HDU-6396：Swordsman（优先队列）
Swordsman
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 131072/131072 K (Java/Others)
Problem Description
Lawson is a magic swordsman with k k kinds of magic attributes v1,v2,v3,…,vk. v 1 , v 2 , v 3 , … , v k . Now Lawson is faced with n monsters and the i-th monster also has k kinds of defensive attributes ai,1,ai,2,ai,3,…,ai,k a i , 1 , a i , 2 , a i , 3 , … , a i , k . If v1≥ai,1 v 1 ≥ a i , 1 and v2≥ai,2 v 2 ≥ a i , 2 and v3≥ai,3 v 3 ≥ a i , 3 and … and vk≥ai,k v k ≥ a i , k , Lawson can kill the i-th monster (each monster can be killed for at most one time) and get EXP from the battle, which means vj v j will increase bi,j b i , j for j=1,2,3,…,k j = 1 , 2 , 3 , … , k .
Now we want to know how many monsters Lawson can kill at most and how much Lawson’s magic attributes can be maximized.
Input
There are multiple test cases. The first line of input contains an integer T T , indicating the number of test cases. For each test case:
The first line has two integers n n and k k (1≤n≤105,1≤k≤5) ( 1 ≤ n ≤ 10 5 , 1 ≤ k ≤ 5 ) .
The second line has k k non-negative integers (initial magic attributes) v1,v2,v3,…,vk. v 1 , v 2 , v 3 , … , v k .
For the next n n lines, the i-th line contains 2k 2 k non-negative integers ai,1,ai,2,ai,3,…,ai,k,bi,1,bi,2,bi,3,…,bi,k a i , 1 , a i , 2 , a i , 3 , … , a i , k , b i , 1 , b i , 2 , b i , 3 , … , b i , k .
It’s guaranteed that all input integers are no more than 109 10 9 and vj+∑ni=1bi,j≤109 v j + ∑ i = 1 n b i , j ≤ 10 9 for j=1,2,3,…,k. j = 1 , 2 , 3 , … , k .
It is guaranteed that the sum of all n≤5×105 n ≤ 5 × 10 5 .
The input data is very large so fast IO (like fread) is recommended.
Output
For each test case:
The first line has one integer which means the maximum number of monsters that can be killed by Lawson.
The second line has k k integers v'1,v'2,v'3,…,v'k v ′ 1 , v ′ 2 , v ′ 3 , … , v ′ k and the i-th integer means maximum of the i-th magic attibute.
Sample Input
1
4 3
7 1 1
5 5 2 6 3 1
24 1 1 1 2 1
0 4 1 5 1 1
6 0 1 5 3 1
Sample Output
3
23 8 4
Hint
For the sample, initial V = [7, 1, 1]
① kill monster #4 (6, 0, 1), V + [5, 3, 1] = [12, 4, 2]
② kill monster #3 (0, 4, 1), V + [5, 1, 1] = [17, 5, 3]
③ kill monster #1 (5, 5, 2), V + [6, 3, 1] = [23, 8, 4]
After three battles, Lawson are still not able to kill monster #2 (24, 1, 1)
because 23 < 24.
思路：我们可以不断从这 n n 个怪物中筛选出满足条件的。
1. 1. 把满足 a[1]≤v[1] a [ 1 ] ≤ v [ 1 ] 的怪物放入一个优先队列 P2 P 2 。
2. 2. 然后从 P2 P 2 中选出所有满足 a[2]≤v[2] a [ 2 ] ≤ v [ 2 ] 的怪物放入一个优先队列 P3 P 3 。
3. 3. 然后从 P3 P 3 中选出所有满足 a[3]≤v[3] a [ 3 ] ≤ v [ 3 ] 的怪物放入一个优先队列 P4 P 4 。
4. 4. 然后从 P4 P 4 中选出所有满足 a[4]≤v[4] a [ 4 ] ≤ v [ 4 ] 的怪物放入一个优先队列 P5 P 5 。
5. 5. 然后从 P5 P 5 中选出所有满足 a[5]≤v[5] a [ 5 ] ≤ v [ 5 ] 的怪物，则这些怪物是可以被杀死的，把杀死这个怪物可以获得的属性更新至 v v 数组，然后重复 1 1 操作。
不过我代码写得很冗长。。题解代码非常简洁
这题还要用快速读入，自己的快读T了，用dls的就A了%%%。。
PS：用dls的快读手动测不了数据，必须用文件输入输出
#include'0'),++i); print(x3);}
}
void println(double x,int y){print(x,y);out('\n');}
void print(char *s){while (*s)out(*s++);}
void println(char *s){while (*s)out(*s++);out('\n');}
void flush(){if (p1!=buf){fwrite(buf,1,p1-buf,stdout);p1=buf;}}
~Ostream_fwrite(){flush();}
}Ostream;
inline void print(int x){Ostream.print(x);}
inline void println(int x){Ostream.println(x);}
inline void print(char x){Ostream.out(x);}
inline void println(char x){Ostream.out(x);Ostream.out('\n');}
inline void print(ll x){Ostream.print(x);}
inline void println(ll x){Ostream.println(x);}
inline void print(double x,int y){Ostream.print(x,y);}
inline void println(double x,int y){Ostream.println(x,y);}
inline void print(char *s){Ostream.print(s);}
inline void println(char *s){Ostream.println(s);}
inline void println(){Ostream.out('\n');}
inline void flush(){Ostream.flush();}
#undef ll
#undef OUT_SIZE
#undef BUF_SIZE
};
int v[6],k;
struct lenka{int a[6],b[6];}a[MAX];
int cmp(const lenka& A,const lenka& B){return A.a[1]1];}
struct lenka2{int a[6],b[6];bool operator<(const lenka2 A)const{return A.a[2]2];}};
struct lenka3{int a[6],b[6];bool operator<(const lenka3 A)const{return A.a[3]3];}};
struct lenka4{int a[6],b[6];bool operator<(const lenka4 A)const{return A.a[4]4];}};
struct lenka5{int a[6],b[6];bool operator<(const lenka5 A)const{return A.a[5]5];}};
lenka2 to2(const lenka& B)
{
lenka2 A;
for(int i=1;i<=k;i++)A.a[i]=B.a[i];
for(int i=1;i<=k;i++)A.b[i]=B.b[i];
return A;
}
lenka3 to3(const lenka2& B)
{
lenka3 A;
for(int i=1;i<=k;i++)A.a[i]=B.a[i];
for(int i=1;i<=k;i++)A.b[i]=B.b[i];
return A;
}
lenka4 to4(const lenka3& B)
{
lenka4 A;
for(int i=1;i<=k;i++)A.a[i]=B.a[i];
for(int i=1;i<=k;i++)A.b[i]=B.b[i];
return A;
}
lenka5 to5(const lenka4& B)
{
lenka5 A;
for(int i=1;i<=k;i++)A.a[i]=B.a[i];
for(int i=1;i<=k;i++)A.b[i]=B.b[i];
return A;
}
void add(const lenka &A){for(int i=1;i<=k;i++)v[i]+=A.b[i];}
void add(const lenka2 &A){for(int i=1;i<=k;i++)v[i]+=A.b[i];}
void add(const lenka3 &A){for(int i=1;i<=k;i++)v[i]+=A.b[i];}
void add(const lenka4 &A){for(int i=1;i<=k;i++)v[i]+=A.b[i];}
void add(const lenka5 &A){for(int i=1;i<=k;i++)v[i]+=A.b[i];}
priority_queuep2;
priority_queuep3;
priority_queuep4;
priority_queuep5;
int ans;
void cal()
{
int tag=1;
while(tag)
{
tag=0;
if(k==1)continue;
while(!p2.empty()&&p2.top().a[2]<=v[2])
{
if(k==2)ans++,add(p2.top()),tag=1;
else p3.push(to3(p2.top()));
p2.pop();
}
if(k==2)continue;
while(!p3.empty()&&p3.top().a[3]<=v[3])
{
if(k==3)ans++,add(p3.top()),tag=1;
else p4.push(to4(p3.top()));
p3.pop();
}
if(k==3)continue;
while(!p4.empty()&&p4.top().a[4]<=v[4])
{
if(k==4)ans++,add(p4.top()),tag=1;
else p5.push(to5(p4.top()));
p4.pop();
}
if(k==4)continue;
while(!p5.empty()&&p5.top().a[5]<=v[5])
{
ans++;
add(p5.top());
p5.pop();
tag=1;
}
}
}
int main()
{
int T;
IO::read(T);
while(T--)
{
int n;
IO::read(n);
IO::read(k);
for(int i=1;i<=k;i++)IO::read(v[i]);
for(int i=1;i<=n;i++)
{
for(int j=1;j<=k;j++)IO::read(a[i].a[j]);
for(int j=1;j<=k;j++)IO::read(a[i].b[j]);
}
sort(a+1,a+n+1,cmp);
ans=0;
for(int i=1;i<=n;i++)
{
cal();
if(a[i].a[1]<=v[1])
{
if(k==1)ans++,add(a[i]);
else p2.push(to2(a[i]));
}
}
cal();
printf("%d\n",ans);
for(int i=1;i<=k;i++)printf("%d%c",v[i],i==k?'\n':' ');
while(!p2.empty())p2.pop();
while(!p3.empty())p3.pop();
while(!p4.empty())p4.pop();
while(!p5.empty())p5.pop();
}
return 0;
}