打怪升级 一开始有N个怪物；主角有K个能力；只有K个能力都击败怪物才能斩杀怪物并获得K个能力的增值；问最多能杀几个怪物；
做法：
用优先队列把怪物能力装进去；能力小放前面；
最重要的是数据量要用读入挂才能过；（读入挂太神奇了！！）
Swordsman
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2049 Accepted Submission(s): 601
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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代码：
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod=998244353;
const int maxn=1e5+90;
const ll inf=0x3f3f3f3f3f3f;
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
};
int v[10];
int value[maxn][10];
int b[maxn][10];
struct node{
int id;
int num;
node(int id,int num){this->id=id;this->num=num;}
bool friend operator<(node a,node b){
return a.num>b.num;
}
};
priority_queue<node>pq[10];
void init(){
for(int i=1;i<=5;i++){
while(!pq[i].empty())pq[i].pop();
}
}
int n,k,t;
int main()
{
// std::ios::sync_with_stdio(false);
//std::cin.tie(0);
IO::read(t);
while(t--){
init();
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(value[i][j]);
}
pq[1].push(node(i,value[i][1]));
for(int j=1;j<=k;j++)IO::read(b[i][j]);
}
int ans=0,ans1=-1;
while(ans!=ans1){
ans1=ans;
for(int i=1;i<k;i++){
while(!pq[i].empty()&&pq[i].top().num<=v[i]){
int now=pq[i].top().id;
pq[i].pop();
pq[i+1].push(node(now,value[now][i+1]));
}
}
while(!pq[k].empty()&&pq[k].top().num<=v[k]){
ans++;
int now=pq[k].top().id;
pq[k].pop();
for(int i=1;i<=k;i++)v[i]+=b[now][i];
}
}
IO::println(ans);
for(int i=1;i<=k;i++){
if(i!=1)IO::print(‘ ‘);
IO::print(v[i]);
}
IO::print(‘\n‘);
}
return 0;
}
原文地址：https://www.cnblogs.com/luowentao/p/9478962.html