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最新推荐文章于 2020-09-23 12:02:18 发布

转载最新推荐文章于 2020-09-23 12:02:18 发布 · 170 阅读

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原文链接：http://www.cnblogs.com/naturepengchen/articles/4375110.html

本文回顾了一次编程比赛的经历，详细解析了两道题目，包括利用数学方法解决的A题和采用图论与优先队列策略解答的B题。

2015-03-29 00:12:33

总结：这场的状态和前两场一样糟糕... （话说自从bc发钱以来，参赛者实力高了一个档次...）

　　和前几场一样，第一题总是被卡... 不过后来沉下心耐心地搞掉了第二题...

A题：方法一：暴力打表 / 打表找规律。方法二：数学搞之，贴一下题解：

　　考虑期望的可加性。第

（1）规律/数学：

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <cstdlib>
 4 #include <cmath>
 5 #include <vector>
 6 #include <map>
 7 #include <set>
 8 #include <stack>
 9 #include <queue>
10 #include <string>
11 #include <iostream>
12 #include <algorithm>
13 using namespace std;
14 
15 #define MEM(a,b) memset(a,b,sizeof(a))
16 #define REP(i,n) for(int i=0;i<(n);++i)
17 #define FOR(i,a,b) for(int i=(a);i<=(b);++i)
18 #define getmid(l,r) ((l) + ((r) - (l)) / 2)
19 #define MP(a,b) make_pair(a,b)
20 
21 typedef long long ll;
22 typedef pair<int,int> pii;
23 const int INF = (1 << 30) - 1;
24 
25 int n,m;
26 
27 int Gcd(int a,int b){
28     while(a > 0 && b > 0){
29         if(a > b) a %= b;
30         else b %= a;
31     }
32     return a + b;
33 }
34 
35 int main(){
36     while(scanf("%d%d",&n,&m) != EOF){
37         int g = Gcd(n * m,n + m);
38         printf("%d/%d\n",n * m / g,(n + m) / g);
39     }
40     return 0;
41 }

View Code

（2）打表：

  1 #include <cstdio>
  2 #include <cstring>
  3 #include <cstdlib>
  4 #include <cmath>
  5 #include <vector>
  6 #include <map>
  7 #include <set>
  8 #include <stack>
  9 #include <queue>
 10 #include <string>
 11 #include <iostream>
 12 #include <algorithm>
 13 using namespace std;
 14 
 15 #define MEM(a,b) memset(a,b,sizeof(a))
 16 #define REP(i,n) for(int i=0;i<(n);++i)
 17 #define FOR(i,a,b) for(int i=(a);i<=(b);++i)
 18 #define getmid(l,r) ((l) + ((r) - (l)) / 2)
 19 #define MP(a,b) make_pair(a,b)
 20 
 21 typedef long long ll;
 22 typedef pair<int,int> pii;
 23 const int INF = (1 << 30) - 1;
 24 
 25 string ans[200] = {
 26 "",
 27 "1/2",
 28 "2/3",
 29 "3/4",
 30 "4/5",
 31 "5/6",
 32 "6/7",
 33 "7/8",
 34 "8/9",
 35 "9/10",
 36 "10/11",
 37 "11/12",
 38 "12/13",
 39 "2/3",
 40 "1/1",
 41 "6/5",
 42 "4/3",
 43 "10/7",
 44 "3/2",
 45 "14/9",
 46 "8/5",
 47 "18/11",
 48 "5/3",
 49 "22/13",
 50 "12/7",
 51 "3/4",
 52 "6/5",
 53 "3/2",
 54 "12/7",
 55 "15/8",
 56 "2/1",
 57 "21/10",
 58 "24/11",
 59 "9/4",
 60 "30/13",
 61 "33/14",
 62 "12/5",
 63 "4/5",
 64 "4/3",
 65 "12/7",
 66 "2/1",
 67 "20/9",
 68 "12/5",
 69 "28/11",
 70 "8/3",
 71 "36/13",
 72 "20/7",
 73 "44/15",
 74 "3/1",
 75 "5/6",
 76 "10/7",
 77 "15/8",
 78 "20/9",
 79 "5/2",
 80 "30/11",
 81 "35/12",
 82 "40/13",
 83 "45/14",
 84 "10/3",
 85 "55/16",
 86 "60/17",
 87 "6/7",
 88 "3/2",
 89 "2/1",
 90 "12/5",
 91 "30/11",
 92 "3/1",
 93 "42/13",
 94 "24/7",
 95 "18/5",
 96 "15/4",
 97 "66/17",
 98 "4/1",
 99 "7/8",
100 "14/9",
101 "21/10",
102 "28/11",
103 "35/12",
104 "42/13",
105 "7/2",
106 "56/15",
107 "63/16",
108 "70/17",
109 "77/18",
110 "84/19",
111 "8/9",
112 "8/5",
113 "24/11",
114 "8/3",
115 "40/13",
116 "24/7",
117 "56/15",
118 "4/1",
119 "72/17",
120 "40/9",
121 "88/19",
122 "24/5",
123 "9/10",
124 "18/11",
125 "9/4",
126 "36/13",
127 "45/14",
128 "18/5",
129 "63/16",
130 "72/17",
131 "9/2",
132 "90/19",
133 "99/20",
134 "36/7",
135 "10/11",
136 "5/3",
137 "30/13",
138 "20/7",
139 "10/3",
140 "15/4",
141 "70/17",
142 "40/9",
143 "90/19",
144 "5/1",
145 "110/21",
146 "60/11",
147 "11/12",
148 "22/13",
149 "33/14",
150 "44/15",
151 "55/16",
152 "66/17",
153 "77/18",
154 "88/19",
155 "99/20",
156 "110/21",
157 "11/2",
158 "132/23",
159 "12/13",
160 "12/7",
161 "12/5",
162 "3/1",
163 "60/17",
164 "4/1",
165 "84/19",
166 "24/5",
167 "36/7",
168 "60/11",
169 "132/23",
170 "6/1"
171 };
172 
173 int main(){
174     int n,m;
175     while(scanf("%d%d",&n,&m) != EOF){
176         cout << ans[(n - 1) * 12 + m] << endl;
177     }
178     return 0;
179 }

View Code

B题：按顺序不断地确定序列里的数，可以发现：每轮取数都是取入度<=k的最大的数。那么我们只要维护一个优先队列，把当前入度<=k的点全部加进来，然后取数，

　　注意取完数后k要被更新，表示剩下还能删多少条边，同时要根据取的数的出边去更新其他相关数的入度，如果其他数的入度更新完后<=k，那么入队。

　　这么做难以计算时间复杂度... bc并没有卡掉我的程序... Lucky。

　　题解给的是线段树+二分的做法... 就测试数据而言，两种做法的效率差不多。

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <cstdlib>
 4 #include <cmath>
 5 #include <vector>
 6 #include <map>
 7 #include <set>
 8 #include <stack>
 9 #include <queue>
10 #include <ctime>
11 #include <string>
12 #include <iostream>
13 #include <algorithm>
14 using namespace std;
15 
16 #define MEM(a,b) memset(a,b,sizeof(a))
17 #define REP(i,n) for(int i=0;i<(n);++i)
18 #define FOR(i,a,b) for(int i=(a);i<=(b);++i)
19 #define getmid(l,r) ((l) + ((r) - (l)) / 2)
20 #define MP(a,b) make_pair(a,b)
21 
22 typedef long long ll;
23 typedef pair<int,int> pii;
24 const int INF = (1 << 30) - 1;
25 const int MAXN = 200010;
26 
27 int n,m,k;
28 int first[MAXN],ecnt;
29 int deg[MAXN];
30 int vis[MAXN];
31 int ans[MAXN],sz;
32 
33 struct edge{
34     int v,next;
35 }e[MAXN * 2];
36 
37 void Add_edge(int u,int v){
38     e[++ecnt].next = first[u];
39     e[ecnt].v = v;
40     first[u] = ecnt;
41 }
42 
43 void Topo(){
44     sz = 0;
45     priority_queue<int> Q;
46     while(!Q.empty()) Q.pop();
47     for(int i = 1; i <= n; ++i) if(deg[i] <= k) Q.push(i);
48     while(!Q.empty()){
49         int cur = Q.top(); Q.pop();
50         if(!vis[cur] && deg[cur] <= k){
51             k -= deg[cur];
52             deg[cur] = 0;
53             vis[cur] = 1;
54         }
55         else continue;
56         ans[sz++] = cur;
57         for(int i = first[cur]; i != -1; i = e[i].next) if(!vis[e[i].v]){
58             int v = e[i].v;
59             deg[v]--;
60             if(deg[v] <= k) Q.push(v);
61         }
62     }
63 }
64 
65 int main(){
66     int a,b;
67     while(scanf("%d%d%d",&n,&m,&k) != EOF){
68         //clock_t st,ed;
69         //st = clock();
70         MEM(first,-1);
71         ecnt = 0;
72         MEM(deg,0);
73         MEM(vis,0);
74         REP(i,m){
75             scanf("%d%d",&a,&b);
76             Add_edge(a,b);
77             deg[b]++;
78         }
79         Topo();
80         //ed = clock();
81         printf("%d",ans[0]);
82         for(int i = 1; i < sz; ++i) printf(" %d",ans[i]);
83         puts("");
84         //printf("time : %.2f\n",(double)(ed - st) / CLOCKS_PER_SEC);
85     }
86     return 0;
87 }

View Code

转载于:https://www.cnblogs.com/naturepengchen/articles/4375110.html