浙江财经大学
信息管理与工程学院

HDU 5813 Elegant Construction

本文由 Ocrosoft 于 2016-08-11 9:39:21 发表

Elegant Construction

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 665    Accepted Submission(s): 346
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
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
 

Author
SYSU
 

Solution

题意:

要求构造一个有向图,使得点i能够恰好到达Ai个点.(直接间接皆可)
输出任意满足条件的图即可,没有要求最小.

思路:

先对Ai数组升序排序。
对于Ai != 0的点, 要连接m个点,那么可以跟它之前的点连接,这样每条边只会增加一个可到达点。
至于无法成树的情况,判断一下某个点前面是否有Ai个点就可以了。

#include <set>
#include <map>
#include <list>
#include <cmath>
#include <stack>
#include <queue>
#include <ctime>
#include <string>
#include <cstdio>
#include <vector>
#include <cctype>
#include <climits>
#include <sstream>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <functional>
#define ms(a) memset(a,0,sizeof(a))
typedef long long ll;
const int INF=INT_MAX;
const int MAXN=4000+10;
using namespace std;
int n,m;
int main()
{
    int T,cas=1;
    cin>>T;
    while(T--)
    {
        pair<int,int> v[MAXN];
        cin>>n;
        for(int i=1;i<=n;i++)
        {
            int t;cin>>t;
            v[i]=make_pair(t,i);
        }
        sort(v+1,v+1+n);
        int cnt=0,flag=1;
        for(int i=1;i<=n;i++)
        {
            if(v[i].first>=i)
            {flag=0;break;}
            cnt+=v[i].first;
        }
        if(!flag)printf("Case #%d: No\n",cas++);
        else
        {
            printf("Case #%d: Yes\n%d\n",cas++,cnt);
            for(int i=1;i<=n;i++)
                for(int j=1;j<=v[i].first;j++)
                    printf("%d %d\n",v[i].second,v[j].second);
        }
    }
    return 0;
}

 

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