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题目链接:
Time Limit: 4000/2000 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others)
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>
using namespace std;
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef long long LL;
template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<‘0‘||CH>‘9‘;F= CH==‘-‘,CH=getchar());
for(num=0;CH>=‘0‘&&CH<=‘9‘;num=num*10+CH-‘0‘,CH=getchar());
F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
if(!p) { puts("0"); return; }
while(p) stk[++ tp] = p%10, p/=10;
while(tp) putchar(stk[tp--] + ‘0‘);
putchar(‘\n‘);
}
const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e18;
const int N=1e5+10;
const int maxn=5e3+4;
const double eps=1e-12;
struct node
{
int a,id;
}po[N];
int cmp(node x,node y)
{
if(x.a==y.a)return x.id<y.id;
return x.a<y.a;
}
int main()
{
int t,Case=0,n;
read(t);
while(t--)
{
printf("Case #%d: ",++Case);
read(n);
int sum=0;
For(i,1,n)
{
read(po[i].a);
sum+=po[i].a;
po[i].id=i;
}
int flag=0;
sort(po+1,po+n+1,cmp);
For(i,1,n)
{
if(po[i].a>=i)
{
flag=1;
break;
}
}
if(flag)printf("No\n");
else
{
printf("Yes\n");
printf("%d\n",sum);
int num=0;
For(i,1,n)
{
for(int j=1;j<=po[i].a;j++)
{
printf("%d %d\n",po[i].id,po[j].id);
}
}
}
}
return 0;
}
hdu-5813 Elegant Construction(贪心)
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原文地址:http://www.cnblogs.com/zhangchengc919/p/5759375.html