POJ-2762 Going from u to v or from v to u? （强连通分量[Tarjan]&&(拓扑排序||树形DP)）
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Going from u to v or from v to u?
http://poj.org/problem?id=2762
| Time Limit: 2000MS | Memory Limit: 65536K | |
Description
In order to make their sons brave, Jiajia and Wind take them to a big cave. The cave has n rooms, and one-way corridors connecting some rooms. Each time, Wind choose two rooms x and y, and ask one of their little sons go from one to the other. The son can either go from x to y, or from y to x. Wind promised that her tasks are all possible, but she actually doesn't know how to decide if a task is possible. To make her life easier, Jiajia decided to choose a cave in which every pair of rooms is a possible task. Given a cave, can you tell Jiajia whether Wind can randomly choose two rooms without worrying about anything?
Input
The first line contains a single integer T, the number of test cases. And followed T cases.
The first line for each case contains two integers n, m(0 < n < 1001,m < 6000), the number of rooms and corridors in the cave. The next m lines each contains two integers u and v, indicating that there is a corridor connecting room u and room v directly.
The first line for each case contains two integers n, m(0 < n < 1001,m < 6000), the number of rooms and corridors in the cave. The next m lines each contains two integers u and v, indicating that there is a corridor connecting room u and room v directly.
Output
The output should contain T lines. Write 'Yes' if the cave has the property stated above, or 'No' otherwise.
Sample Input
1 3 3 1 2 2 3 3 1
Sample Output
Yes
题目大意：给定一个有向图，判断是否对任意两点u，v，有u可达v 或 v可达u？
先找出所有的强连通分量，则各强连通分量内是相互可达的，只用判断强连通分量之间是否至少单向可达
将各强连通分量缩成一点，建立新图，则新图是DAG
刚开始也认为只要从一个入度为0的点到出度为0的点的长度为总点数，就满足题意，否则不行，但是看见有人说不是只有一条链的情况下会输出Yes，于是放弃了
最终用拓扑排序AC，若某次队列中点的数目大于1（则这些点之间不能单向可达），则输出No，否则输出Yes
没有初始化indeg数组，导致WA了很久都没看出来...
解法一：Tarjan&&拓扑排序
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
const int MAXN=1005;
int n,m,num,cnt,ans,top,head,tail;
int stak[MAXN],que[MAXN];
int low[MAXN],dfn[MAXN],color[MAXN],indeg[MAXN];//biocks[i]表示删掉i点后，能形成的连通块数
vector<int> g[MAXN];
bool isIn[MAXN],mp[MAXN][MAXN];
void Tarjan(int u,int p) {
stak[++top]=u;
isIn[u]=true;
dfn[u]=low[u]=++num;
int v;
for(int i=0;i<g[u].size();++i) {
v=g[u][i];
if(dfn[v]==0) {
Tarjan(v,u);
low[u]=min(low[u],low[v]);
}
else if(isIn[v]&&dfn[v]<low[u])
low[u]=dfn[v];
}
if(dfn[u]==low[u]) {
++cnt;
do {
v=stak[top--];
isIn[v]=false;
color[v]=cnt;//同一个强连通分量缩成一个点
} while(v!=u);
}
}
bool TopoSort() {
head=tail=0;
for(int i=1;i<=cnt;++i)
if(indeg[i]==0)
que[tail++]=i;
if(tail-head>1)//如果入度为0的点超过一个，则这些点不能相互单向可达
return false;
int u;
while(head!=tail) {
u=que[head++];
for(int i=1;i<=cnt;++i) {
if(mp[u][i]) {
--indeg[i];
if(indeg[i]==0)//入度为0的点入队
que[tail++]=i;
}
}
if(tail-head>1)//如果入度为0的点超过一个，则这些点不能相互单向可达
return false;
}
return true;
}
int main() {
int T,s,e;
scanf("%d",&T);
while(T-->0) {
scanf("%d%d",&n,&m);
for(int i=1;i<=n;++i) {
g[i].clear();
dfn[i]=indeg[i]=0;
isIn[i]=false;
}
while(m-->0) {
scanf("%d%d",&s,&e);
g[s].push_back(e);
}
cnt=num=top=0;//cnt表示强连通分量
for(int i=1;i<=n;++i) {
if(dfn[i]==0)
Tarjan(i,0);
}
memset(mp,false,sizeof(mp));
for(int i=1;i<=n;++i) {
for(int j=0;j<g[i].size();++j) {
s=color[i];
e=color[g[i][j]];
if(s!=e&&!mp[s][e]) {//建立新图，不考虑重边
mp[s][e]=true;
++indeg[e];
}
}
}
printf("%s\n",TopoSort()?"Yes":"No");
}
return 0;
}
解法二：Tarjan&&树形DP
又想了想，即使新图不是一条链，但是满足题意的情况下，必定有一条路径满足：入度为0的点到出度为0的点的所经过的点的个数等于所有点的个数
dp[i]表示从i点出发，一条路径上能经过的点数的最大值
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
const int MAXN=1005;
int n,m,num,cnt,ans,top;
int stak[MAXN];
int low[MAXN],dfn[MAXN],color[MAXN],dp[MAXN];//biocks[i]表示删掉i点后，能形成的连通块数
vector<int> g[MAXN],mp[MAXN];
bool isIn[MAXN],haveIndeg[MAXN];
void Tarjan(int u,int p) {
stak[++top]=u;
isIn[u]=true;
dfn[u]=low[u]=++num;
int v;
for(int i=0;i<g[u].size();++i) {
v=g[u][i];
if(dfn[v]==0) {
Tarjan(v,u);
low[u]=min(low[u],low[v]);
}
else if(isIn[v]&&dfn[v]<low[u])
low[u]=dfn[v];
}
if(dfn[u]==low[u]) {
++cnt;
do {
v=stak[top--];
isIn[v]=false;
color[v]=cnt;//同一个强连通分量缩成一个点
} while(v!=u);
}
}
void dfs(int u) {
if(dp[u]!=-1)
return ;
if(mp[u].size()==0) {//如果点u是叶子节点，则其能经过的点数为1
dp[u]=1;
return ;
}
for(int i=0;i<mp[u].size();++i) {
dfs(mp[u][i]);
dp[u]=max(dp[u],dp[mp[u][i]]+1);//点u的经过的点的个数为其各子结点经过的点的个数+1
}
}
bool Judge() {
cnt=num=top=0;//cnt表示强连通分量
for(int i=1;i<=n;++i) {
if(dfn[i]==0)
Tarjan(i,0);
}
int s,e,sta;
memset(mp,false,sizeof(mp));
for(int i=1;i<=n;++i) {
for(int j=0;j<g[i].size();++j) {
s=color[i];
e=color[g[i][j]];
if(s!=e) {//建立新图
mp[s].push_back(e);
haveIndeg[e]=true;;
}
}
}
num=0;
for(int i=1;i<=cnt;++i) {
dp[i]=-1;
if(!haveIndeg[i]) {
sta=i;
++num;
}
}
if(num>1)
return false;
dfs(sta);
return dp[sta]==cnt;//判断入度为0的点能达到的最大深度是否等于点的个数
}
int main() {
int T,s,e;
scanf("%d",&T);
while(T-->0) {
scanf("%d%d",&n,&m);
for(int i=1;i<=n;++i) {
g[i].clear();
mp[i].clear();
dfn[i]=0;
isIn[i]=haveIndeg[i]=false;
}
while(m-->0) {
scanf("%d%d",&s,&e);
g[s].push_back(e);
}
printf("%s\n",Judge()?"Yes":"No");
}
return 0;
}