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思路很妙……

有個前提條件：血量無限，這樣話肯定先打會回血的怪，再打會掉血的怪

對於會回血的怪，按照受到傷害的順序從小往大打

對於會掉血的怪似乎並不是很好搞，考慮：將每一時刻的血量函數畫出來，然後反過來看（從右往左看這個函數），就相當於回血量和掉血量互換，會掉血的怪會變成會回血的怪。因為最終的血量是已知的，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;
}
 

BZOJ3709 Bohater 貪心