("%d", &matrix[i][j]); } } // 初始化第一行和第一列的dp值 dp[0][0] = matrix[0][0]; for (int i = 1; i < n; i++) { dp[i][0] = dp[i - 1][0] + matrix[i][0]; } for (int j = 1; j < m; j++) { dp[0][j] = dp[0][j - 1] + matrix[0][j]; } // 状态转移方程 for (int i = 1; i < n; i++) { for (int j = 1; j < m; j++) { dp[i][j] = max(dp[i][j - 1], dp[i - 1][j]) + matrix[i][j]; } } printf("%d", dp[n - 1][m - 1]); return 0; } 
#include #include using namespace std; const int MAXN = 10000; int a[MAXN], b[MAXN], c[MAXN]; int dp[MAXN][3]; int main() { int n; scanf("%d", &n); for (int i = 0; i < n; i++) { scanf("%d%d%d", &a[i], &b[i], &c[i]); } dp[0][0] = a[0]; dp[0][1] = b[0]; dp[0][2] = c[0]; for (int i = 1; i < n; i++) { dp[i][0] = min(dp[i - 1][1], dp[i - 1][2]) + a[i]; dp[i][1] = min(dp[i - 1][0], dp[i - 1][2]) + b[i]; dp[i][2] = min(dp[i - 1][0], dp[i - 1][1]) + c[i]; } printf("%d", min(min(dp[n - 1][0], dp[n - 1][1]), dp[n - 1][2])); return 0; } 
#include #include #include using namespace std; const int MAXN = 100 + 1; int dp[MAXN][MAXN]; int main() { string s, t; cin >> s >> t; for (int i = 0; i <= s.length(); i++) { dp[i][0] = i; } for (int j = 0; j <= t.length(); j++) { dp[0][j] = j; } for (int i = 1; i <= s.length(); i++) { for (int j = 1; j <= t.length(); j++) { dp[i][j] = min(dp[i - 1][j] + 1, dp[i][j - 1] + 1); dp[i][j] = min(dp[i][j], dp[i - 1][j - 1] + (s[i - 1] == t[j - 1] ? 0 : 1)); } } cout << dp[s.length()][t.length()]; return 0; } 
#include #include #include using namespace std; const int mod =1e4+7; vector dp(2,0),tmp(2,0); int n; //dp[0] 表示前一位选 0，dp[1]表示前一位不选0 int main(){ cin>>n; dp[0]=1,dp[1]=9; for(int i=2;i<=n;i++){ tmp[0] = dp[1]; tmp[1] = (dp[0]+dp[1])*9 % mod; dp = tmp; } cout<<(dp[0]+dp[1])%mod< 相关内容
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