[POJ3177]Redundant Paths（双连通图，割边，桥，重边）

题目链接：http://poj.org/problem?id=3177

和上一题一样，只是有重边。

如何解决重边的问题？

1、  构造图G时把重边也考虑进来，然后在划分边双连通分量时先把桥删去，再划分，其中桥的一端的割点归入当前正在划分的边双连通分量。这个处理比较麻烦；

2、  在输入图G的边时，若出现重边，则不把重边放入图G，然后在划分边双连通分量时依然用Low划分。

  1 /*
  2 ━━━━━┒ギリギリ♂ eye！
  3 ┓┏┓┏┓┃キリキリ♂ mind！
  4 ┛┗┛┗┛┃＼○／
  5 ┓┏┓┏┓┃ /
  6 ┛┗┛┗┛┃ノ)
  7 ┓┏┓┏┓┃
  8 ┛┗┛┗┛┃
  9 ┓┏┓┏┓┃
 10 ┛┗┛┗┛┃
 11 ┓┏┓┏┓┃
 12 ┛┗┛┗┛┃
 13 ┓┏┓┏┓┃
 14 ┃┃┃┃┃┃
 15 ┻┻┻┻┻┻
 16 */
 17 #include <algorithm>
 18 #include <iostream>
 19 #include <iomanip>
 20 #include <cstring>
 21 #include <climits>
 22 #include <complex>
 23 #include <fstream>
 24 #include <cassert>
 25 #include <cstdio>
 26 #include <bitset>
 27 #include <vector>
 28 #include <deque>
 29 #include <queue>
 30 #include <stack>
 31 #include <ctime>
 32 #include <set>
 33 #include <map>
 34 #include <cmath>
 35 using namespace std;
 36 #define fr first
 37 #define sc second
 38 #define cl clear
 39 #define BUG puts("here!!!")
 40 #define W(a) while(a--)
 41 #define pb(a) push_back(a)
 42 #define Rint(a) scanf("%d", &a)
 43 #define Rll(a) scanf("%lld", &a)
 44 #define Rs(a) scanf("%s", a)
 45 #define Cin(a) cin >> a
 46 #define FRead() freopen("in", "r", stdin)
 47 #define FWrite() freopen("out", "w", stdout)
 48 #define Rep(i, len) for(int i = 0; i < (len); i++)
 49 #define For(i, a, len) for(int i = (a); i < (len); i++)
 50 #define Cls(a) memset((a), 0, sizeof(a))
 51 #define Clr(a, x) memset((a), (x), sizeof(a))
 52 #define Full(a) memset((a), 0x7f7f, sizeof(a))
 53 #define lp p << 1
 54 #define rp p << 1 | 1
 55 #define pi 3.14159265359
 56 #define RT return
 57 typedef long long LL;
 58 typedef long double LD;
 59 typedef unsigned long long ULL;
 60 typedef pair<int, int> pii;
 61 typedef pair<string, int> psi;
 62 typedef map<string, int> msi;
 63 typedef vector<int> vi;
 64 typedef vector<LL> vl;
 65 typedef vector<vl> vvl;
 66 typedef vector<bool> vb;
 67 
 68 typedef struct Edge {
 69     int v;
 70     bool cut;
 71     Edge() {}
 72     Edge(int vv) : v(vv) { cut = 0; }
 73 }Edge;
 74 
 75 const int maxn = 5500;
 76 const int maxm = 555011;
 77 int n, m;
 78 int dig[maxn];
 79 int dfn[maxn], low[maxn], idx;
 80 vector<Edge> G[maxn];
 81 bool vis[maxn];
 82 int st[maxn], top;
 83 int belong[maxn], bcnt;
 84 
 85 void tarjan(int u, int p) {
 86     int v;
 87     low[u] = dfn[u] = ++idx;
 88     vis[u] = 1;
 89     st[top++] = u;
 90     Rep(i, G[u].size()) {
 91         v = G[u][i].v;
 92         if(v == p) continue;
 93         if(!dfn[v]) {
 94             tarjan(v, u);
 95             low[u] = min(low[u], low[v]);
 96             if(low[v] > dfn[u]) {
 97                 G[u][i].cut = 1;
 98                 Rep(j, G[v].size()) {
 99                     if(G[v][j].v== u) {
100                         G[v][j].cut = 1;
101                         break;
102                     }
103                 }
104             }
105         }
106         else if(vis[v]) low[u] = min(low[u], dfn[v]);
107     }
108     if(low[u] == dfn[u]) {
109         bcnt++;
110         do {
111             v = st[--top];
112             vis[v] = 0;
113             belong[v] = bcnt;
114         } while(v != u);
115     }
116 }
117 
118 int main() {
119     FRead();
120     int u, v;
121     while(~Rint(n) && ~Rint(m)) {
122         Rep(i, n+50) G[i].cl();
123         Cls(vis); Cls(dig); Cls(dfn); Cls(low);
124         top = 0; idx = 0; bcnt = 0;
125         Rep(i, m) {
126             Rint(u); Rint(v);
127             G[u].pb(Edge(v)); G[v].pb(Edge(u));
128         }
129         tarjan(1, 0);
130         int ret = 0;
131         For(u, 1, n+1) {
132             printf("%d ", belong[u]);
133             Rep(i, G[u].size()) {
134                 if(G[u][i].cut) {
135                     dig[belong[u]]++;
136                 }
137             }
138         }
139         printf("\n");
140         For(i, 1, bcnt+1) {
141             if(dig[i] == 1) ret++;
142         }
143         printf("%d\n", (ret+1)>>1);
144     }
145     RT 0;
146 }

 

posted @ 2016-05-23 16:10  Kirai  阅读(213)  评论(0)  编辑  收藏  举报