Codeforces140CNew Year Snowmen

一道优先队列好题

不难发现，每次取最多的几个雪球能保证凑出来的雪人最多，然后我们就先离散化一下，然后封装丢到优先队列里，每次取出队首的三个，将数量各自减一，表示凑了一个雪人，然后将数量大于一的再丢回去，直到剩余种数小于3为止

代码

//By AcerMo
#include<queue>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int M=100500;
int n,a[M],b[M],c[M];
struct emm
{int v,id;bool friend operator < (emm a,emm b){return a.v<b.v;}
}ad;
struct an{int a,b,c;}e[M];
priority_queue<emm>q;
inline int read()
{int x=0;char ch=getchar();while (ch>'9'||ch<'0') ch=getchar();while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();return x;
}
signed main()
{n=read();int cnt=0,s=0,tot=0;for (int i=1;i<=n;i++) a[i]=read();sort(a+1,a+n+1);for (int i=1;i<=n;i++)if (a[i]==a[i-1]) b[cnt]++;else b[++cnt]++,c[cnt]=a[i];for (int i=1;i<=cnt;i++)ad.id=c[i],ad.v=b[i],q.push(ad);if (q.size()<3) return puts("0"),0;while (q.size()>=3){emm x=q.top();q.pop();emm y=q.top();q.pop();emm z=q.top();q.pop();s++;x.v--;y.v--;z.v--;e[++tot].a=x.id;e[tot].b=y.id;e[tot].c=z.id;if (e[tot].a<e[tot].b) swap(e[tot].a,e[tot].b);if (e[tot].a<e[tot].c) swap(e[tot].a,e[tot].c);if (e[tot].b<e[tot].c) swap(e[tot].b,e[tot].c);if (x.v>0) q.push(x);if (y.v>0) q.push(y);if (z.v>0) q.push(z);}printf("%d\n",s);for (int i=1;i<=tot;i++)printf("%d %d %d\n",e[i].a,e[i].b,e[i].c);return 0;
}

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Codeforces140CNew Year Snowmen

一道优先队列好题

代码

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