CodeForces-1082D Maximum Diameter Graph

Topic links: codeforces the Maximum-1082D a Diameter Graph

The meaning of problems

There are n points, the i-th point of maximum degree $ $ a_i, seeking to make the composition of FIG communication scheme largest diameter.

---

Thinking

Determining whether a connected graph can be constructed, if possible, all points greater than 1 degree together into one line, and the rest of it is even possible at the leftmost and rightmost points, and even then can easily, so it must be greedy optimal.

---

 Implementation code

 1 #include <cstdio>
 2 const int maxn = 510;
 3 struct Node
 4 {
 5     int deg, idx;
 6 } no[maxn];
 7 struct Edge
 8 {
 9     int u, v;
10 } ed[maxn];
11 int no1[maxn];
12 
13 int main() {
14     int n;
15     while (~scanf("%d", &n)) {
16         int cnt = 0, cnt1 = 0, tot = 0, sum = 0, ai;
17         for (int i = 1; i <= n; i++) {
18             scanf("%d", &ai);
19             sum += ai;
20             if (ai > 1) no[cnt++] = {ai, i};
21             else no1[cnt1++] = i;
22         }
23         if (sum < (n - 1) * 2) {
24             puts("NO");
25             continue;
26         }
27         int ans1 = 0, ans2 = 0;
28         for (int i = 0; i < cnt - 1; i++) {
29             ed[ans2++] = {no[i].idx, no[i+1].idx};
30             no[i].deg--, no[i+1].deg--;
31             ans1++;
32         }
33         if (cnt1) {
34             d [ANS2 ++] = {No1 [cnt1- 1 ], in [ 0 ]} .idx;
35              in [ 0 ] .deg-- ;
36              ANS1 ++ ;
37              cnt1-- ;
38          }
 39          if (cm -1) {
 40              d [ANS2 ++] = {No1 [cnt1- 1 ], in [cm -1 ]} .idx;
41              in [cnt- 1 ] .deg-- ;
42              ANS1 ++ ;
43              cnt1-- ;
44          }
 45          while (cnt1) {
46             for (int i = 0; i < cnt; i++) {
47                 if (no[i].deg) {
48                     ed[ans2++] = {no[i].idx, no1[cnt1-1]};
49                     no[i].deg--;
50                     cnt1--;
51                 }
52                 if (cnt1 == 0) break;
53             }
54         }
55         printf("YES %d\n", ans1);
56         printf("%d\n", ans2);
57         for (int i = 0; i < ans2; i++) printf("%d %d\n", ed[i].u, ed[i].v);
58     }
59     return 0;
60 }

View Code