2016HDU多校联赛-HDU-5813-Elegant Construction(构造联通图)
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Elegant Construction
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 301 Accepted Submission(s): 158
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
题意:第一行给出T,代表有T组数据,每组数据首先给出N,代表有1~N个点,接着一行给出N个数,第i个数代表第i个点能到达Ai个点,直接间接都行。
当时不会做,后来看题解又补的。
官方题解:将顶点按能到达的点数从小到大排序,排好序之后每个点只能往前面的点连边. 因而如果存在一个排在第i位的点,要求到达的点数大于i-1,则不可行;否则就可以按照上述方法构造出图. 复杂度O(N^2).
代码
#include<iostream>#include<algorithm>#include<stdio.h>#include<string.h>using namespace std;const int maxn=1005;struct node{ int num; int point;} num[maxn];bool cmp(node x,node y){ return x.num<y.num;}int main(){ int T; scanf("%d",&T); for(int casen=1; casen<=T; casen++) { int N; scanf("%d",&N); bool flag=0;//满足条件标记为1,否则为0 for(int i=0; i<N; i++) { scanf("%d",&num[i].num); num[i].point=i+1; if(num[i].num==0) flag=1; } sort(num,num+N,cmp); int sum=0;//边的数量 for(int i=0; i<N; i++) { if(num[i].num>=i+1)//存在一个排在第i位的点,要求到达的点数大于i-1,则不可行 {// printf("***\n");// printf("%d %d\n",num[i].num,num[i].point); flag=0; break; } sum+=num[i].num; } if(flag==0) { printf("Case #%d: No\n",casen); continue; } printf("Case #%d: Yes\n%d\n",casen,sum); for(int i=0; i<N; i++) for(int j=0; j<num[i].num; j++) printf("%d %d\n",num[i].point,num[j].point); } return 0;}
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