Accepted

foo.cpp:28:0: warning: ignoring #pragma comment [-Wunknown-pragmas]
# pragma comment(linker, "/STACK:1024000000,1024000000")

状态 耗时 内存占用

#1 Accepted 0ms 1.098MiB
#2 Accepted 0ms 1.094MiB
#3 Accepted 0ms 1.09MiB
#4 Accepted 0ms 1.09MiB
#5 Accepted 0ms 1.094MiB
#6 Accepted 0ms 1.094MiB
#7 Accepted 0ms 1.09MiB
#8 Accepted 15ms 1.094MiB
#9 Accepted 15ms 1.086MiB
#10 Accepted 15ms 1.098MiB
代码

include <cstdio>

include <cstring>

include <cstdlib>

include <iostream>

include <vector>

include <queue>

include <stack>

include <map>

include <set>

include <cmath>

include <algorithm>

using namespace std;

define lowbit(x) ((x)&(-x))

define pi acos(-1.0)

define eps 1e-8

define MOD 1000000007

define INF 1000000000

define mem(a,b) memset(a,b,sizeof(a))

define FOR(i,a,n) for(int i=a; i<=n; ++i)

define FO(i,a,n) for(int i=a; i<n; ++i)

define bug puts("H");

define lch p<<1,l,mid

define rch p<<1|1,mid+1,r

define mp make_pair

define pb push_back

typedef pair<int,int> PII;
typedef vector<int> VI;

pragma comment(linker, "/STACK:1024000000,1024000000")

typedef long long LL;
int Scan() {
int res=0, flag=0;
char ch;
if((ch=getchar())=='-') flag=1;
else if(ch>='0'&&ch<='9') res=ch-'0';
while((ch=getchar())>='0'&&ch<='9') res=res*10+(ch-'0');
return flag?-res:res;
}
void Out(int a) {
if(a<0) {putchar('-'); a=-a;}
if(a>=10) Out(a/10);
putchar(a%10+'0');
}
const int N=105;
//Code begin...

int a[N][3], dp[N][N][3];

bool check(int i, int b, int j, int c)
{
int a1, a2, b1, b2;
if (b==0) a1=1, a2=2;
else if (b==1) a1=0, a2=2;
else a1=0, a2=1;
if (c==0) b1=1, b2=2;
else if (c==1) b1=0, b2=2;
else b1=0, b2=1;
return (a[i][a1]<=a[j][b1]&&a[i][a2]<=a[j][b2])||(a[i][a2]<=a[j][b1]&&a[i][a1]<=a[j][b2]);
}
int main ()
{
int n, m;
scanf("%d%d",&n,&m);
FOR(i,1,n) scanf("%d%d%d",&a[i][0],&a[i][1],&a[i][2]);
FOR(i,1,n) FOR(j,1,i) {
int ma=0;
FO(k,0,i) FO(l,0,3) ma=max(ma,dp[k][j-1][l]);
FO(l,0,3) dp[i][j][l]=ma+a[i][l];
FO(k,j,i) FO(l1,0,3) FO(l2,0,3) if (check(i,l1,k,l2)) dp[i][j][l1]=max(dp[i][j][l1],dp[k][j][l2]+a[i][l1]);
}
int ans=0;
FOR(i,m,n) FO(j,0,3) ans=max(ans,dp[i][m][j]);
printf("%d\n",ans);
return 0;
}

97年NOI二试的题 貌似放错地方了

第一个点没秒杀

编译通过...

├ 测试数据 01：答案正确... 56ms

├ 测试数据 02：答案正确... 0ms

├ 测试数据 03：答案正确... 0ms

├ 测试数据 04：答案正确... 0ms

├ 测试数据 05：答案正确... 0ms

├ 测试数据 06：答案正确... 0ms

├ 测试数据 07：答案正确... 0ms

├ 测试数据 08：答案正确... 0ms

├ 测试数据 09：答案正确... 0ms

├ 测试数据 10：答案正确... 0ms

以第i个第j组第k条边作高划分状态 DP