poj3045 Cow Acrobats (思维，贪心)

题目：

poj3045 Cow Acrobats

解析：

贪心题，类似于国王游戏

考虑两个相邻的牛\(i\),\(j\)

设他们上面的牛的重量一共为\(sum\)

把\(i\)放在上面，危险值分别为\(x_1=sum-s_i\),$ x_2=sum+w_i-s_j\(
把\)j\(放在上面，危险值分别为\)x_3=sum-s_j$, \(x_4=sum+w_j-s_i\)

若把j放在上面更优，则有\(max(x_3,x_4)<max(x_1,x_2)\)

有四种情况

\(x_3<x_1\)

\(x_3<x_2\)

\(x_4<x_1\)

\(x_4<x_2\)

显然的是\(x_3<x_2\)，\(x_1>x_4\)

而\(s_i\)和\(s_j\)关系不确定

所以一定有\(w_i+s_i>w_j+w_j\)

按\(w+s\)排序从小到大排序，大的在下面

代码：

很简单

#include <iostream>
#include <algorithm>
#include <cstdio>
#define int long long
using namespace std;

const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;

int n;

struct node {
	int w, s;
	bool operator <(const node &oth) const {
		return w + s < oth.w + oth.s;
	}
} e[N];

template<class T>inline void read(T &x) {
	x = 0; char ch = getchar();
	while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
	return;
}

signed main() {
	read(n);
	for (int i = 1; i <= n; ++i) read(e[i].w), read(e[i].s);
	sort(e + 1, e + 1 + n);
	int sum = 0, ans = -INF;
	for (int i = 1; i <= n; ++i) ans = max(ans, sum - e[i].s), sum += e[i].w;
	cout << ans << endl;
}