wdcxc的专栏

思路：记忆dp，注意数据类型#include<iostream> #include<string.h>using namespace std;int input[50][50]; long long table[50][50];long long dp(int x, int y, int n) { int rx = x - 9 > 0 ? x - 9 : 0; int ry = y

思路：记忆dp,用前一个月的所有可取状态更新后一个月的所有可取状态，结果是最后一个月的所有可取状态的最小值（可取状态即雇佣worker数>=所需worker数）#include<iostream> #include<vector> #include<algorithm> #include<utility> using namespace std;vector<int> wn; //worke

思路：记忆dp#include<iostream> #include<string.h>using namespace std;int w[100][100][100];int solve(int a, int b, int c) { if (a <= 0 || b <= 0 || c <= 0) return 1; if (a > 20 || b > 20 || c

思路：状态压缩dp#include<iostream> #include<vector> #include<algorithm> #include<string> using namespace std;//状态结构体 struct state{ int prestate,precourse, minreduce, usedtime; };//课程结构体 struct course{

思路：每一种方块可以多次利用，不过最终只可以利用6种姿势，全部输入排序，然后从大到小dp#include<iostream> #include<vector> #include<algorithm> #include<string.h> using namespace std;struct rec{ int l, w,h; };bool cmp(rec r1, rec r2) { i

思路：4个游标打表#include<iostream> #include<string.h> #include<vector> #include<algorithm> using namespace std;unsigned long long table[5843];void gettable() { int a = 1, b = 1, c = 1, d = 1; table[1]

题目中先后的意思是顺序前后的意思#include<iostream> #include<string.h> #include<algorithm>using namespace std;long long num[10100], len[100], table[10100][100], sum[10100];long long dp(int n,int m) { memset(table,

注意table初始化的值，尽量大，不过不能用INT_MAX，会溢出；不能用系统自带pow，会TLE#include<iostream> #include<climits> #include<cmath> #include<algorithm> #include<cstdio>using namespace std;int input[2001], table[2001][1001];void dp(

#include<iostream> #include<cstdio> #include<algorithm> using namespace std;int input[1000000];int main() { int n; while (scanf("%d",&n)!=-1) { for (int i = 0; i < n; i++) {

思路：多重背包#include<iostream> #include<cstdio> #include<algorithm> #include<string.h> #include<vector>using namespace std;int input[7]; vector<int> v; int table[60010];bool dp(int sum) { v.clear();

思路：bfs从后往前求路径长度，dfs从前往后求路径数，记忆dp#include<iostream> #include<vector> #include<string.h> #include<queue> #include<utility> using namespace std;int map[60][60]; int dir[4][2] = { { -1, 0 }, { 0, 1 }, { 1,

#include<iostream> #include<string.h> #include<cstring> #include<string> #include<algorithm>using namespace std;int table[1000][1000]; string s1, s2;int dp() { memset(table, 0, sizeof(table));

思路：01背包#include<iostream> #include<string.h> #include<algorithm>using namespace std;int val[1001], wei[1001], table[1001];int dp(int n,int w) { memset(table, 0, sizeof(table)); for (int i = 0;

思路：从后向前更新 #include #include #include #include #include using namespace std; int map[100010][11]; int solve(int maxtime) { int res[11], tres[11]; memset(tres, 0, sizeof(tres)); for (int i

#include<iostream> #include<vector>using namespace std;int input[30001]; vector<int> no;int solve(int n) { no.clear(); for (int i = 0; i < n; i++) { int j = 0; for (; j < no

思路：记忆dp#include<iostream> #include<string.h> #include<algorithm>using namespace std;bool used[500][500]; int chess[500][500]; long long table[500][500];long long dp(int x, int y, int n, int k) { us

#include<iostream> #include<string.h> using namespace std;int table[80];void dp(int n,int m) { memset(table, 0, sizeof(table)); table[0] = 1; for (int i = 0; i < n; i++) { for (

思路：记忆dp，先排序#include<iostream> #include<vector> #include<algorithm> #include<utility>using namespace std;struct node{ int w, s, sum, pre, ini_no; };bool cmp(node n1, node n2) { if (n1.w != n2.w)

#include<cstdio> #include<iostream> #include<algorithm> #include<string.h> #include<utility> #include<vector> using namespace std;int dp[10010], coinw[10010], coinp[10010];void solve(int w, int n) {