POJ 1144 Tarjan 割点 解题报告
2017-10-13 14:40
405 查看
Network
Description
A Telephone Line Company (TLC) is establishing a new telephone cable network. They are connecting several places numbered by integers from 1 to N . No two places have the same number. The lines are bidirectional and always connect together two places and in each place the lines end in a telephone exchange. There is one telephone exchange in each place. From each place it is
possible to reach through lines every other place, however it need not be a direct connection, it can go through several exchanges. From time to time the power supply fails at a place and then the exchange does not operate. The officials from TLC realized that in such a case it can happen that besides the fact that the place with the failure is unreachable, this can also cause that some other places cannot connect to each other. In such a case we will say the place (where the failure
occured) is critical. Now the officials are trying to write a program for finding the number of all such critical places. Help them.
Input
The input file consists of several blocks of lines. Each block describes one network. In the first line of each block there is the number of places N < 100. Each of the next at most N lines contains the number of a place followed by the numbers of some places to which there is a direct line from this place. These at most N lines completely describe the network, i.e., each direct connection of two places in the network is contained at least in one row. All numbers in one line are separated
by one space. Each block ends with a line containing just 0. The last block has only one line with N = 0;
Output
The output contains for each block except the last in the input file one line containing the number of critical places.
Sample Input
5
5 1 2 3 4
0
6
2 1 3
5 4 6 2
0
0
Sample Output
1
2
【解题报告】
代码如下:
Description
A Telephone Line Company (TLC) is establishing a new telephone cable network. They are connecting several places numbered by integers from 1 to N . No two places have the same number. The lines are bidirectional and always connect together two places and in each place the lines end in a telephone exchange. There is one telephone exchange in each place. From each place it is
possible to reach through lines every other place, however it need not be a direct connection, it can go through several exchanges. From time to time the power supply fails at a place and then the exchange does not operate. The officials from TLC realized that in such a case it can happen that besides the fact that the place with the failure is unreachable, this can also cause that some other places cannot connect to each other. In such a case we will say the place (where the failure
occured) is critical. Now the officials are trying to write a program for finding the number of all such critical places. Help them.
Input
The input file consists of several blocks of lines. Each block describes one network. In the first line of each block there is the number of places N < 100. Each of the next at most N lines contains the number of a place followed by the numbers of some places to which there is a direct line from this place. These at most N lines completely describe the network, i.e., each direct connection of two places in the network is contained at least in one row. All numbers in one line are separated
by one space. Each block ends with a line containing just 0. The last block has only one line with N = 0;
Output
The output contains for each block except the last in the input file one line containing the number of critical places.
Sample Input
5
5 1 2 3 4
0
6
2 1 3
5 4 6 2
0
0
Sample Output
1
2
【解题报告】
代码如下:
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; #define N 110 int n,cnt,head ; struct Edge{int to,nxt;}e[N*N]; int dfn ,low ,vis ,cut ; int indexx,root,ans; void init() { cnt=-1;indexx=root=ans=0; memset(head,-1,sizeof(head)); memset(dfn,-1,sizeof(dfn)); memset(low,-1,sizeof(low)); memset(vis,0,sizeof(vis)); memset(cut,0,sizeof(cut)); } void adde(int u,int v) { e[++cnt].to=v;e[cnt].nxt=head[u];head[u]=cnt; e[++cnt].to=u;e[cnt].nxt=head[v];head[v]=cnt; } void tarjan(int u) { dfn[u]=low[u]=++indexx; vis[u]=1; for(int i=head[u];~i;i=e[i].nxt) { int v=e[i].to; if(!vis[v]) { tarjan(v); low[u]=min(low[u],low[v]); if(low[v]>=dfn[u]&&u!=1) cut[u]=1; else if(u==1) root++; } else low[u]=min(low[u],dfn[v]); } } int main() { while(scanf("%d",&n)&&n) { init();int u,v; while(scanf("%d",&u)&&u) { while(getchar()!='\n') { scanf("%d",&v); adde(u,v); } } for(int i=1;i<=n;++i) if(dfn[i]==-1) tarjan(i); if(root>1) ans++; for(int i=2;i<=n;++i) if(cut[i]) ans++; printf("%d\n",ans); } return 0; }
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; #define N 110 int n,cnt=-1,head ; struct Edge{int to,nxt;}e[N*N]; int idc,dfn ,low ,vis ,cut ; void init() { cnt=-1;idc=0; memset(head,-1,sizeof(head)); memset(dfn,0,sizeof(dfn)); memset(low,0,sizeof(low)); memset(vis,0,sizeof(vis)); memset(cut,0,sizeof(cut)); } void adde(int u,int v) { e[++cnt].to=v;e[cnt].nxt=head[u];head[u]=cnt; e[++cnt].to=u;e[cnt].nxt=head[v];head[v]=cnt; } void Tarjan(int u) { dfn[u]=low[u]=++idc;vis[u]=1; for(int i=head[u];~i;i=e[i].nxt) { int v=e[i].to; if(!vis[v]) { Tarjan(v); low[u]=min(low[u],low[v]); if(low[v]>=dfn[u]) cut[u]++; } else low[u]=min(low[u],dfn[v]); } } int main() { while(~scanf("%d",&n)&&n) { init(); int u,v; while(scanf("%d",&u)&&u) { while(getchar()!='\n') { scanf("%d",&v); adde(u,v); } } for(int i=1;i<=n;++i) if(!vis[i]) Tarjan(i); int ans=0; cut[1]--; for(int i=1;i<=n;++i) printf("%d ",dfn[i]);puts(""); for(int i=1;i<=n;++i) printf("%d ",low[i]);puts(""); for(int i=1;i<=n;++i) { printf("%d ",cut[i]); if(cut[i]>0) ans+=1; } puts(""); printf("%d\n",ans); } return 0; }
相关文章推荐
- POJ2553 The Bottom of a Graph Tarjan 矩阵 pascal 解题报告
- POJ 2186 Tarjan 缩点 解题报告
- Tarjan算法求解桥和边双连通分量(附POJ 3352 Road Construction解题报告)
- POJ2186 Popular Cows Tarjan 链接表 pascal解题报告
- POJ 2942 Tarjan双联通分量+二分图 解题报告
- Tarjan算法求解桥和边双连通分量(附POJ 3352 Road Construction解题报告
- poj 2105 解题报告
- POJ 1014 解题报告
- poj2533解题报告
- Poj 2184解题报告(01背包变种)
- 广大暑假训练1 E题 Paid Roads(poj 3411) 解题报告
- POJ 1017 解题报告
- poj1741 Tree解题报告
- poj 1142Smith Numbers(解题报告)
- POJ 1904 King's Quest 解题报告
- poj 1207 The 3n + 1 problem 解题报告
- poj解题报告——2356
- 【原】 POJ 1088 滑雪 递归+memoization 解题报告
- 【原】 POJ 1157 LITTLE SHOP OF FLOWERS 动态规划 解题报告
- 【原】 POJ 2352 Stars 树状数组 解题报告