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此题关键在于维护点的连通性以及连通块的信息,容易想到并查集,但是并查集却不支持删边操作,于是考虑逆序处理,这样删边就变成了加边操作,每一个连通块的信息可以用stl中的multiset来维护,注意集合合并的时候要启发式合并(这里是按照集合的大小来合并,每次小的集合合并到大的集合里),不然会超时。
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 #include <set> 5 using namespace std; 6 7 const int N = 20001; 8 const int M = 300001; 9 int f[N]; 10 int v[N]; 11 multiset<int> s[N]; 12 multiset<int> edge[N]; 13 int n, m, q; 14 15 struct Op 16 { 17 char cmd[2]; 18 int x, y; 19 } op[M]; 20 21 int findf( int x ) 22 { 23 if ( f[x] != x ) f[x] = findf( f[x] ); 24 return f[x]; 25 } 26 27 void union_set( int x, int y ) 28 { 29 x = findf(x), y = findf(y); 30 if ( x != y ) 31 { 32 if ( s[x].size() > s[y].size() ) swap( x, y ); 33 f[x] = y; 34 for ( multiset<int>::iterator it = s[x].begin(); it != s[x].end(); it++ ) 35 { 36 s[y].insert( *it ); 37 } 38 s[x].clear(); 39 } 40 } 41 42 int main () 43 { 44 int _case = 1; 45 while ( scanf("%d%d%d", &n, &m, &q) != EOF ) 46 { 47 for ( int i = 1; i <= n; i++ ) 48 { 49 scanf("%d", &v[i]); 50 edge[i].clear(); 51 } 52 for ( int i = 1; i <= m; i++ ) 53 { 54 int u, v; 55 scanf("%d%d", &u, &v); 56 if ( u > v ) swap( u, v ); 57 edge[u].insert(v); 58 } 59 for ( int i = 1; i <= q; i++ ) 60 { 61 scanf("%s%d%d", &op[i].cmd, &op[i].x, &op[i].y); 62 if ( op[i].cmd[0] == ‘U‘ ) 63 { 64 int tn = op[i].y; 65 op[i].y = v[op[i].x]; 66 v[op[i].x] = tn; 67 } 68 else if ( op[i].cmd[0] == ‘E‘ ) 69 { 70 if ( op[i].x > op[i].y ) swap( op[i].x, op[i].y ); 71 edge[op[i].x].erase( edge[op[i].x].find(op[i].y) ); 72 } 73 } 74 for ( int i = 1; i <= n; i++ ) 75 { 76 f[i] = i; 77 s[i].clear(); 78 s[i].insert(v[i]); 79 } 80 for ( int i = 1; i <= n; i++ ) 81 { 82 for ( multiset<int>::iterator it = edge[i].begin(); it != edge[i].end(); it++ ) 83 { 84 union_set( i, (*it) ); 85 } 86 } 87 int cnt = 0, sum = 0; 88 for ( int i = q; i >= 1; i-- ) 89 { 90 if ( op[i].cmd[0] == ‘F‘ ) 91 { 92 cnt++; 93 int x = op[i].x, y = op[i].y; 94 x = findf(x); 95 multiset<int>::iterator it = s[x].lower_bound(y); 96 if ( it != s[x].end() ) 97 { 98 sum += (*it); 99 } 100 } 101 else if ( op[i].cmd[0] == ‘U‘ ) 102 { 103 int x = op[i].x, y = op[i].y; 104 int fx = findf(x); 105 multiset<int>::iterator it = s[fx].find(v[x]); 106 s[fx].erase(it); 107 s[fx].insert(y); 108 v[x] = y; 109 } 110 else 111 { 112 union_set( op[i].x, op[i].y ); 113 } 114 } 115 double avg = ( sum * 1.0 ) / ( cnt * 1.0 ); 116 printf("Case %d: %.3f\n", _case++, avg); 117 } 118 return 0; 119 }
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原文地址:http://www.cnblogs.com/huoxiayu/p/4777607.html