Cerc 2012 J (topological sorting)

Some nodes with topological order need to be colored. These nodes are divided into two types. Ask how many switches are required to dye the end point in the given topological order.

Idea: we set up two stacks to store two different types of nodes according to the topological sorting method. Each time we finish processing all the nodes in one stack, we can process the other, in this way, the number of switches is the final answer.

The Code is as follows:

  1 /**************************************************  2  * Author     : xiaohao Z  3  * Blog     : http://www.cnblogs.com/shu-xiaohao/  4  * Last modified : 2014-06-30 12:30  5  * Filename     : cerc_j.cpp  6  * Description     :   7  * ************************************************/  8   9 #include <iostream> 10 #include <cstdio> 11 #include <cstring> 12 #include <cstdlib> 13 #include <cmath> 14 #include <algorithm> 15 #include <queue> 16 #include <stack> 17 #include <vector> 18 #include <set> 19 #include <map> 20 #define MP(a, b) make_pair(a, b) 21 #define PB(a) push_back(a) 22  23 using namespace std; 24 typedef long long ll; 25 typedef pair<int, int> pii; 26 typedef pair<unsigned int,unsigned int> puu; 27 typedef pair<int, double> pid; 28 typedef pair<ll, int> pli; 29 typedef pair<int, ll> pil; 30  31 const int INF = 0x3f3f3f3f; 32 const double eps = 1E-6; 33 const int LEN = 100000+10; 34 int n, m, vex[LEN], ind[LEN], tmp_ind[LEN]; 35 bool vis[LEN]; 36 struct E{ 37     int to, next; 38 }edge[LEN*10]; 39 int head[LEN], top; 40  41 void init(){ 42     top = 0; 43     memset(head, -1, sizeof head); 44 } 45  46 void addedge(int u, int v){ 47     edge[top].next = head[u]; 48     edge[top].to = v; 49     head[u] = top++; 50 } 51  52 int solve(int flag){ 53     for(int i=0; i<n; i++) 54         ind[i] = tmp_ind[i]; 55     memset(vis, 0, sizeof vis); 56     queue<int> q1, q0; 57     for(int i=0; i<n; i++){ 58         if(!ind[i]){ 59             vis[i] = 1; 60             if(!vex[i]) q0.push(i); 61             else q1.push(i); 62         } 63     } 64     int ret = 0, cnt = 0; 65     for(;;ret++){ 66         if(flag){ 67             if(q1.empty()) { 68                 ret--; break; 69             } 70             while(!q1.empty()){ 71                 int nv = q1.front(); q1.pop(); 72                 cnt ++; 73                 for(int i=head[nv]; i!=-1; i=edge[i].next){ 74                     int nx = edge[i].to; 75                     ind[nx] --; 76                     if(!ind[nx] && !vis[nx]){ 77                         vis[nx] = 1; 78                         if(vex[nx]) q1.push(nx); 79                         else q0.push(nx); 80                     } 81                 } 82             } 83         }else{ 84             if(q0.empty()) { 85                 ret--; break; 86             } 87             while(!q0.empty()){ 88                 int nv = q0.front(); q0.pop(); 89                 cnt ++; 90                 for(int i=head[nv]; i!=-1; i=edge[i].next){ 91                     int nx = edge[i].to; 92                     ind[nx] --; 93                     if(!ind[nx] && !vis[nx]){ 94                         vis[nx] = 1; 95                         if(vex[nx]) q1.push(nx); 96                         else q0.push(nx); 97                     } 98                 } 99             }100         }101         flag = !flag;102     }103     if(cnt != n) return INF;104     return ret;105 }106 107 int main()108 {109 //    freopen("in.txt", "r", stdin);110 111     int a, b, T;112     scanf("%d", &T);113     while(T--){114         init();115         memset(tmp_ind, 0, sizeof tmp_ind);116         scanf("%d%d", &n, &m);117         for(int i=0; i<n; i++){118             scanf("%d", &vex[i]);119             vex[i] --;120         }121         for(int i=0; i<m; i++){122             scanf("%d%d", &a, &b);123             a--, b--;124             addedge(a, b);125             tmp_ind[b] ++;126         }127         int ans = min(solve(0), solve(1));128         printf("%d\n", ans);129     }130     return 0;131 }

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