题解：
贪心,用(力量+重量)排序然后\(dfs\);
\(code\):
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
struct node{
int h,w,s;
}a[30];
int n,h;
bool c[30];
long long ans=-1;
bool cmp(const node &x,const node &y)
{
return x.w+x.s>y.w+y.s;
}
void dfs(int x,int ch)
{
c[x]=ch;
if(x==n)
{
long long tmp=1e9,len=0;
for(int i=1;i<=n;i++)
if(c[i])
{
len+=a[i].h;
long long sum=0;
for(int j=i+1;j<=n;j++)
sum+=c[j]?a[j].w:0;
tmp=min(tmp,a[i].s-sum);
}
if(len>=h)ans=max(ans,tmp);return;}
dfs(x+1,1);
dfs(x+1,0);
}
int main()
{
scanf("%d%d",&n,&h);
for(int i=1;i<=n;i++)
scanf("%d%d%d",&a[i].h,&a[i].w,&a[i].s);
sort(a+1,a+n+1,cmp);
dfs(0,0);
if(ans==-1) printf("Mark is too tall\n");
else printf("%lld",ans);
}
题解：
\(dfs\)二进制状压记录状态,标记状态
\(code:\)
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include<ctype.h>
#define bin(x) (1<<x)
#define inf 1e9+9
#define ll long long
using namespace std;
char buf[1<<20],*p1,*p2;
inline char gc()
{
// return getchar();
return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<20,stdin))==p1?0:*p1++;
}
template<typename T>
inline void read(T &x)
{
char tt;
bool flag=0;
while(!isdigit(tt=gc())&&tt!='-');
tt=='-'?(flag=1,x=0):(x=tt-'0');
while(isdigit(tt=gc())) x=x*10+tt-'0';
if(flag) x=-x;
}
int n,tot;
int a[25][25];
int w[25],ans;
bool book[1<<21];
void dfs(int s,int rest) {
if(book[s]) return;
book[s]=1;
if (rest==1)
{
ans|=s;
return;
}
for(int i=0;i<n;i++)
if((bin(i)&s)&&w[i+1]>0) {
for(int j=1;j<=n;j++)w[j]-=a[j][i+1];
dfs(s^bin(i),rest-1);
for(int j=1;j<=n;j++)w[j]+=a[j][i+1];
}
}
int t;
int main()
{
// freopen("1.txt","w",stdout);
read(t);
while(t--)
{
read(n);
for(int i=1;i<=n;i++)
{
w[i]=0;
for(int j=1;j<=n;j++)
read(a[i][j]),w[i]+=a[i][j];
}
ans=0;
tot=bin(n)-1;
for(int i=0;i<=tot;i++) book[i]=0;
dfs(tot,n);
bool flag=0;
for(int i=0;i<n;i++)
if(bin(i)&ans)
printf("%d ",i+1),flag=1;
if (!flag) putchar('0');
putchar(10);
}
}
题解：
同原先\(THH\)学长出的数数\(......\)
\(code:\)
#include<stdio.h>
#include<string.h>
#define ll long long
#define mod 1000000007
ll f[31][31][31][1<<9];
ll n,m,lim;
void inc(ll &a,ll b) {
a+=b;
a-=a<mod?0:mod;
}
ll dfs(ll u,ll l,ll v,ll s) {
if(l==0) return s==0;
if(u==1) return 0;
if(f[u][l][v][s]!=-1) return f[u][l][v][s];
ll ans=0;
inc(ans,dfs(u,l-1,v,s^1^(1<<(u-v))));
if(v>1&&u-v+1<=lim) inc(ans,dfs(u,l,v-1,s));
else if(!(s&1)) inc(ans,dfs(u-1,l,u-2,s>>1));
return f[u][l][v][s]=ans;
}
int main() {
memset(f,-1,sizeof(f));
scanf("%lld%lld%lld",&n,&m,&lim);
printf("%lld",dfs(n,m,n-1,0));
}
