POJ 3045 Cow Acrobats 贪心\二分

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题目链接： 点我
题目分析： 本来做的是二分的练习题，但在想的过程中发现贪心就好了。每次只考虑最下面选哪个牛，策略是选择w+v最大的那个，设其他任一牛为w’，v’，所有牛总重为sum，那么两头牛分别在最下层的风险是sum-w-v和sum-w’-v’，所以每次都选择w+v最大的牛放在最下面。将w+v排序后贪心即可。
附：二分和贪心的代码

贪心：

Problem: 3045       User: ChenyangDu
Memory: 544K        Time: 94MS
Language: C++       Result: Accepted

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;

const int maxn = 50000+5,INF = 1000000000+7;
int n,sum;

struct cow{int w,v;}in[maxn];

bool cmp(cow a,cow b){
    return a.w+a.v<b.w+b.v;
}

int main(){
    //freopen("in.txt","r",stdin);
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%d%d",&in[i].w,&in[i].v);
    }
    sort(in,in+n,cmp);
    int ans = -INF;
    for(int i=0;i<n;i++){
        ans = max(ans,sum - in[i].v);
        sum += in[i].w;
    }
    printf("%d",ans);
    return 0;
}

二分：

Problem: 3045       User: ChenyangDu
Memory: 612K        Time: 125MS
Language: C++       Result: Accepted
Source Code
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;

const int maxn = 50000+5,INF = 1000000000+7;
int n,sum,sum_back;

struct cow{int w,v;}in[maxn];

bool cmp(cow a,cow b){
    return a.w+a.v>b.w+b.v;
}

bool ok(int x){
    sum = sum_back;
    for(int i=0;i<n;i++){
        cow a = in[i];
        if(sum > a.w + a.v + x){
            return false;
        }
        sum -= a.w;
    }
    return true;
}

int main(){
    //freopen("in.txt","r",stdin);
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%d%d",&in[i].w,&in[i].v);
        sum += in[i].w;
    }
    sum_back = sum;
    sort(in,in+n,cmp);
    int l = -INF,r = INF;
    while(r-l>1){
        int mid = (l+r)>>1;
        if(ok(mid))r = mid;
        else l = mid;
    }
    cout<<r;
    return 0;
}