codeforces 404C - Restore Graph
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题目链接:http://codeforces.com/problemset/problem/404/C
题目大意:n个点,每个点上最多有m个点与其相连,某点到所有点的最短路,判断是否能还原这张图,如果可以输出相连的边。
题目分析:首先需要找到原点,如果一个点的距离是0的话,则是起点,但是肯定只有一个。然后判断一下其余距离的点是不是都是在[1,k]范围就行了。
代码参考:
#include<map>#include<set>#include<cmath>#include<queue>#include<stack>#include<cstdio>#include<vector>#include<string>#include<cstring>#include<sstream>#include<iostream>#include<algorithm>#include<functional>using namespace std;typedef long long LL;const int N = 1 << 17;vector<int> v[N];int main(){int n, m, i, j, k, x, y, z, r;while (~scanf("%d%d", &n, &k)){for (i = 0; i < N; ++i)//初始化{v[i].clear();}int M = -1;//求最远的距离for (i = 1; i <= n; ++i){scanf("%d", &x);v[x].push_back(i);//保存第i个点的距离为xM = max(M, x);}if (v[0].size() != 1 || v[1].size() > k)//如果起点不唯一或者和起点的点数大于k{puts("-1");continue;}bool f = true;for (i = 2; i <= M; ++i)//枚举每个距离{if (v[i - 1].size() * (k - 1LL) < v[i].size())//每个点与前面的点相连之后还剩下k-1个位置可以和后面的点相连{//v[i - 1].size() * (k - 1LL)可能超出int范围puts("-1");f = false;break;}}if (f){printf("%d\n", n - 1);//最小生成树由n-1条边连接for (i = 0; i < M; ++i)//枚举每个距离{int cnt = 1;//统计已经和几条边相连j = r = 0;while (j < v[i].size() && r < v[i + 1].size()){if (i > 0 && cnt < k){printf("%d %d\n", v[i][j], v[i + 1][r]);cnt++;r++;}else if (i == 0 && cnt <= k)//i==0时特判{printf("%d %d\n", v[i][j], v[i + 1][r]);cnt++;r++;}else{j++;//如果超过k了,就将剩下的点接在另外的等价的点上cnt = 1;}}}}}return 0;}
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