DAG上的dp 因为本身升序就是拓扑序,所以建出图来直接从1到ndp即可,设f[i][j]为到i花费了j #include<iostream> #include<cstdio> using namespace std; const int N=1005,inf=1e9+7; int n,m,b,h[N],cnt,f[N][N],ans=-inf; struct qwe { int ne,to,va,c; }e[N*10]; int read() { int r=0,f=1; cha…

有点类似背包 , 就是那样子搞... ------------------------------------------------------------------------------------ #include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #define rep( i , n ) for( int i = 0 ;  i < n ; ++i…

http://www.lydsy.com/JudgeOnline/problem.php?id=1649 又是题解... 设f[i][j]表示费用i长度j得到的最大乐趣 f[i][end[a]]=max{f[i-cost[a][begin[a]]+w[a]} 当f[i-cost[a][begin[a]]可行时 初始化f=-1 f[0][0]=0 #include <cstdio> #include <cstring> #include <cmath> #include…

http://www.lydsy.com/JudgeOnline/problem.php?id=1649 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 743  Solved: 370[Submit][Status][Discuss] Description The cows are building a roller coaster! They want your help to design as fun a roller coaster as…

Description The cows are building a roller coaster! They want your help to design as fun a roller coaster as possible, while keeping to the budget. The roller coaster will be built on a long linear stretch of land of length L (1 <= L <= 1,000). The…

很像背包. 这种在一个数轴上进行操作的题常常需要对区间排序. f[i][j]表示距离到i时,花费为j时的权值之和. f[x[i]+l[i]][j+c[i]]=max{f[x[i]][j]+w[i]}(1<=i<=n,0<=j<=B) #include<cstdio> #include<cstring> #include<algorithm> using namespace std; struct Line{int x,l,w,c;}a[10001…

1648: [Usaco2006 Dec]Cow Picnic 奶牛野餐 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 554  Solved: 346[Submit][Status][Discuss] Description The cows are having a picnic! Each of Farmer John's K (1 <= K <= 100) cows is grazing in one of N (1 <= N <…

直接从每个奶牛所在的farm dfs , 然后算一下.. ---------------------------------------------------------------------------------------- #include<cstdio> #include<algorithm> #include<cstring> #include<iostream> #include<vector>   #define rep( i…

从每个奶牛所在草场dfs,把沿途dfs到的草场的con都+1,最后符合条件的草场就是con==k的,扫一遍统计一下即可 #include<iostream> #include<cstdio> using namespace std; const int K=105,N=1005; int k,n,m,p[K],h[N],cnt,c[N],v[N],ti,ans; struct qwe { int ne,to; }e[N*10]; int read() { int r=0,f=1;…

Description The cows are having a picnic! Each of Farmer John's K (1 <= K <= 100) cows is grazing in one of N (1 <= N <= 1,000) pastures, conveniently numbered 1...N. The pastures are connected by M (1 <= M <= 10,000) one-way paths (no p…

严格次短路模板,用两个数组分别维护最短路和次短路,用dijskstra,每次更新的时候先更新最短路再更新次短路 写了spfa版的不知道为啥不对-- #include<iostream> #include<cstdio> #include<queue> using namespace std; const int N=5005,inf=1e9; int n,m,h[N],cnt,d1[N],d2[N]; struct qwe { int ne,to,va; }e[2000…

每个ai在最后sum中的值是本身值乘上组合数,按这个dfs一下即可 #include<iostream> #include<cstdio> using namespace std; int n,s,ans[15],c[20][20]; bool u[15],f=0; int dfs(int a,int b) { if(b==n) { if(a==s) f=1; return 0; } for(int i=1;i<=n;i++) if(!u[i]) { u[i]=1,ans[b…

在洛谷上被卡了一个点开了O2才过= = bfs即可,为方便存储,把所有坐标+500 #include<iostream> #include<cstdio> #include<queue> using namespace std; const int N=1005,dx[]={-1,1,0,0},dy[]={0,0,-1,1}; int n,sx,sy; bool a[N][N],v[N][N]; struct qwe { int x,y,b; qwe(int X=0,i…

传送门 先按照起点 sort 一遍. 这样每一个点的只由前面的点决定. f[i][j] 表示终点为 i,花费 j 的最优解 状态转移就是一个01背包. ——代码 #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> int L, N, B; ][]; inline int read() { , f = ; char ch = getchar(); ; ) +…

有点神奇的dp 首先注意到任意一个数都能被表示成若干个斐波那契数的和的形式 先求出n可以字典序最大的表示 设f[i][0/1]表示第i个斐波那契数选或者不选 如果当前数不选,那就选比他小的两个数,否则,需要不选比他小的两个数(连续的影响) #include<iostream> #include<cstdio> using namespace std; const int N=105; long long n,a[N],s[N],top,f[N][2]; int main() { s…

完了不会dp了 设f[i][j]为以i结尾,有j个时的最优值,辅助数组g[i][j]为s选了i和j,i~j中的误差值 转移是f[j][i]=min(f[k][i-1]+g[k][j]) #include<iostream> #include<cstdio> #include<algorithm> using namespace std; const long long N=105; long long n,m,a[N],g[N][N],f[N][N]; int main…

很像贪心的dp啊 这个定金尾款的设定让我想起了lolita和jk制服的尾款地狱-- 设f[i][j]为从j到i的付定金的最早月份然后从f[k][j-1]转移来,两种转移f[i][j]=min(f[i][j],f[j-1][k]+1)是当前这个月付[k-1,j-1]的尾款和[j,i]的定金,f[i][j]=min(f[i][j],f[j-1][k]+2)是先付[k-1,j-1]的尾款,下个月再付[j,i]的定金 然后答案要+2,是最后一次付定金的尾款加上第一个月没有工资 #include<iost…

参考:https://blog.csdn.net/cgh_andy/article/details/52506738 没有get到什么重点的dp--做的莫名其妙 注意滑雪一个坡可以滑很多次 设f[i][j]为时间为i能力为j的最大滑雪次数,预处理l[i][j]为在i时获得j能力的最晚开始时间,w[i]为有能力j时最短的能滑雪时间 模拟转移即可 #include<iostream> #include<cstdio> #include<cstring> using name…

P2854 [USACO06DEC]牛的过山车Cow Roller Coaster 题目描述 The cows are building a roller coaster! They want your help to design as fun a roller coaster as possible, while keeping to the budget. The roller coaster will be built on a long linear stretch of land o…

1648: [Usaco2006 Dec]Cow Picnic 奶牛野餐 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 432  Solved: 270[Submit][Status] Description The cows are having a picnic! Each of Farmer John's K (1 <= K <= 100) cows is grazing in one of N (1 <= N <= 1,000)…

P2854 [USACO06DEC]牛的过山车Cow Roller Coaster dp 对铁轨按左端点排个序,蓝后就是普通的二维dp了. 设$d[i][j]$为当前位置$i$,成本为$j$的最小花费 $d[i+a[u].w][j+a[u].c]=max(d[i+a[u].w][j+a[u].c],d[i][j]+a[u].f)$ #include<iostream> #include<cstdio> #include<cstring> #include<algo…

P2854 [USACO06DEC]牛的过山车Cow Roller Coaster 题目描述 The cows are building a roller coaster! They want your help to design as fun a roller coaster as possible, while keeping to the budget. The roller coaster will be built on a long linear stretch of land o…

Kattis - honey[DP] 题意 有一只蜜蜂,在它的蜂房当中,蜂房是正六边形的,然后它要出去,但是它只能走N步,第N步的时候要回到起点,给出N, 求方案总数 思路 用DP 因为N == 14 所以 最多走7步 我们不妨设 (7, 7) 为原点,然后 dp[0][7][7] = 1 因为 N == 0 的时候 方案数只有一个 那就是 不动吧.. dp[i][j][k] i 代表第几步 j k 分别表示 目前的位置 一个点 在一张图里面本来有八个方向可以走 这里六边形 我们只取六个方向就可…

HDOJ 1423 Greatest Common Increasing Subsequence [DP][最长公共上升子序列] Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 8768 Accepted Submission(s): 2831 Problem Description This is a problem from ZOJ 24…

HDOJ 1501 Zipper [DP][DFS+剪枝] Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 10886 Accepted Submission(s): 3925 Problem Description Given three strings, you are to determine whether the third str…

HDOJ 1257 最少拦截系统 [DP] Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 50753 Accepted Submission(s): 19895 Problem Description 某国为了防御敌国的导弹袭击,发展出一种导弹拦截系统.但是这种导弹拦截系统有一个缺陷:虽然它的第一发炮弹能够到达任意的高度,但是以后每一发炮弹…

HDOJ 1159 Common Subsequence[DP] Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 44280 Accepted Submission(s): 20431 Problem Description A subsequence of a given sequence is the given sequence wit…

HDOJ_1087_Super Jumping! Jumping! Jumping! [DP] Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 44670 Accepted Submission(s): 20693 Problem Description Nowadays, a kind of chess game called "Super…

POJ_2533 Longest Ordered Subsequence[DP][最长递增子序列] Longest Ordered Subsequence Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 58448 Accepted: 26207 Description A numeric sequence of ai is ordered if a1 < a2 < - < aN. Let the subsequenc…

HackerRank - common-child[DP] 题意 给出两串长度相等的字符串,找出他们的最长公共子序列e 思路 字符串版的LCS AC代码 #include <iostream> #include <cstdio> #include <algorithm> #include <cmath> #include <deque> #include <vector> #include <queue> #include…