程序员人生网站导航

HDU-5813-Elegant Construction【多校2016】【贪心】

栏目：php教程时间：2016-10-07 10:37:56

5813-Elegant Construction

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

题目链接：HDU⑸813

题目大意：给出n个点（1~n），和每一个点能到达的点的个数（直接或间接）。问是不是存在这类可能性

题目思路：贪心，先给点的出度排序。

eg : 0 1 2 3
比如3这个点出度为2：先连1，再连2.如果不能连了则这是不可能存在的出度值。（每一个点只能向前面的点连，由于不存在回路）

以下是代码：

#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，这个0输出不输出都可以
        for (int i = 0; i < len; i++)
        {
            printf("%d %d\n",ans[i].first,ans[i].second);
        }
    }
}

------分隔线----------------------------

上一篇 考研为了什么

下一篇 SpringMVC视频学习笔记

------分隔线----------------------------