hdu 5813 Elegant Construction 构造

Elegant Construction

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 185    Accepted Submission(s): 99
Special Judge

Problem Description

Being an ACMer requires knowledge in many fields, because problems in this contest may use physics, biology, and even musicology as background. And now in this problem, you are being a city architect!
A city with N towns (numbered 1 through N) is under construction. You, the architect, are being responsible for designing how these towns are connected by one-way roads. Each road connects two towns, and passengers can travel through in one direction.

For business purpose, the connectivity between towns has some requirements. You are given N non-negative integers a1 .. aN. For 1 <= i <= N, passenger start from town i, should be able to reach exactly ai towns (directly or indirectly, not include i itself). To prevent confusion on the trip, every road should be different, and cycles (one can travel through several roads and back to the starting point) should not exist.

Your task is constructing such a city. Now it's your showtime!

 

Input

The first line is an integer T (T <= 10), indicating the number of test case. Each test case begins with an integer N (1 <= N <= 1000), indicating the number of towns. Then N numbers in a line, the ith number ai (0 <= ai < N) has been described above.

 

Output

For each test case, output "Case #X: Y" in a line (without quotes), where X is the case number starting from 1, and Y is "Yes" if you can construct successfully or "No" if it's impossible to reach the requirements.

If Y is "Yes", output an integer M in a line, indicating the number of roads. Then M lines follow, each line contains two integers u and v (1 <= u, v <= N), separated with one single space, indicating a road direct from town u to town v. If there are multiple possible solutions, print any of them.

 

Sample Input

332 1 021 143 1 1 0

 

Sample Output

Case #1: Yes21 22 3Case #2: NoCase #3: Yes41 21 32 43 4

 

Author

SYSU

 

Source

2016 Multi-University Training Contest 7

 

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题意：现在需要连一些有向边，使得从所有点i出发能够到达的点数为ai(不包括自身)。

解法：

按照点的权值排序，依次为点i填充i点出发的边。

所以只用考虑前i-1个点。

只用判断i-1是否大于等于a[i]即可，如果是，连上a[i]条边，否则"No"

(就是从ai值最小的点，一个一个加进去)

#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>#include<vector>using namespace std;#define all(x) (x).begin(), (x).end()#define for0(a, n) for (int (a) = 0; (a) < (n); (a)++)#define for1(a, n) for (int (a) = 1; (a) <= (n); (a)++)#define mes(a,x,s)  memset(a,x,(s)*sizeof a[0])#define mem(a,x)  memset(a,x,sizeof a)#define ysk(x)  (1<<(x))typedef long long ll;typedef pair<int, int> pii;const int INF =0x3f3f3f3f;const int maxn=1000;int n,ans;vector<int >fa[maxn+10];struct Node{    int id,all;    Node(){}    bool operator<(const Node c)const    {        return all<c.all;    }}a[maxn+10];bool work(){    ans=0;   for(int i=0;i<=n;i++)  fa[i].clear();   for1(i,n)   {       int sum=i-1;       if(sum<a[i].all)  return false;       for(int j=1;j<=a[i].all;j++)       {           fa[ a[j].id ].push_back(a[i].id );       }       ans+=a[i].all;   }   return true;}int main(){   std::ios::sync_with_stdio(false);   int T,kase=0;   cin>>T;   while(T--)   {       cin>>n;       for1(i,n)       {         a[i].id=i;         cin>>a[i].all;       }        sort(a+1,a+1+n);       printf("Case #%d: ",++kase);       bool ok;       puts( (ok=work())?"Yes":"No");       if(!ok)  continue;       printf("%d\n",ans);       for(int i=1;i<=n;i++)       {           int siz=fa[i].size();           for(int j=0;j<siz;j++)           {               int x=fa[i][j];               printf("%d %d\n",x,i);           }       }   }   return 0;}