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HDU 4035 - Maze

时间:2014-11-06 23:32:09      阅读:247      评论:0      收藏:0      [点我收藏+]

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  1 /*
  2 ID:esxgx1
  3 LANG:C++
  4 PROG:hdu4035
  5 */
  6 #include <cstdio>
  7 #include <cstring>
  8 #include <iostream>
  9 #include <algorithm>
 10 using namespace std;
 11 
 12 template<int maxn, int maxe>
 13 class graph {
 14     int fw[maxe], t[maxe];
 15     int h[maxn], p;
 16 
 17 public:
 18     graph() :p(0) {memset(h, -1, sizeof h);}
 19     void clear() {p = 0, memset(h, -1, sizeof h);}
 20     
 21     void addedge(int s, int _t) {
 22         fw[p] = h[s], t[p] = _t;
 23         h[s] = p++;
 24     }
 25     
 26     int begin(int s) {return h[s];}
 27     int for_each(int &e, int &_t) {
 28         int ret = e;
 29         if (~e) {
 30             _t = t[e];
 31             e = fw[e];    
 32         }
 33         return ret;
 34     } 
 35     
 36 };
 37 
 38 typedef double DB;
 39 const int maxn = 10007;
 40 graph <maxn, maxn*2> g;
 41 
 42 int adjn[maxn];
 43 int K[maxn], E[maxn];
 44 
 45 DB A[maxn], B[maxn], C[maxn];
 46 
 47 #define eps        1e-9
 48 
 49 int dfs(int s, int fa)
 50 {
 51     DB bfa = (1.0 - (K[s] + E[s])/100.0) / adjn[s];
 52     C[s] = 1.0 - (E[s]+K[s])/100.0, A[s] = K[s]/100.0, B[s] = bfa;
 53     DB div = 1.0;
 54     for(int e = g.begin(s), t; ~g.for_each(e, t); ) {
 55         if (t != fa) {
 56             dfs(t, s);
 57             A[s] += bfa * A[t], div -= bfa * B[t],
 58             C[s] += bfa * C[t];
 59         }
 60     }
 61     if (div < eps) return 0;
 62     A[s] /= div; B[s] /= div; C[s] /= div;
 63     return 1;
 64 }
 65 
 66 int dfs0(int s, DB &ans)
 67 {
 68     DB bfa = (1.0 - (K[s] + E[s])/100.0) / adjn[s];
 69     DB div = 1.0;
 70     C[s] = 1.0 - (E[s]+K[s])/100.0;
 71     for(int e = g.begin(s), t; ~g.for_each(e, t); ) {
 72         if (!dfs(t, s)) return 0;
 73         div -= bfa * (A[t] + B[t]),
 74         C[s] += bfa * C[t];
 75     }
 76     if (div < eps) return 0;
 77     ans = C[s] / div;
 78     return 1;
 79 }
 80 
 81 int main(void)
 82 {
 83     #ifndef ONLINE_JUDGE
 84     freopen("in.txt", "r", stdin);
 85     #endif
 86     int T;
 87     scanf("%d", &T);
 88     for(int t = 1; t<=T; ++t, g.clear()) {
 89         memset(adjn, 0, sizeof adjn);
 90         int N;
 91         scanf("%d", &N);
 92         for(int i=1; i<N; ++i) {
 93             int s, t;
 94             scanf("%d%d", &s, &t); --s, --t;
 95             g.addedge(s, t);
 96             g.addedge(t, s);
 97             ++adjn[s], ++adjn[t];
 98         }
 99         for(int i=0; i<N; ++i)
100             scanf("%d%d", &K[i], &E[i]);
101         
102         double ans;
103         if (dfs0(0, ans)) printf("Case %d: %.6f\n", t, ans);
104         else printf("Case %d: impossible\n", t);
105     }
106     return 0;
107 }

 

2014-11-06 22:24:38 Accepted 4035 203MS 1308K 2056 B G++

 

 

===========================================

再简短一点:

 1 /*
 2 ID:esxgx1
 3 LANG:C++
 4 PROG:hdu4035
 5 */
 6 #include <cstdio>
 7 #include <cstring>
 8 #include <iostream>
 9 #include <algorithm>
10 using namespace std;
11 
12 template<int maxn, int maxe>
13 class graph {
14     int fw[maxe], t[maxe];
15     int h[maxn], p;
16 
17 public:
18     graph() :p(0) {memset(h, -1, sizeof h);}
19     void clear() {p = 0, memset(h, -1, sizeof h);}
20     
21     void addedge(int s, int _t) {
22         fw[p] = h[s], t[p] = _t;
23         h[s] = p++;
24     }
25     
26     int begin(int s) {return h[s];}
27     int for_each(int &e, int &_t) {
28         int ret = e;
29         if (~e) {
30             _t = t[e];
31             e = fw[e];    
32         }
33         return ret;
34     } 
35     
36 };
37 
38 typedef double DB;
39 const int maxn = 10007;
40 graph <maxn, maxn*2> g;
41 
42 int adjn[maxn];
43 int K[maxn], E[maxn];
44 
45 DB A[maxn], B[maxn], C[maxn];
46 
47 #define eps        1e-9
48 
49 int dfs(int s, int fa)
50 {
51     DB bfa = (1.0 - (K[s] + E[s])/100.0) / adjn[s];
52     C[s] = 1.0 - (E[s]+K[s])/100.0, A[s] = K[s]/100.0, B[s] = bfa;
53     DB div = 1.0;
54     for(int e = g.begin(s), t; ~g.for_each(e, t); ) {
55         if (t != fa) {
56             dfs(t, s);
57             A[s] += bfa * A[t], div -= bfa * B[t],
58             C[s] += bfa * C[t];
59         }
60     }
61     if (div < eps) return 0;
62     A[s] /= div; B[s] /= div; C[s] /= div;
63     return 1;
64 }
65 
66 int main(void)
67 {
68     #ifndef ONLINE_JUDGE
69     freopen("in.txt", "r", stdin);
70     #endif
71     int T;
72     scanf("%d", &T);
73     for(int t = 1; t<=T; ++t, g.clear()) {
74         memset(adjn, 0, sizeof adjn);
75         int N;
76         scanf("%d", &N);
77         for(int i=1; i<N; ++i) {
78             int s, t;
79             scanf("%d%d", &s, &t); --s, --t;
80             g.addedge(s, t);
81             g.addedge(t, s);
82             ++adjn[s], ++adjn[t];
83         }
84         for(int i=0; i<N; ++i)
85             scanf("%d%d", &K[i], &E[i]);
86 
87         if (dfs(0, -1) && 1-A[0] >= eps)
88             printf("Case %d: %.6f\n", t, C[0] / (1-A[0]));    
89         else printf("Case %d: impossible\n", t);
90     }
91     return 0;
92 }

 

HDU 4035 - Maze

标签:style   blog   http   io   color   ar   os   sp   for   

原文地址:http://www.cnblogs.com/e0e1e/p/hdu_4035.html

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