BZOJ3709 Bohater 贪心

传送门

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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_JUDGE
	freopen("in","r",stdin);
	//freopen("out","w",stdout);
#endif
	N = 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;
}