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POJ 1985 Cow Marathon(树的直径&BFS+BFS)

2016-04-22 11:04 411 查看
大意:找出图中的相距最远两个点的距离。

思路:由于从任意一个点出发找到最远的一个点,一定在相距最远两点的其中一个端点,所以两次BFS找最远距离。

vector:

#include<map>
#include<queue>
#include<cmath>
#include<cstdio>
#include<stack>
#include<iostream>
#include<cstring>
#include<algorithm>
#define ll long long
#define inf 0x3f3f3f3f
#define eps 1e-8
#include<vector>
#define ls l,mid,rt<<1
#define rs mid+1,r,rt<<1|1
using namespace std;

struct node{
int to,w;
};
vector<node>q[100010];
char str[2];
int ans,dis[100010];
bool vis[100010];

int bfs(int u){
memset(dis,0,sizeof(dis));
memset(vis,false,sizeof(vis));
queue<int>Q;
while(!Q.empty())
Q.pop();
Q.push(u);
vis[u] = true;
int bj = u;
while(!Q.empty()){
int t = Q.front();
Q.pop();
if(dis[t] > ans){
ans = dis[t];
bj = t;
}
int len = q[t].size();
for(int i = 0;i < len; ++ i){
node tmp = q[t][i];
if(vis[tmp.to]) continue;
vis[tmp.to] = true;
dis[tmp.to ] = dis[t] + tmp.w;
Q.push(tmp.to);
}
}
return bj;
}

int main(){
int n,m,i,j,k,a,b,c;
while(~scanf("%d%d",&n,&m)){

for(i = 0;i <= n;++ i){
q[i].clear();
}
int star;
for(i = 0;i < m;++ i){
scanf("%d%d%d%s",&a,&b,&c,str);
star = a;
q[a].push_back({b,c});
q[b].push_back({a,c});
}
ans = 0;
int cnt = bfs(star);
ans = 0;
bfs(cnt);
printf("%d\n",ans);
}
return 0;
}


前向星:

#include<map>
#include<queue>
#include<cmath>
#include<cstdio>
#include<stack>
#include<iostream>
#include<cstring>
#include<algorithm>
#define ll long long
#define inf 0x3f3f3f3f
#define eps 1e-8
#include<vector>
#define ls l,mid,rt<<1
#define rs mid+1,r,rt<<1|1
using namespace std;

struct node{
int to,w,next;
}q[100010];
char str[2];
int ans,dis[100010],cnt,head[100010];
bool vis[100010];
void bu(int a,int b,int c){
q[cnt].to = b;
q[cnt].w = c;
q[cnt].next=head[a];
head[a] = cnt++;

q[cnt].to = a;
q[cnt].w = c;
q[cnt].next=head[b];
head[b] = cnt++;
}
int bfs(int u){
memset(dis,0,sizeof(dis));
memset(vis,false,sizeof(vis));
queue<int>Q;
while(!Q.empty())
Q.pop();
Q.push(u);
vis[u] = true;
int bj = u;
while(!Q.empty()){
int t = Q.front();
Q.pop();
if( dis[t] > ans){
ans = dis[t];
bj = t;
}
for(int i = head[t] ;~i;i = q[i].next){
int tmp = q[i].to;
if(vis[tmp]) continue;
vis[tmp] = true;
dis[tmp] = dis[t] + q[i].w;
Q.push(tmp);
}
}
return bj;
}

int main(){
int n,m,i,j,k,a,b,c;
while(~scanf("%d%d",&n,&m)){
cnt = 0;
for(i = 1;i <= n;++ i){
head[i] = -1;
}
int star;
for(i = 0;i < m;++ i){
scanf("%d%d%d%s",&a,&b,&c,str);
star = a;
bu(a,b,c);
}
ans = 0;
int pp = bfs(star);
ans = 0;
bfs(pp);
printf("%d\n",ans);
}
return 0;
}
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标签:  树的直径 bfs