loj 1429(可相交的最小路径覆盖)
题目链接：http://lightoj.com/volume_showproblem.php?problem=1429
思路：这道题还是比较麻烦的，对于求有向图的可相交的最小路径覆盖，首先要解决成环问题，可以先染色缩点重建图，然后就是如何来处理这个路径可以相交这个问题，这里可以用bfs求出任意两点之间是否可达，如果可达，就连边，然后就是HK算法求最大匹配了，最小路径覆盖 = 顶点数 - 最大匹配。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <vector> 6 #include <queue> 7 #include <stack> 8 using namespace std; 9 10 const int MAXN = (1000 + 10); 11 const int MAXM = (10000 + 10); 12 int n, m; 13 int cnt, scc_count; 14 bool Instack[MAXN]; 15 int low[MAXN], dfn[MAXN], color[MAXN]; 16 vector<int > g[MAXN]; 17 stack<int > S; 18 19 void Tarjan(int u) 20 { 21 low[u] = dfn[u] = ++cnt; 22 Instack[u] = true; 23 S.push(u); 24 for (int i = 0; i < (int)g[u].size(); i++) { 25 int v = g[u][i]; 26 if (dfn[v] == 0) { 27 Tarjan(v); 28 low[u] = min(low[u], low[v]); 29 } else if (Instack[v]) { 30 low[u] = min(low[u], dfn[v]); 31 } 32 } 33 if (low[u] == dfn[u]) { 34 scc_count++; 35 int v; 36 do { 37 v = S.top(); 38 S.pop(); 39 Instack[v] = false; 40 color[v] = scc_count; 41 } while (u != v); 42 } 43 } 44 45 bool Isok[MAXN][MAXN]; 46 bool mark[MAXN]; 47 vector<int > reg[MAXN]; 48 49 void bfs(int st) 50 { 51 memset(mark, false, sizeof(mark)); 52 queue<int >que; 53 que.push(st); 54 mark[st] = true; 55 while (!que.empty()) { 56 int u = que.front(); 57 que.pop(); 58 for (int i = 0; i < (int)reg[u].size(); i++) { 59 int v = reg[u][i]; 60 if (!mark[v]) { 61 mark[v] = true; 62 que.push(v); 63 } 64 } 65 } 66 } 67 68 void Build() 69 { 70 for (int i = 1; i <= scc_count; i++) { 71 reg[i].clear(); 72 } 73 for (int i = 1; i <= scc_count; i++) { 74 for (int j = 1; j <= scc_count; j++) { 75 if (i != j && Isok[i][j]) { 76 reg[i].push_back(j); 77 } 78 } 79 } 80 } 81 82 int lx[MAXN], ly[MAXN]; 83 int distx[MAXN], disty[MAXN]; 84 85 bool MaxMatch_bfs() 86 { 87 bool flag = false; 88 memset(distx, 0, sizeof(distx)); 89 memset(disty, 0, sizeof(disty)); 90 queue<int > que; 91 for (int i = 1; i <= scc_count; i++) { 92 if (lx[i] == -1) que.push(i); 93 } 94 while (!que.empty()) { 95 int u = que.front(); 96 que.pop(); 97 for (int i = 0; i < (int)reg[u].size(); i++) { 98 int v = reg[u][i]; 99 if (disty[v] == 0) { 100 disty[v] = distx[u] + 1; 101 if (ly[v] == -1) flag = true; 102 else { 103 distx[ly[v]] = disty[v] + 1; 104 que.push(ly[v]); 105 } 106 } 107 } 108 } 109 return flag; 110 } 111 112 int dfs(int u) 113 { 114 for (int i = 0; i < (int)reg[u].size(); i++) { 115 int v = reg[u][i]; 116 if (disty[v] == distx[u] + 1) { 117 disty[v] = 0; 118 if (ly[v] == -1 || dfs(ly[v])) { 119 ly[v] = u; 120 lx[u] = v; 121 return 1; 122 } 123 } 124 } 125 return 0; 126 } 127 128 int MaxMatch() 129 { 130 memset(lx, -1, sizeof(lx)); 131 memset(ly, -1, sizeof(ly)); 132 int res = 0; 133 while (MaxMatch_bfs()) { 134 for (int i = 1; i <= scc_count; i++) { 135 if (lx[i] == -1) res += dfs(i); 136 } 137 } 138 return res; 139 } 140 141 142 int main() 143 { 144 int _case, t = 1; 145 scanf("%d", &_case); 146 while (_case--) { 147 scanf("%d %d", &n, &m); 148 for (int i = 1; i <= n; i++) { 149 g[i].clear(); 150 reg[i].clear(); 151 } 152 while (m--) { 153 int u, v; 154 scanf("%d %d", &u, &v); 155 g[u].push_back(v); 156 } 157 //强联通缩点重建图 158 cnt = scc_count = 0; 159 memset(dfn, 0, sizeof(dfn)); 160 for (int i = 1; i <= n; i++) { 161 if (dfn[i] == 0) Tarjan(i); 162 } 163 for (int u = 1; u <= n; u++) { 164 for (int i = 0; i < (int)g[u].size(); i++) { 165 int v = g[u][i]; 166 if (color[u] != color[v]) { 167 reg[color[u]].push_back(color[v]); 168 } 169 } 170 } 171 //bfs求出新图中的任意两点之间是否可达 172 memset(Isok, false, sizeof(Isok)); 173 for (int i = 1; i <= scc_count; i++) { 174 bfs(i); 175 for (int j = 1; j <= scc_count; j++) { 176 if (mark[j]) { 177 Isok[i][j] = true; 178 } 179 } 180 } 181 //对于那些可达的点重新连边 182 Build(); 183 //bfs求解最大匹配； 184 //最小路径覆盖 = 顶点数 - 最大匹配数 185 int ans = MaxMatch(); 186 printf("Case %d: %d\n", t++, scc_count- ans); 187 } 188 return 0; 189 }