传送门

思路很妙……

有个前提条件：血量无限，这样话肯定先打会回血的怪，再打会掉血的怪

对于会回血的怪，按照受到伤害的顺序从小往大打

对于会掉血的怪似乎并不是很好搞，考虑：将每一时刻的血量函数画出来，然后反过来看（从右往左看这个函数），就相当于回血量和掉血量互换，会掉血的怪会变成会回血的怪。因为最终的血量是已知的，所以按照上面的方法再做一遍就可以了。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<ctime>
#include<cctype>
#include<algorithm>
#include<cstring>
#include<iomanip>
#include<queue>
#include<map>
#include<set>
#include<bitset>
#include<vector>
#include<stack>
#include<cmath>
//This code is written by Itst
using namespace std;inline int read(){int a = 0;char c = getchar();bool f = 0;while(!isdigit(c) && c != EOF){if(c == '-')f = 1;c = getchar();}if(c == EOF)exit(0);while(isdigit(c)){a = a * 10 + c - 48;c = getchar();}return f ? -a : a;
}#define PII pair < int , int >
#define st first
#define nd second
#define PIII pair < PII , int >
#define auto vector < PIII > :: iterator
vector < PIII > mons[2];
long long hp , End;
int N;signed main(){
#ifndef ONLINE_JUDGEfreopen("in","r",stdin);//freopen("out","w",stdout);
#endifN = read(); End = hp = read();for(int i = 1 ; i <= N ; ++i){int d = read() , a = read();a >= d ?mons[0].push_back(PIII(PII(d , a) , i)) :mons[1].push_back(PIII(PII(a , d) , i));End = End - d + a;}if(End <= 0){puts("NIE"); return 0;}sort(mons[0].begin() , mons[0].end());for(auto t = mons[0].begin() ; t != mons[0].end() ; ++t){if(hp - (*t).st.st <= 0){puts("NIE"); return 0;}hp = hp - (*t).st.st + (*t).st.nd;}sort(mons[1].begin() , mons[1].end());for(auto t = mons[1].begin() ; t != mons[1].end() ; ++t){if(End - (*t).st.st <= 0){puts("NIE"); return 0;}End = End - (*t).st.st + (*t).st.nd;}puts("TAK");for(auto t = mons[0].begin() ; t != mons[0].end() ; ++t)printf("%d " , (*t).nd);reverse(mons[1].begin() , mons[1].end());for(auto t = mons[1].begin() ; t != mons[1].end() ; ++t)printf("%d " , (*t).nd);return 0;
}