BZOJ3709 [PA2014]Bohater 【贪心】

题目链接

题解

贪心很显然

我们先干掉能回血的怪，当然按照\(d\)升序顺序，因为打得越多血越多，\(d\)大的尽量往后打

然后再干掉会扣血的怪，当然按照\(a\)降序顺序，因为最后受的伤害一定，回的血也一定，先尽量回多的血以尽量承受住当前伤害

#include<algorithm>

#include<iostream>

#include<cstring>

#include<cstdio>

#include<queue>

#include<cmath>

#include<map>

#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)

#define REP(i,n) for (int i = 1; i <= (n); i++)

#define mp(a,b) make_pair<int,int>(a,b)

#define cls(s) memset(s,0,sizeof(s))

#define cp pair<int,int>

#define LL long long int

using namespace std;

const int maxn = 100005,maxm = 100005,INF = 1000000000;

inline int read(){

	int out = 0,flag = 1; char c = getchar();

	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}

	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}

	return out * flag;

}

struct node{int a,d,id;}a[maxn],b[maxn];

inline bool cmp1(const node& a,const node& b){

	return a.d < b.d;

}

inline bool cmp2(const node& a,const node& b){

	return a.a > b.a;

}

int n,ai,bi;

LL z;

int main(){

	n = read(); z = read();

	int x,y;

	REP(i,n){

		x = read(); y = read();

		if (y >= x) a[++ai] = (node){y,x,i};

		else b[++bi] = (node){y,x,i};

	}

	sort(a + 1,a + 1 + ai,cmp1);

	for (int i = 1; i <= ai; i++){

		if (z <= a[i].d){puts("NIE"); return 0;}

		z = z - a[i].d + a[i].a;

	}

	sort(b + 1,b + 1 + bi,cmp2);

	for (int i = 1; i <= bi; i++){

		if (z <= b[i].d){puts("NIE"); return 0;}

		z = z - b[i].d + b[i].a;

	}

	puts("TAK");

	for (int i = 1; i <= ai; i++)

		printf("%d ",a[i].id);

	for (int i = 1; i <= bi; i++)

		printf("%d ",b[i].id);

	return 0;

}

巴特西

BZOJ3709 [PA2014]Bohater 【贪心】

题目链接

题解

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