loj 1429(可相交的最小路径覆盖)

来源：互联网发布：dj香烟淘宝链接编辑：程序博客网时间：2024/06/03 15:31

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 }

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