Swordsman（输入优化模板）

Swordsman

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1096    Accepted Submission(s): 309

 

Problem Description

Lawson is a magic swordsman with k kinds of magic attributes v1,v2,v3,…,vk. 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. If v1≥ai,1 and v2≥ai,2 and v3≥ai,3 and … and vk≥ai,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 will increase bi,j for 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, indicating the number of test cases. For each test case:
The first line has two integers n and k (1≤n≤105,1≤k≤5).
The second line has k non-negative integers (initial magic attributes) v1,v2,v3,…,vk.
For the next n lines, the i-th line contains 2k non-negative integers ai,1,ai,2,ai,3,…,ai,k,bi,1,bi,2,bi,3,…,bi,k.
It's guaranteed that all input integers are no more than 109 and vj+∑i=1nbi,j≤109 for j=1,2,3,…,k.

It is guaranteed that the sum of all n ≤5×105.
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 integers 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.

 

 

Source

2018 Multi-University Training Contest 7

 

 

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思路：思路很简单，就是各个属性值按照由小到大依次排序，按照属性值能吃就吃，当每一次循环没有全部被吃完的情况就跳出while。但是这道题要用到强大的输入优化。（贴一份强大的dls代码+读入优化模板）

当然这道题也让让我见识了强有力的转化，每一个数组都有独特的意义，想明白了就不绕了。

代码：

#include<bits/stdc++.h>
using namespace std;
namespace IO{
        #define BUF_SIZE 100000
        #define OUT_SIZE 100000
        #define ll long long
        //fread->read

        bool IOerror=0;
        inline char nc(){
            static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
            if (p1==pend){
                p1=buf; pend=buf+fread(buf,1,BUF_SIZE,stdin);
                if (pend==p1){IOerror=1;return -1;}
                //{printf("IO error!\n");system("pause");for (;;);exit(0);}
            }
            return *p1++;
        }
        inline bool blank(char ch){return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';}
        inline void read(int &x){
            bool sign=0; char ch=nc(); x=0;
            for (;blank(ch);ch=nc());
            if (IOerror)return;
            if (ch=='-')sign=1,ch=nc();
            for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';
            if (sign)x=-x;
        }
        inline void read(ll &x){
            bool sign=0; char ch=nc(); x=0;
            for (;blank(ch);ch=nc());
            if (IOerror)return;
            if (ch=='-')sign=1,ch=nc();
            for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';
            if (sign)x=-x;
        }
        inline void read(double &x){
            bool sign=0; char ch=nc(); x=0;
            for (;blank(ch);ch=nc());
            if (IOerror)return;
            if (ch=='-')sign=1,ch=nc();
            for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';
            if (ch=='.'){
                double tmp=1; ch=nc();
                for (;ch>='0'&&ch<='9';ch=nc())tmp/=10.0,x+=tmp*(ch-'0');
            }
            if (sign)x=-x;
        }
        inline void read(char *s){
            char ch=nc();
            for (;blank(ch);ch=nc());
            if (IOerror)return;
            for (;!blank(ch)&&!IOerror;ch=nc())*s++=ch;
            *s=0;
        }
        inline void read(char &c){
            for (c=nc();blank(c);c=nc());
            if (IOerror){c=-1;return;}
        }
        //fwrite->write
        struct Ostream_fwrite{
            char *buf,*p1,*pend;
            Ostream_fwrite(){buf=new char[BUF_SIZE];p1=buf;pend=buf+BUF_SIZE;}
            void out(char ch){
                if (p1==pend){
                    fwrite(buf,1,BUF_SIZE,stdout);p1=buf;
                }
                *p1++=ch;
            }
            void print(int x){
                static char s[15],*s1;s1=s;
                if (!x)*s1++='0';if (x<0)out('-'),x=-x;
                while(x)*s1++=x%10+'0',x/=10;
                while(s1--!=s)out(*s1);
            }
            void println(int x){
                static char s[15],*s1;s1=s;
                if (!x)*s1++='0';if (x<0)out('-'),x=-x;
                while(x)*s1++=x%10+'0',x/=10;
                while(s1--!=s)out(*s1); out('\n');
            }
            void print(ll x){
                static char s[25],*s1;s1=s;
                if (!x)*s1++='0';if (x<0)out('-'),x=-x;
                while(x)*s1++=x%10+'0',x/=10;
                while(s1--!=s)out(*s1);
            }
            void println(ll x){
                static char s[25],*s1;s1=s;
                if (!x)*s1++='0';if (x<0)out('-'),x=-x;
                while(x)*s1++=x%10+'0',x/=10;
                while(s1--!=s)out(*s1); out('\n');
            }
            void print(double x,int y){
                static ll mul[]={1,10,100,1000,10000,100000,1000000,10000000,100000000,
                    1000000000,10000000000LL,100000000000LL,1000000000000LL,10000000000000LL,
                    100000000000000LL,1000000000000000LL,10000000000000000LL,100000000000000000LL};
                if (x<-1e-12)out('-'),x=-x;x*=mul[y];
                ll x1=(ll)floor(x); if (x-floor(x)>=0.5)++x1;
                ll x2=x1/mul[y],x3=x1-x2*mul[y]; print(x2);
                if (y>0){out('.'); for (size_t i=1;i<y&&x3*mul[i]<mul[y];out('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
};
const int N=100005,K=6;
int n,k,id[K][N],a[N][K],b[N][K],Num[N],pos[K],val[K],Ans,Ind;
int cmp(int x,int y)
{
    return a[x][Ind]<a[y][Ind];///ind是属性。同一个属性这样子排序
}
void solve()
{
        IO::read(n);
        IO::read(k);
        for(int i=1;i<=n;i++)
            Num[i]=0;
        for(int i=1;i<=k;i++)
            IO::read(val[i]),pos[i]=1;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=k;j++)
                IO::read(a[i][j]);
            for(int j=1;j<=k;j++)
                IO::read(b[i][j]);
        }
        
        for(Ind=1;Ind<=k;Ind++)
        {
            for(int i=1;i<=n;i++)
                id[Ind][i]=i;///id 【ind】 【i】表示第ind的这一
                
            sort(id[Ind]+1,id[Ind]+n+1,cmp);
        }
        Ans=0;
        while(1)
        {
            int last=Ans;
            for(int i=1;i<=k;i++)///按照属性
            {
                while(pos[i]<=n&&a[id[i][pos[i]]][i]<=val[i])///说明这个点是可以的
                {
                    int t=id[i][pos[i]];///t代表的是i这个属性序号为pos[i]的序号值
                    Num[t]++;///
                    if(Num[t]==k)
                    {
                        Ans++;
                        for(int j=1;j<=k;j++)
                            val[j]+=b[t][j];
                    }
                    pos[i]++;    ///说明i这个属性目前到了第几个序号
                }
            }
            if(Ans==last)///当这种循环不行的时候就停止。
                break;
        }
        IO::println(Ans);
        for(int i=1;i<=k;i++)
            IO::print(val[i]),IO::print(i!=k?' ':'\n');
}
int main()
{
    int t;
    cin>>t;
    while(t--)
        solve();
    return 0;
}