HDU-2018多校7th-1011-Swordsman 附杜教fread代码
题意：
背景是打怪升级的故事，有k个不同属性的初始的能力值，每只怪物也有相同个数的能力值，你只能放倒k个能力值都比怪物大的，每放倒一个怪物，都可以得到相应的k个能力值。
思路：
根据k个能力值得到k个不同的排序，设立k个不同的指针从小到大开始移动，对满足被放倒的条件的属性进行标记，当某只monster的所有防御属性都被标记时，更新打怪者的能力值。
这里的移动操作，可以通过判断cnt是否更新 来结束循环。而每次移动操作都使得指针跑到打怪者能力小于monster的位子。
自己在练习的时候想的思路有点不周全，我们先对k个能力单独排序，单独给能放倒的怪兽打标记，最后看标记个数是否等于k个。反例是：对某种属性值，你只有打败了A号怪物，才能通过得到的能力值打败其他怪物，而在另一种属性下，A好怪物不能被放倒，但其他怪物能被放倒，这就造成了误判。
代码：
//#include<bits/stdc++.h> //#include<unordered_map> //#include<unordered_set> #include<functional> #include<algorithm> //#include<ext/rope> #include<iomanip> #include<climits> #include <iostream> #include<cstring> #include<cstdlib> #include<cstddef> #include<stdio.h> #include<memory> #include<vector> #include<cctype> #include<string> #include<cmath> #include<queue> #include<deque> #include<ctime> #include<stack> #include<map> #include<set> #define fi first #define se second #define pb push_back #define INF 0x3f3f3f3f #define pi 3.1415926535898 #define lowbit(a) (a&(-a)) #define lson l,(l+r)/2,rt<<1 #define rson (l+r)/2+1,r,rt<<1|1 #define Min(a,b,c) min(a,min(b,c)) #define Max(a,b,c) max(a,max(b,c)) using namespace std; //using namespace __gnu_cxx; typedef long long ll; typedef pair<int,int> P; typedef unsigned long long ull; const ll LLMAX=2e18; const int MOD=1e9+7; const int MAXN=1e6+10; 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 }; using namespace IO; template<class T> inline void read(T &DataIn) { DataIn=0; T Flag=0; char c=getchar(); while(!isdigit(c)){ Flag|=c=='-'; c=getchar(); } while(isdigit(c)){ DataIn=DataIn*10+c-'0'; c=getchar(); } DataIn= Flag? -DataIn: DataIn; } template<class T> inline void write(T DataOut,char EndChar='\n') { T lenth=0,number[30]; if(DataOut==0){ putchar(48); return; } while(DataOut>0){ number[++lenth]=DataOut%10; DataOut/=10;} for(int i=lenth;i>=1;i--) putchar(number[i]+48); putchar(EndChar); } int a[10],pos[10]; struct node{ ll hit[10],exp[10],id; }b[MAXN]; int p[10][MAXN]; inline bool cmp1(int x,int y){ return b[x].hit[0]<b[y].hit[0]; } inline bool cmp2(int x,int y){ return b[x].hit[1]<b[y].hit[1]; } inline bool cmp3(int x,int y){ return b[x].hit[2]<b[y].hit[2]; } inline bool cmp4(int x,int y){ return b[x].hit[3]<b[y].hit[3]; } inline bool cmp5(int x,int y){ return b[x].hit[4]<b[y].hit[4]; } int vis[MAXN]; int main() { int T; cin>>T; while(T--){ int n,k,cnt=0; read(n);read(k); //scanf("%d%d",&n,&k); for(int i=0;i<k;i++) read(a[i]); for(int i=0;i<n;i++){ vis[i] = 0; b[i].id=i; for(int j=0;j<k;j++) read(b[i].hit[j]),p[j][i] = i;; for(int j=0;j<k;j++) read(b[i].exp[j]); } pos[0] = pos[1] = pos[2] =pos[3] = pos[4] = 0; for(int i=0; i<k; i++){ if(i==0) sort(p[0],p[0]+n,cmp1); else if(i==1) sort(p[1],p[1]+n,cmp2); else if(i==2) sort(p[2],p[2]+n,cmp3); else if(i==3) sort(p[3],p[3]+n,cmp4); else if(i==4) sort(p[4],p[4]+n,cmp5); } while(true){ int c = cnt; for(int i=0; i<k; i++){ while(pos[i] < n && b[p[i][pos[i]]].hit[i] <= a[i]){ int id = p[i][pos[i]]; vis[id]++; if(vis[id] == k){ cnt++; for(int j=0; j<k; j++){ a[j] += b[id].exp[j]; } } pos[i]++; } } if(cnt==c){ break; } } //write(cnt); IO::println(cnt); for(int i=0;i<k;i++) IO::print(a[i]),IO::print(i!=k-1?' ':'\n'); } return 0; }
skr
