yyg的专栏

思路：转化为DAG上的动态规划，长方体的“可堆叠”关系是一个典型的二元关系，用图来建模。如果一个长方体X可以放在另外一个长方体Y的上面我们就从X到Y连一条边。这个有向图在缩点后是无环的。是一个DAG，求DAG上的最大高度即可。代码：#include#include#includeusing namespace std;const int MAXN = 1000+5;long long

#include#include#includeusing namespace std;const int maxn = 100005;const int inf = 0x3f3f3f3f;int t , n , m , a , b , e , k;int head[maxn] , dp[maxn] , dis[maxn] , vis[maxn];struct T{

#include#include#includeusing namespace std;int dp[109];int cost[109];int value[109];int country[109];int volume;void ZeroOne(int cost,int value){ for(int i=volume;i>=cost;i--) d

#include#includeint a[105] , p[105] , n , dp[105] , sum[105];void f(){ for(int i = 1 ; i <= n ; i ++){ dp[i] = (sum[i] + 10) * p[i]; for(int j = 1 ; j <= i ; j ++){

#include#include#include#includeusing namespace std;const int inf = 0x3f3f3f3f;int n , k;int dis[205] , cost[205][205] , dp[205][205];int main(){ int cases = 1; while(~scanf("%d%d"

#include#include#includeusing namespace std;const int maxn = 505;const int inf = 0x3f3f3f3f;int n;int w[maxn] , c[maxn] , E , F , V , dp[maxn * 10000];int main(){ int t; scanf("%d"

#include#include#includeusing namespace std;const long long mod=1e9+7;const int MAXN = 1000005;int max(int a ,int b){ return a > b ? a : b;}int n;int num[1000005];int dp[2][MAXN];int

#include#includeint n , str[500004];int stack[500004];int top;const int inf = 0x3f3f3f3f;void push(int n){ stack[top] = n; top ++; return;}int pop(){ top --; return stack[

#include#includeint n , k , ans;const int maxn = 105;int map[maxn][maxn];int dp[maxn][maxn];int dir[4][2] = {1 , 0 , -1 , 0 , 0 , 1 , 0 , -1};int f(int x , int y){ int tx , ty; int ret

#include#include#includeusing namespace std;const int maxn = 105;const int inf = 0x3f3f3f3f;char s1[maxn] , s2[maxn];int dp[maxn][maxn] ,l1 , l2;int map[6][6] = {{inf , inf ,inf ,inf ,inf ,

#include#include#includeusing namespace std;const int inf = 0x3f3f3f3f;int n;struct T{ char sub[104]; int cost; int dead;}wk[20];struct T1{ char s[20]; int mcost; i

代码：#include#include//using namespace std;int num[205] , n;char op[205];double p[205];double dp[205][205];int cases = 1;int main(){ while(~scanf("%d" , &n)){ for(int i = 0 ; i

这个不能过，找不出错。。。有知道的告诉一下。。#pragma comment(linker, "/STACK:1024000000,1024000000")#include#include#include#include#includeusing namespace std;const int maxn = 6010;struct T{ int u; in

思路：DP，仍然是求DAG图上的最长路径，枚举每一个入度为0的点，算出它可以到达的最长路径并且记录最大值即可。代码：#include#include#includeusing namespace std;struct T{ int u; int v; int next;}edge[1000010];int val[100010] , head[10000

思路：简单DP代码：#include#includeint r , c;int h[105][105];int dp[105][105];int f(int i , int j){ int max = -1; if(dp[i][j] != -1) return dp[i][j]; if(h[i - 1][j] = 0){ if(max

最简单的树形dp#include#include#includeusing namespace std;int n , m , e , a , b , val[205];int head[205];int dp[205][205];struct T{ int u , v , next;}edge[41000];void addedge(int u , int