题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4786
先跑一遍最小生成树，注意判断是否已全部联通（用一个记号来统计最后生成树中有多少条边）。再记下最小生成树的权值和A。
再反向排序，求一遍最大生成树。记下权值和B。问题转换成求[A,B]内是否有斐波那契数存在。
1 #include <algorithm> 2 #include <iostream> 3 #include <iomanip> 4 #include <cstring> 5 #include <climits> 6 #include <complex> 7 #include <fstream> 8 #include <cassert> 9 #include <cstdio> 10 #include <bitset> 11 #include <vector> 12 #include <deque> 13 #include <queue> 14 #include <stack> 15 #include <ctime> 16 #include <set> 17 #include <map> 18 #include <cmath> 19 20 using namespace std; 21 22 typedef struct P { 23 int u; 24 int v; 25 int w; 26 }P; 27 28 bool cmp1(P p1, P p2) {return p1.w < p2.w;} 29 bool cmp2(P p1, P p2) {return p1.w > p2.w;} 30 int A, B; 31 32 const int maxn = 100010; 33 int n, m, cnt; 34 int pre[maxn]; 35 int fib[maxn]; 36 P e[maxn]; 37 38 int find(int x) { 39 return x == pre[x] ? pre[x] : pre[x] = find(pre[x]); 40 } 41 42 bool unite(int x, int y) { 43 x = find(x); 44 y = find(y); 45 if(x != y) { 46 pre[x] = y; 47 return 1; 48 } 49 return 0; 50 } 51 52 void init() { 53 for(int i = 0; i < maxn; i++) { 54 pre[i] = i; 55 } 56 } 57 58 void clear() { 59 init(); 60 memset(e, 0, sizeof(e)); 61 A = 0, B = 0; 62 cnt = 0; 63 } 64 65 int main() { 66 // freopen("in", "r", stdin); 67 int T; 68 fib[0] = 0, fib[1] = 1, fib[2] = 1; 69 for(int i = 3; i <= 26; i++) { 70 fib[i] = fib[i-1] + fib[i-2]; 71 } 72 scanf("%d", &T); 73 for(int _ = 1; _ <= T; _++) { 74 printf("Case #%d: ", _); 75 clear(); 76 scanf("%d %d", &n, &m); 77 for(int i = 0; i < m; i++) { 78 scanf("%d %d %d", &e[i].u, &e[i].v, &e[i].w); 79 } 80 sort(e, e+m, cmp1); 81 for(int i = 0; i < m; i++) { 82 if(unite(e[i].u, e[i].v)) { 83 A += e[i].w; 84 cnt++; 85 } 86 } 87 if(cnt != n - 1) printf("No "); 88 else { 89 init(); 90 sort(e, e+m, cmp2); 91 for(int i = 0; i < m; i++) { 92 if(unite(e[i].u, e[i].v)) { 93 B += e[i].w; 94 cnt++; 95 } 96 } 97 int flag = 0; 98 for(int i = 1; i <= 25; i++) { 99 if(fib[i] <= B && fib[i] >= A) { 100 flag = 1; 101 break; 102 } 103 } 104 if(flag) printf("Yes "); 105 else printf("No "); 106 } 107 } 108 }
还排序个毛啊，直接反着跑一遍QAQ，反正有序的，二分查找一下不好吗。。。肥不拉几序列只要前25个，打一个小表就行了QAQ。。
1 #include <algorithm> 2 #include <iostream> 3 #include <iomanip> 4 #include <cstring> 5 #include <climits> 6 #include <complex> 7 #include <fstream> 8 #include <cassert> 9 #include <cstdio> 10 #include <bitset> 11 #include <vector> 12 #include <deque> 13 #include <queue> 14 #include <stack> 15 #include <ctime> 16 #include <set> 17 #include <map> 18 #include <cmath> 19 20 using namespace std; 21 22 typedef struct P { 23 int u; 24 int v; 25 int w; 26 }P; 27 28 bool cmp1(P p1, P p2) {return p1.w < p2.w;} 29 bool cmp2(P p1, P p2) {return p1.w > p2.w;} 30 int A, B; 31 32 const int maxn = 100010; 33 int n, m, cnt; 34 int pre[maxn]; 35 int fib[maxn]; 36 P e[maxn]; 37 38 int find(int x) { 39 return x == pre[x] ? pre[x] : pre[x] = find(pre[x]); 40 } 41 42 bool unite(int x, int y) { 43 x = find(x); 44 y = find(y); 45 if(x != y) { 46 pre[x] = y; 47 return 1; 48 } 49 return 0; 50 } 51 52 void init() { 53 for(int i = 0; i < maxn; i++) { 54 pre[i] = i; 55 } 56 } 57 58 void clear() { 59 init(); 60 memset(e, 0, sizeof(e)); 61 A = 0, B = 0; 62 cnt = 0; 63 } 64 65 void __FIB__() { 66 fib[0]=0,fib[1]=1,fib[2]=1,fib[3]=2,fib[4]=3,fib[5]=5, 67 fib[6]=8,fib[7]=13,fib[8]=21,fib[9]=34,fib[10]=55, 68 fib[11]=89,fib[12]=144,fib[13]=233,fib[14]=377, 69 fib[15]=610,fib[16]=987,fib[17]=1597,fib[18]=2584, 70 fib[19]=4181,fib[20]=6765,fib[21]=10946,fib[22]=17711, 71 fib[23]=28657,fib[24]=46368,fib[25]=75025; 72 } 73 74 int main() { 75 // freopen("in", "r", stdin); 76 // freopen("out", "w", stdout); 77 int T; 78 __FIB__(); 79 scanf("%d", &T); 80 for(int _ = 1; _ <= T; _++) { 81 printf("Case #%d: ", _); 82 clear(); 83 scanf("%d %d", &n, &m); 84 for(int i = 0; i < m; i++) { 85 scanf("%d %d %d", &e[i].u, &e[i].v, &e[i].w); 86 } 87 sort(e, e+m, cmp1); 88 for(int i = 0; i < m; i++) { 89 if(unite(e[i].u, e[i].v)) { 90 A += e[i].w; 91 cnt++; 92 } 93 } 94 if(cnt != n - 1) printf("No "); 95 else { 96 init(); 97 for(int i = m; i >= 0; i--) { 98 if(unite(e[i].u, e[i].v)) { 99 B += e[i].w; 100 cnt++; 101 } 102 } 103 int flag = 0; 104 int ll = 1, rr = 25; 105 while(ll < rr) { 106 int mm = (ll + rr) >> 1; 107 if(fib[mm] <= B && fib[mm] >= A) { 108 flag = 1; 109 break; 110 } 111 else if(fib[mm] > B) rr = mm; 112 else if(fib[mm] < A) ll = mm + 1; 113 } 114 if(flag) printf("Yes "); 115 else printf("No "); 116 } 117 } 118 }