潮鸣り

~~传送门：http://acm.hdu.edu.cn/showproblem.php?pid=1231最大连续子序列 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 20626 Accepted Submission(s): 9

从里到外逆推 //#pragma commmpnt(linkmpr, "/STACK:102400000,102400000") #include #include #include #include #include #include using namespace std; typedef long long LL; const int M = 100010; const int

最长上升子序列 O(n*logn) //#pragma commmpnt(linkmpr, "/STACK:102400000,102400000") #include #include #include #include #include #include using namespace std; typedef long long LL; const int M = 10001

#include "string" #include "iostream" #include "cstdio" #include "cmath" #include "set" #include "queue" #include "vector" #include "cctype" #include "sstream" #include "cstdlib" #include "c

完全背包 for i=[0,n)     for(j=weight[i]; j         tab[j] = max(tab[j-weight[i]]+value[i],tab[j]) //#include #include #include #include using namespace std; typedef long long LL; const int

//#pragma commmpnt(linkmpr, "/STACK:102400000,102400000") #include #include #include #include #include #include using namespace std; typedef long long LL; const int M = 100010; const int INF =0x

#include #include #include #include #include #include using namespace std; typedef long long LL; const int M = 10000; const int INF = 0x3f3f3f3f; int n; int dir[][2] = { 1,0,-1,0,0,-1,0,1 }; int mp[5

dp[i]= i-1时间段产奶量 + 现在的产奶量 #include #include #include #include #include using namespace std; typedef long long LL; const int M = 10000; const int INF = 0x3f3f3f3f; struct node { int st, et, w; boo

#include #include #include #include #include using namespace std; typedef long long LL; const int M = 10000; const int INF = 0x3f3f3f3f; int n, k; int dir[][2] = { 1,0,-1,0,0,1,0,-1 }; int dp[105][10

基础DP #include #include #include #include #include using namespace std; typedef long long LL; const int M = 10000; const int INF = 0x3f3f3f3f; string s[1050]; int dp[1050][1050]; //char s[1050][1050]

思路：反面考虑，用总的方案数减去A，B，C三点在同一路径上的方案数。于是我们可以确定中间点B，在当前以B为根求得的son中任选一个，在剩下的节点n-tmp-1（tmp为已经求得的B的儿子的个数）中任选一个,产生tmp*(n-tmp-1)中组合。 //#pragma comment(linker, "/STACK:102400000,102400000") #pragma comment(link

求最长上升子序列的长度以及个数，由于数据水，可以直接找  正解是DP+网络流 #include "cstring" #include "iostream" #include "algorithm" #include "cstdio" #include "queue" #include "set" using namespace std; typedef long long LL; const i

int a[M],b[M]; int search(int num,int low,int high) { int mid; while(low<=high) { mid=(low+high)/2; if(num>=b[mid]) low=mid+1; else high=mid

龟兔赛跑 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 12597    Accepted Submission(s): 4735 Problem Description 据说在很久很久以前，可怜的兔子经历了人生中最

dp[i][j], j = 0, 1 (dp[i][0]表示以i号平台左边为起点到地面的最短时间，dp[i][1]]表示以i号平台右边为起点到地面的最短时间) 状态转移方程： dp[i][0] = H[i] - H[m] + Min (dp[m][0] + X1[i] - X1[m], dp[m][1] + X2[m] - X1[i]);  m为i左边下面的平台的编号 dp[i][1]