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图论题：一开始我是用tarjan算法做的，wrong answer 了很多次，然后又用了floyd-warshell算法，也wa了
最后找了题解，原来最后的dataset后面不是组数，是样例的编号，题根本就没说，让人怎么理解。。。
tarjan
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<map>
#include<algorithm>
#define M 100
using namespace std;
int qis;
int maps[50][50];
int dfn[M],low[M],vis[M],stacks[M];
int belongs[25][50];
int cnt ,scnt,begin,ans;
void init(){
memset(dfn,0,sizeof(dfn));
memset(low,0,sizeof(low));
memset(stacks,0,sizeof(stacks));
memset(vis,0,sizeof(vis));
memset(belongs,0,sizeof(belongs));
memset(maps,0,sizeof(maps));
qis = cnt = scnt = begin = ans =0;
}
void tarjan(int x){
int v;
dfn[x] = low [x] = ++cnt ;
stacks[++begin] = x;
for(int i = 1;i<=qis;i++){
if(maps[x][i]){
if(!dfn[i]){
tarjan(i);
low[x] = min(low[x],low[i]);
}
else if(!vis[i]){
low[x] = min(low[x],dfn[i]);
}
}
}
if(low[x] == dfn[x]){
scnt ++;
do{
v = stacks[begin--];
belongs[scnt][++belongs[scnt][0]] = v;//顶点处理
vis[v] = 1;
}while(v != x);
if(belongs[scnt][0]>1)ans ++;
}
}
int main(){
int flag = 1;
int n,m;
int cases = 1;
while(scanf("%d%d",&n,&m),n||m){
string a,b;
init();
map<string,int>ms;
map<int,string>_ms;
ms.clear();_ms.clear();
for(int i = 0;i<m;i++){
cin>>a>>b;
if(ms[a] == 0)ms[a] = ++qis;
if(ms[b] == 0)ms[b] = ++qis;
_ms[ms[a]] = a;
_ms[ms[b]] = b;
maps[ms[a]][ms[b]] = 1;
}
for(int i =1;i<= qis;i++){
if(!dfn[i])tarjan(i);
}
if(flag == 1)flag = 0;
else puts("");
printf("Calling circles for data set %d:\n",cases++);
for(int i = 1;i<=scnt;i++){
int j;
for(j = 1;j<belongs[i][0];j++)
cout<<_ms[belongs[i][j]]<<", ";
cout<<_ms[belongs[i][j]]<<endl;
}
}
}
floyd-warshell:
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<map>
#include<algorithm>
#define M 100
using namespace std;
int qis;
int maps[50][50];
int vis[M];
int belongs[25][50];
int scnt ,ans;
int cases = 1;
int main(){
int flag = 1;
int n,m;
while(scanf("%d%d",&n,&m),n||m){
qis =ans = scnt = 0;
string a,b;
map<string,int>ms;
map<int,string>_ms;
ms.clear();_ms.clear();
memset(vis,0,sizeof(vis));
memset(maps,0,sizeof(maps));
memset(belongs,0,sizeof(belongs));
for(int i = 1;i<=n;i++)maps[i][i] = 1;
for(int i = 0;i<m;i++){
cin>>a>>b;
if(ms[a] == 0)ms[a] = ++qis;
if(ms[b] == 0)ms[b] = ++qis;
_ms[ms[a]] = a;
_ms[ms[b]] = b;
maps[ms[a]][ms[b]] = 1;
}
//floyd
for(int k = 1;k <= n;k++)
for(int i = 1;i <= n;i++)
for(int j = 1;j <= n;j++){
maps[i][j] = maps[i][j] || (maps[i][k] && maps[k][j]);
}
for(int i = 1;i<=n;i++){
if(vis[i] == 0){
scnt++;
for(int j = 1;j<=n;j++){
if(maps[i][j]&&maps[j][i]){
belongs[scnt][++belongs[scnt][0]] = j;
vis[j] = 1;
}
}
if(belongs[scnt][0]>1)ans++;
}
}
if(flag == 1)flag = 0;
else puts("");
printf("Calling circles for data set %d:\n",cases++);
for(int i = 1;i<=scnt;i++){
int j;
for(j = 1;j<belongs[i][0];j++)
cout<<_ms[belongs[i][j]]<<", ";
cout<<_ms[belongs[i][j]]<<endl;
}
}
}
--uva247（calling circles）强联通与floyd-warshell
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原文地址：http://www.cnblogs.com/lovelystone/p/4737750.html
