POJ 3045 Cow Acrobats 贪心\二分

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: ChenyangDuMemory: 544K        Time: 94MSLanguage: 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: ChenyangDuMemory: 612K        Time: 125MSLanguage: C++       Result: AcceptedSource 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;}