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
3
3
2 1 0
2
1 1
4
3 1 1 0
Sample Output
Case #1: Yes
2
1 2
2 3
Case #2: No
Case #3: Yes
4
1 2
1 3
2 4
3 4
題目鏈接：HDU⑸813
題目大意：給出n個點（1~n），和每一個點能到達(dá)的點的個數(shù)（直接或間接）。問是不是存在這類可能性
題目思路：貪心，先給點的出度排序。
eg : 0 1 2 3
比如3這個點出度為2：先連1，再連2.如果不能連了則這是不可能存在的出度值。（每一個點只能向前面的點連，由于不存在回路）
以下是代碼：
#include <iostream>
#include <iomanip>
#include <fstream>
#include <sstream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <cctype>
#include <algorithm>
#include <functional>
#include <numeric>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <vector>
#include <queue>
#include <deque>
#include <list>
using namespace std;
struct node
{
int a,pos;
}p[2005];
bool cmp(node a,node b)
{
return a.a < b.a;
}
vector <pair<int,int> > ans;
int n;
int solve()
{
for (int i = 1; i <= n; i++)
{
int num = p[i].a;
for (int j = 1; j < i; j++)
{
if (p[i].a == p[j].a) break;
num--;
}
if (num > 0) return 0;
}
return 1;
}
int main()
{
int t;
cin >> t;
int cas = 1;
while(t--)
{
cin >> n;
ans.clear();
for (int i = 1; i <= n; i++)
{
cin >> p[i].a;
p[i].pos = i;
}
sort(p + 1,p + n + 1,cmp);
int flag = solve();
if (!flag)
{
printf("Case #%d: No\n",cas++);
continue;
}
for (int i = 1; i <= n; i++)
{
int num = p[i].a;
for (int j = 1; j < i; j++)
{
if (p[i].a == p[j].a) break;
if (num <= 0) break;
num--;
ans.push_back(make_pair(p[i].pos,p[j].pos));
}
}
int len = ans.size();
printf("Case #%d: Yes\n",cas++);
printf("%d\n",len); //如果len等于0，這個0輸出不輸出都可以
for (int i = 0; i < len; i++)
{
printf("%d %d\n",ans[i].first,ans[i].second);
}
}
}
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