文章目录
- 0x01 Telephone Lines POJ - 3662
- 0x02 P1073 [NOIP2009 提高组] 最优贸易
- 0x03 道路和航线 BZOJ2200
- 0x04 Sorting It All Out POJ - 1094 topo
- 0x05 Sightseeing trip POJ - 1734 最小环问题
- 0x06 Cow Relays POJ - 3613 S到E经过k条边的最短路
- 0x07 走廊泼水节 （Kruskal）
0x01 Telephone Lines POJ - 3662
Telephone Lines
题意：从1到N修一条电缆，有p对电线杆之间是可以连接的，电信公司可以提供k条电缆，其他的由John提供，求john提供的电缆的最长的那根的长度（ret）。
思路：实则是求最短最长的边。
二分结果(sum)。对于 边值>sum, 电信公司需要提供电缆。
用djk 计算 1->n 路径上的最短路径。 满足d[n]< k ，sum是一个符合的结果。
代码如下：
#include<iostream>
#include<cstring>
#include<queue>
#define ll long long
#define ld long double
#define ull unsigned long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
ll gcd(ll a,ll b){ return b? gcd(b,a%b):a;}
const int N=2e4+10;
const ll P=1e9+7;
ll read(){
ll s = 0, f = 1; char ch = getchar();
while(!isdigit(ch)){
if(ch == '-') f = -1;
ch = getchar();
}
while(isdigit(ch)) s = (s << 3) + (s << 1) + (ch ^ 48), ch = getchar();
return s * f;
}
using namespace std;
ll n,p,k;
ll head[N],ver[N],nex[N],edgevalue[N];
ll cnt;
void addedge(ll u,ll v,ll value)
{
ver[++cnt]=v;
edgevalue[cnt]=value;
nex[cnt]=head[u];
head[u]=cnt;
}
priority_queue<pair<int,int>> q;
ll d[N],v[N];
bool ischeck(ll sum)
{
memset(d,0x3f,sizeof(d));
memset(v,0,sizeof(v));
while(!q.empty())
{
q.pop();
}
d[1]=0;
q.push(make_pair(0,1)); // 距离 点
while(!q.empty())
{
ll x=q.top().second;
q.pop();
if(v[x]) continue;
v[x]=1;
for(int i=head[x];i;i=nex[i])
{
int y=ver[i],z=edgevalue[i];
if(z> sum) z=1;
else z=0;
if(d[y]>d[x]+z)
{
d[y]=d[x]+z;
q.push(make_pair(-d[y],y));
}
}
if(v[n]==1)
break;
}
// printf("%lld\n",d[n]);
// getchar();
if(d[n]<=k)
return true;
else
return false;
}
void solve(){
n=read();
p=read();
k=read();
ll u,v,value;
rep(i,1,p)
{
u=read();
v=read();
value=read();
addedge(u,v,value);
addedge(v,u,value);
}
ll l=0,r=1000010;
ll mid;
while(l<r)
{
mid=(l+r)>>1;
// printf("l:%lld r:%lld mid:%lld\n",l,r,mid);
if(ischeck(mid))
{
r=mid;
}
else
{
l=mid+1;
}
}
if(r==1000010) printf("-1\n");
else
printf("%lld\n",r);
return ;
}
int main (){
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
solve();
getchar();
getchar();
return 0;
}
0x02 P1073 [NOIP2009 提高组] 最优贸易
最优贸易
题意：从1到达n号，在沿途中可选一点购买水晶球，在此之后可卖出去水晶球。商人阿龙只能购买一次，
问商人阿龙 旅游结束后，通过这样的方式最多赚取多少旅费？途中 一座城市可被重复经过。
数据范围：
1
<
=
n
<
=
1
0
5
,
1
<
=
m
<
=
5
∗
1
0
5
1<=n<=10^5,1<=m<=5*10^5
1<=n<=105,1<=m<=5∗105
思路：
代码如下：
#include<bits/stdc++.h>
#define ll long long
#define ld long double
#define ull unsigned long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
ll gcd(ll a,ll b){ return b? gcd(b,a%b):a;}
const int N=1e6+10;
const ll P=1e9+7;
ll read(){
ll s = 0, f = 1; char ch = getchar();
while(!isdigit(ch)){
if(ch == '-') f = -1;
ch = getchar();
}
while(isdigit(ch)) s = (s << 3) + (s << 1) + (ch ^ 48), ch = getchar();
return s * f;
}
using namespace std;
ll n,m;
ll vVAlue[N];
ll head[N],nex[N],ver[N];
ll cnt;
ll addedge(ll u,ll v)
{
ver[++cnt]=v;
nex[cnt]=head[u];
head[u]=cnt;
}
ll vis[N];
void djk(ll u,ll *D,ll flag)
{
priority_queue< pair<int,int> > q;
memset(vis,0,sizeof(vis));
if(flag==1)
rep(i,1,n) D[i]=200;
else
rep(i,1,n) D[i]=0;
// rep(i,1,n) cout<<D[i]<<" "; cout<<endl;
D[u]=vVAlue[u];
q.push(make_pair(D[u],u));
while(!q.empty())
{
ll x=q.top().second;
q.pop();
for(int i=head[x];i;i=nex[i])
{
// cout<<"321"<<endl;
ll y=ver[i];
if(flag==1)
{
if(D[y] > D[x])
{
// cout<<vVAlue[y]<<" "<<D[x]<<endl;
D[y]=min(vVAlue[y],D[x]);
q.push(make_pair(-D[y],y));
}
}
else
{
if(D[y] < D[x])
{ //cout<<vVAlue[y]<<" "<<D[x]<<endl;
D[y]=max(vVAlue[y],D[x]);
q.push(make_pair(D[y],y));
}
}
}
}
// cout<<flag<<endl;
// rep(i,1,n)
// {
// printf("u:%lld value:%lld\n",i,D[i]);
// }
return ;
}
struct zw
{
ll x,y,z;
}a[N];
ll D[N],F[N];
void solve(){
n=read();
m=read();
rep(i,1,n) vVAlue[i]=read();
ll x,y,z;
rep(i,1,m)
{
a[i].x=read();
a[i].y=read();
a[i].z=read();
addedge(a[i].x,a[i].y);
if(a[i].z==2)
{
addedge(a[i].y,a[i].x);
}
}
djk(1,D,1);
cnt=0;
memset(head,0,sizeof(head));
rep(i,1,m)
{
addedge(a[i].y,a[i].x);
if(a[i].z==2)
{
addedge(a[i].x,a[i].y);
}
}
djk(n,F,-1);
ll sum=0;
rep(i,1,n) sum=max(sum,F[i]-D[i]);
printf("%lld\n",sum);
return ;
}
int main (){
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
solve();
getchar();
getchar();
return 0;
}
0x03 道路和航线 BZOJ2200
拓扑排序，djk，
道路和航线
题意：
思路：
代码如下：
#include<bits/stdc++.h>
#define ll long long
#define ld long double
#define ull unsigned long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
ll gcd(ll a,ll b){ return b? gcd(b,a%b):a;}
const int N=2e5+10;
// const ll P=1e9+7;
ll read(){
ll s = 0, f = 1; char ch = getchar();
while(!isdigit(ch)){
if(ch == '-') f = -1;
ch = getchar();
}
while(isdigit(ch)) s = (s << 3) + (s << 1) + (ch ^ 48), ch = getchar();
return s * f;
}
using namespace std;
ll n,R,P,S;
ll cnt;
ll head[N],nex[N],edgevalue[N],var[N];
ll D[N],vis[N];
ll num[N],in[N];
ll ans; // 编号个数
queue<int> c; //存放编号
priority_queue< pair<ll,int>> q;
void addedge(ll u,ll v,ll value)
{
var[++cnt]=v;
edgevalue[cnt]=value;
nex[cnt]=head[u];
head[u]=cnt;
}
void dfs(ll u)
{
num[u]=ans;
for(int i=head[u];i;i=nex[i])
{
int v=var[i];
if(num[v]) continue;
dfs(v);
}
return ;
}
void djk(ll s)
{
ll u,v,value;
memset(D,0x7f,sizeof(D));
memset(vis,0,sizeof(vis));
D[s]=0;
c.push(num[s]);
rep(i,1,ans)
{
if(in[i]==0 and num[s]!=i)
c.push(i);
}
while(!c.empty())
{
ll x=c.front();
c.pop();
rep(i,1,n)
{
if(num[i]==x) q.push(make_pair(-D[i],i));
}
while(!q.empty())
{
u=q.top().second;
q.pop();
if(vis[u]) continue;
vis[u]=1;
for(int i=head[u];i;i=nex[i])
{
v=var[i];
value=edgevalue[i];
if(D[v] > D[u]+value)
{
D[v]=D[u]+value;
if(num[u]==num[v])
{
q.push(make_pair(-D[v],v));
}
}
if(num[u]!=num[v])
{
in[num[v]]--;
if(in[num[v]]==0)
{
c.push(num[v]);
}
}
}
}
}
}
void solve(){
n=read();
R=read();
P=read();
S=read();
ll u,v,value;
rep(i,1,R)
{
u=read(); v=read(); value=read();
addedge(u,v,value);
addedge(v,u,value);
}
rep(i,1,n)
{
if(num[i]==0)
{
ans++;
dfs(i);
}
}
rep(i,1,P)
{
u=read();
v=read();
value=read();
in[num[v]]++;
addedge(u,v,value);
}
djk(S);
rep(i,1,n)
{
if(D[i]>0x3f3f3f3f) printf("NO PATH\n");
else printf("%lld\n",D[i]);
}
return ;
}
int main (){
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
solve();
getchar();
getchar();
return 0;
}
0x04 Sorting It All Out POJ - 1094 topo
Sorting It All Out
题意：
思路：
代码如下：
#include<iostream>
#include<cstring>
#include<queue>
#define ll long long
#define ld long double
#define ull unsigned long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
ll gcd(ll a,ll b){ return b? gcd(b,a%b):a;}
const int N=30;
const ll P=1e9+7;
ll read(){
ll s = 0, f = 1; char ch = getchar();
while(!isdigit(ch)){
if(ch == '-') f = -1;
ch = getchar();
}
while(isdigit(ch)) s = (s << 3) + (s << 1) + (ch ^ 48), ch = getchar();
return s * f;
}
using namespace std;
ll n,m,cnt;
ll in[N],in1[N];
ll head[N],var[1000],nex[1000];
void addedge(ll u,ll v)
{
var[++cnt]=v;
nex[cnt]=head[u];
head[u]=cnt;
}
ll ans=0;
char ch[N];
ll topo()
{
rep(i,0,n-1) in[i]=in1[i];
ans=0;
queue<int> q;
rep(i,0,n-1)
{
if(in[i]==0) q.push(i);
}
bool logo=true;
ll u,v;
while(!q.empty())
{
if(q.size()>1) logo=false;
ll u=q.front();
ch[ans++]=u;
q.pop();
for(int i=head[u];i;i=nex[i])
{
v=var[i];
if((--in[v])==0) q.push(v);
}
}
// cout<<ans<<" "<<logo<<n<<endl;
if(ans<n) return -1;
else
{
if(logo==true) return 1;
else return 0;
}
}
void solve(){
while(1)
{
n=read();
m=read();
if(n==0 and m==0) break;
rep(i,0,n)
{
in1[i]=0;
head[i]=0;
cnt=0;
}
char x,y;
ll u,v;
ll res=0;
ll inde=0;
rep(i,1,m)
{
cin>>x>>y>>y;
u=x-'A';
v=y-'A';
addedge(u,v);
in1[v]++;
if(res==0)
{
res=topo();
// cout<<res<<endl;
inde=i;
}
}
// cout<<res<<endl;
if(res==0) cout<<"Sorted sequence cannot be determined."<<endl;
else if(res==-1) cout<<"Inconsistency found after "<<inde<<" relations."<<endl;
else if(res==1)
{
cout<<"Sorted sequence determined after "<<inde<<" relations: ";
rep(i,0,ans-1) printf("%c",ch[i]+'A');
cout<<"."<<endl;
}
}
}
int main (){
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
solve();
getchar();
getchar();
return 0;
}
0x05 Sightseeing trip POJ - 1734 最小环问题
留下一个小疑问点： int 和 ll a=0x3f3f3f3f 在这道题中 使用ll a=0x3f3f3f3f 会错。
Sightseeing trip
题意：
思路：弗洛伊德， 回溯路径。
外层循环k开始，对所以的路径刷新，添上k和步添上k
这个算法的思想很关键
#include<iostream>
#include<cstring>
#include<vector>
#include<algorithm>
#define ll long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
const int N=300+10;
using namespace std;
int a[N][N],dist[N][N],pos[N][N];
ll n,m;
ll ans=1e8;
vector<int> path;
void get_path(int x,int y){
if(pos[x][y]==0) return ;
get_path(x,pos[x][y]);
path.push_back(pos[x][y]);
get_path(pos[x][y],y);
}
void solve(){
cin>>n>>m;
// memset(a,0x3f,sizeof(a));
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
a[i][j]=1e8;
}
}
for(int i=0;i<=n;i++) a[i][i]=0;
for(int i=1;i<=m;i++)
{
int u,v,w;
cin>>u>>v>>w;
a[u][v]=a[v][u]=min(a[u][v],w);
}
memcpy(dist,a,sizeof(a));
for(int k=1;k<=n;k++){
for(int i=1;i<k;i++){
for(int j=i+1;j<k;j++){
if(dist[i][j]+a[j][k]+a[k][i]<ans){
ans=dist[i][j]+a[j][k]+a[k][i];
path.clear();
path.push_back(i);
get_path(i,j);
path.push_back(j);
path.push_back(k);
}
}
}
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
if(dist[i][j]> dist[i][k]+dist[k][j]){
dist[i][j]=dist[i][k]+dist[k][j];
pos[i][j]=k;
}
}
}
}
if(ans == 1e8){
cout << "No solution." << endl;
}
else
{
for(int i=0;i<(int)path.size();i++){
cout<<path[i]<<" ";
}
}
return ;
}
int main (){
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
solve();
getchar();
getchar();
return 0;
}
0x06 Cow Relays POJ - 3613 S到E经过k条边的最短路
传送门
题意：
思路: 点编号映射，矩阵快速幂
代码如下：
#include<iostream>
#include<cstring>
#define ll long long
#define ld long double
#define ull unsigned long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
ll gcd(ll a,ll b){ return b? gcd(b,a%b):a;}
const int N=2e5+10;
const ll P=1e9+7;
const ll INF=0x3f3f3f3f;
ll read(){
ll s = 0, f = 1; char ch = getchar();
while(!isdigit(ch)){
if(ch == '-') f = -1;
ch = getchar();
}
while(isdigit(ch)) s = (s << 3) + (s << 1) + (ch ^ 48), ch = getchar();
return s * f;
}
using namespace std;
ll p[1010];
ll a[300];
ll cnt;
struct M{
ll edge[210][210];
void init(){
rep(i,1,200){
rep(j,1,200){
edge[i][j]=INF;
}
}
}
};
M add(M a,M b){
M sum;
sum.init();
rep(i,1,cnt){
rep(j,1,cnt){
rep(k,1,cnt){
sum.edge[i][j]=min(sum.edge[i][j],a.edge[i][k]+b.edge[k][j]);
}
}
}
return sum;
}
void solve(){
ll n,t,s,e;
cin>>n>>t>>s>>e;
ll w,u,v;
M sum;
sum.init();
for(int i=1;i<=t;i++){
cin>>w>>u>>v;
if(!p[u]){
p[u]=(++cnt);
a[cnt]=u;
}
if(!p[v]){
p[v]=(++cnt);
a[cnt]=v;
}
sum.edge[p[u]][p[v]]=min(w, sum.edge[p[u]][p[v]]);
sum.edge[p[v]][p[u]]=min(w, sum.edge[p[v]][p[u]]);
}
M ans;
memcpy(ans.edge,sum.edge,sizeof(sum.edge));
// sum=add(sum,sum);
n--;
while(n){
if(n&1) ans=add(ans,sum);
sum=add(sum,sum);
n>>=1;
}
cout<<ans.edge[p[s]][p[e]]<<endl;
}
int main (){
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
solve();
getchar();
getchar();
return 0;
}
0x07 走廊泼水节 （Kruskal）
走廊泼水节
题意：
思路： 克里斯特+模拟
#include<bits/stdc++.h>
#define ll long long
#define ld long double
#define ull unsigned long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
ll gcd(ll a,ll b){ return b? gcd(b,a%b):a;}
const int N=1e4+10;
const ll P=1e9+7;
ll read(){
ll s = 0, f = 1; char ch = getchar();
while(!isdigit(ch)){
if(ch == '-') f = -1;
ch = getchar();
}
while(isdigit(ch)) s = (s << 3) + (s << 1) + (ch ^ 48), ch = getchar();
return s * f;
}
using namespace std;
ll fa[N];
ll num[N];
ll getfa(ll x){
return x==fa[x] ? x: fa[x]=getfa(fa[x]);
}
struct zw{
int x,y,z;
}a[N];
bool operator <(zw a,zw b){
return a.z<b.z;
}
void solve(){
ll n;
cin>>n;
rep(i,1,n-1){
cin>>a[i].x>>a[i].y>>a[i].z;
}
rep(i,1,n){
fa[i]=i;
num[i]=1;
}
sort(a+1,a+n);
ll sum=0;
for(int i=1;i<=n-1;i++){
int x=getfa(a[i].x);
int y=getfa(a[i].y);
if(x==y) continue;
sum+=( num[x]*num[y]-1 )*(a[i].z+1);
fa[y]=x;
num[x]+=num[y];
}
cout<<sum<<endl;
return ;
}
int main (){
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int T=read();
while(T--)
solve();
getchar();
getchar();
return 0;
}
