专题训练集-最短路
POJ2253
本题的意思是求出点1到点2之间所有路间，每条路的最大边的最小值。所以根据这个只需要对每两个点进行松弛操作，然后在松弛操作中进行最大边选取即可。
至于为什么在这里选择了floyd算法，那是因为这两个点之间存在极其多的路径可供选择，而这里需要讨论每一条路的最大边，然后综合求最小值，所以这个很明显不能用普通的最短路算法来实现。这就需要floyd里这种类似动态规划的思想来解决是比较好的。在这里，所谓的松弛操作就是：已知a，b两点间存在一条通路，权值为w，假如在中间插入点c，并且a，c两点间存在一条权值为w1的通路，c，b两点间存在一条权值为w2的通路，且w>w1,w>w2，则w=max（w1,w2），这就是所谓最大边的最小值选取，即认为w是a与b间的最大边，w1与w2之间的最大边是a->c->b这条路的最大边。
初始化的时候认为“最大边”是两点间距。
import java.io.*;
import java.text.*;
import java.math.*;
import java.util.*;
public class Main
{
public static class Point
{
int x,y;
public Point(int a,int b)
{
this.x=a;
this.y=b;
}
}
public static double distance(int x1,int x2,int y1,int y2)
{
return Math.sqrt((double)((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)));
}
public static void floyd(int limit1)
{
for(int k=1;k<=limit1;k++)
{
for(int i=1;i<=limit1;i++)
{
for(int j=1;j<=limit1;j++)
{
if(dis[i][j]>dis[i][k]&&dis[i][j]>dis[k][j])
dis[i][j]=Math.max(dis[i][k],dis[k][j]);
}
}
}
}
static double dis[][]=new double[450][450];
static Point points[]=new Point[1322];
public static void main(String args[])throws IOException
{
BufferedReader bf=new BufferedReader(new InputStreamReader(System.in));
StreamTokenizer st=new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
PrintWriter pw=new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
int cases=0;
while(st.nextToken()!=StreamTokenizer.TT_EOF)
{
int n=(int)st.nval;
if(n==0)
break;
for(int i=1;i<=n;i++)
{
int tmp1,tmp2;
st.nextToken();tmp1=(int)st.nval;
st.nextToken();tmp2=(int)st.nval;
points[i]=new Point(tmp1,tmp2);
}
for(int i=1;i<=n;i++)
for(int j=i+1;j<=n;j++)
dis[i][j]=dis[j][i]=distance(points[i].x,points[j].x,points[i].y,points[j].y);
floyd(n);
DecimalFormat df=new DecimalFormat("0.000");
pw.println("Scenario #"+(++cases));
pw.println("Frog Distance = "+df.format(dis[1][2]));
pw.println("");
}
pw.flush();
}
}POJ1797
本题的要求是求1到n所有路径中最小值的最大值。由于数据较大，不能用floyd进行松弛，所以这里选择了spfa。
这里的松弛操作是对于a->b这样一条路，从1到b点的所有路径中最小值的最大值为从从min(1到a点的所有路径中最小值的最大值,w(a,b))，一直更新这个答案即可。
#include<pch.h>
#include <iostream>
#include <cstdio>
#include <bits/stdc++.h>
#include <queue>
#include <map>
#include <algorithm>
#include <stack>
#include <iomanip>
#include <cstring>
#include <cmath>
#define endl '\n'
#define DETERMINATION main
#define For(a,b,c,d) for(int a=b;a<=c;a+=d)
#pragma GCC optimize(2)
#pragma warning(disable:4996)
#define lldin(a) scanf("%lld", &a)
#define println(a) printf("%lld\n", a)
#define print(a) printf("%lld ", a)
#define reset(a, b) memset(a, b, sizeof(a))
#define debug std::cout<<"procedures above are available"<<"\n";
#define BigInteger __int128
using namespace std;
const long long INF = 2147483647;
const double PI = acos(-1);
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int mod = 1e9 + 7;
//template<typename T>
//inline BigInteger nextBigInteger()
//{
// BigInteger tmp = 0, si = 1;char c; c = getchar();
// while (!isdigit(c))
//{if (c == '-')si = -1;c = getchar();}
// while (isdigit(c))
// {tmp = tmp * 10 + c - '0';c = getchar();}
// return si * tmp;
//}
//std::ostream& operator<<(std::ostream& os, __int128 T)
//{
// if (T<0) os<<"-";if (T>=10 ) os<<T/10;if (T<=-10) os<<(-(T/10));
// return os<<( (int) (T%10) >0 ? (int) (T%10) : -(int) (T%10) ) ;
//}
//void output(BigInteger x)
//{
// if (x < 0)
// {x = -x;putchar('-');}
// if (x > 9) output(x / 10);
// putchar(x % 10 + '0');
// }
/**Operation Overlord 1944.6.6 Daybreak**/
/**Last Remote**/
ll cnt = 0, dis[123122];
ll relation[1500][1500];
ll Min(ll a, ll b)
{
if (b >= a)
return a;
else
return b;
}
bool vis[322122];
void spfa(ll st, ll limit)
{
for (int i = 1; i <= limit; i++)
{
dis[i] = 0;
vis[i] = false;
}
queue<ll>q;
q.push(st);
dis[st] = INF;
vis[st] = true;
while (!q.empty())
{
ll current = q.front();
q.pop();
vis[current] = false;
for (int i = 1; i <= limit; i++)
{
if (relation[current][i])
{
if (dis[i] < min(dis[current], relation[current][i]))
{
dis[i] = min(dis[current], relation[current][i]);
if (!vis[i])
{
vis[i] = true;
q.push(i);
}
}
}
}
}
}
int DETERMINATION()
{
//std::ios::sync_with_stdio(false);
//std::cin.tie(0), std::cout.tie(0);
ll t;
lldin(t);
for (int y = 1; y <= t; y++)
{
ll n, m;
lldin(n), lldin(m);
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
relation[i][j] = 0;
for (int i = 1; i <= m; i++)
{
ll tmp1, tmp2, tmp3;
lldin(tmp1), lldin(tmp2), lldin(tmp3);
relation[tmp1][tmp2] = tmp3;
relation[tmp2][tmp1] = tmp3;
}
spfa(1, n);
printf("Scenario #%d:\n", y);
println(dis[n]);
puts("");
}
return 0;
}POJ3268
正向建边，可以把从x到每个点的返回距离求出来，并且开一个专门的数组把他们储存好。
反向建边，可以把从每个点到x的前往距离求出来（都是以x为起点），与上组内的数据对应相加，然后求最大值即可。
//#include<pch.h>
#include <iostream>
#include <cstdio>
//#include <bits/stdc++.h>
#include<queue>
#include <map>
#include <algorithm>
#include <stack>
#include <iomanip>
#include <cstring>
#include <cmath>
#define DETERMINATION main
#define lldin(a) scanf("%lld", &a)
#define println(a) printf("%lld\n", a)
#define print(a) printf("%lld ", a)
#define reset(a, b) memset(a, b, sizeof(a))
#define debug cout<<"procedures above are available"<<endl;
using namespace std;
const int INF = 2147183646;
//const double PI = acos(-1);
typedef long long ll;
//typedef unsigned long long ull;
//typedef long double ld;
//const int mod = 10000000;
//const int tool_const = 19991126;
//const int tool_const2 = 33;
inline ll nextLong()
{
ll tmp = 0, si = 1;
char c;
c = getchar();
while (c > '9' || c < '0')
{
if (c == '-')
si = -1;
c = getchar();
}
while (c >= '0' && c <= '9')
{
tmp = tmp * 10 + c - '0';
c = getchar();
}
return si * tmp;
}
/**Maintain your determination.Nobody knows the magnificent landscape
at his destination before the arrival with stumble.**/
/**Last Remote**/
struct node
{
ll next,to,value;
}nodes[1231222];
ll heads[1312];
ll cnt=0;
void cons(ll from,ll to,ll value)
{
nodes[++cnt]=node{heads[from],to,value};
heads[from]=cnt;
}
struct status
{
ll pos,dis;
};
bool operator<(status a,status b)
{
return a.dis>b.dis;
}
ll dis[5400];
void spfa(ll st,ll limit)
{
for(int i=1;i<=limit;i++)
dis[i]=INF;
priority_queue<status>pq;
pq.push(status{st,0});
dis[st]=0;
while(!pq.empty())
{
status current=pq.top();
pq.pop();
if(current.dis>dis[current.pos])
continue;
else
{
for(int i=heads[current.pos];i!=-1;i=nodes[i].next)
{
ll t=nodes[i].to;
if(dis[t]>dis[current.pos]+nodes[i].value)
{
dis[t]=dis[current.pos]+nodes[i].value;
pq.push(status{t,dis[t]});
}
}
}
}
}
struct tmpnode
{
ll from,to,value;
}tmpnodes[2321112];
ll tmpans[12322];
int DETERMINATION()
{
ios::sync_with_stdio(false);
cin.tie(0),cout.tie(0);
ll n,m,x;
reset(heads,-1);
cin>>n>>m>>x;
for(int i=1;i<=m;i++)
{
ll tmp1,tmp2,tmp3;
cin>>tmp1>>tmp2>>tmp3;
tmpnodes[i].from=tmp1;
tmpnodes[i+m].from=tmp2;
tmpnodes[i].to=tmp2;
tmpnodes[i+m].to=tmp1;
tmpnodes[i].value=tmp3;
tmpnodes[i+m].value=tmp3;
}
for(int i=1;i<=m;i++)
cons(tmpnodes[i].from,tmpnodes[i].to,tmpnodes[i].value);
spfa(x,n);
for(int i=1;i<=n;i++)
tmpans[i]+=dis[i];
reset(heads,-1);
for(int i=m+1;i<=2*m;i++)
cons(tmpnodes[i].from,tmpnodes[i].to,tmpnodes[i].value);
spfa(x,n);
ll ans=0;
for(int i=1;i<=n;i++)
{
tmpans[i]+=dis[i];
ans=max(tmpans[i],ans);
}
cout<<ans<<endl;
return 0;
}POJ3259
求出一个负环就可以判断YES了
//#include<pch.h>
#include <iostream>
#include <cstdio>
//#include <bits/stdc++.h>
#include<queue>
#include <map>
#include <algorithm>
#include <stack>
#include <iomanip>
#include <cstring>
#include <cmath>
#define DETERMINATION main
#define lldin(a) scanf("%lld", &a)
#define println(a) printf("%lld\n", a)
#define print(a) printf("%lld ", a)
#define reset(a, b) memset(a, b, sizeof(a))
#define debug cout<<"procedures above are available"<<endl;
using namespace std;
const int INF = 2147183646;
const double PI = acos(-1);
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int mod = 10000000;
const int tool_const = 19991126;
const int tool_const2 = 33;
inline ll nextLong()
{
ll tmp = 0, si = 1;
char c;
c = getchar();
while (c > '9' || c < '0')
{
if (c == '-')
si = -1;
c = getchar();
}
while (c >= '0' && c <= '9')
{
tmp = tmp * 10 + c - '0';
c = getchar();
}
return si * tmp;
}
/**Maintain your determination.Nobody knows the magnificent landscape
at his destination before the arrival with stumble.**/
/**Last Remote**/
struct node
{
ll next,to,value;
}nodes[324323];
ll heads[234323],cnt=0;
void cons(ll from,ll to,ll value)
{
nodes[++cnt]=node{heads[from],to,value};
heads[from]=cnt;
}
ll dis[341222],cnt2[342322];
bool vis[3421122];
bool spfa(ll st,ll limit)
{
for(int i=1;i<=limit;i++)
{
dis[i]=INF;
vis[i]=false;
cnt2[i]=0;
}
queue<ll>q;
q.push(st);
dis[st]=0;
vis[st]=true;
while(q.empty()==false)
{
ll current=q.front();
q.pop();
vis[current]=false;
for(int i=heads[current];i!=-1;i=nodes[i].next)
{
ll t=nodes[i].to;
if(dis[t]>dis[current]+nodes[i].value)
{
dis[t]=dis[current]+nodes[i].value;
if(cnt2[t]>limit)
{
//cout<<t<<endl;
vis[t]=true;
return true;
}
if(!vis[t])
{
cnt2[t]++;
q.push(t);
vis[t]=true;
}
}
}
}
return false;
}
int DETERMINATION()
{
ios::sync_with_stdio(false);
cin.tie(0),cout.tie(0);
ll t;
cin>>t;
for(int y=1;y<=t;y++)
{
ll n,m,w;
reset(heads,-1);
cnt=0;
cin>>n>>m>>w;
for(int i=1;i<=m;i++)
{
ll tmp1,tmp2,tmp3;
cin>>tmp1>>tmp2>>tmp3;
cons(tmp1,tmp2,tmp3);
cons(tmp2,tmp1,tmp3);
//debug;
}
for(int i=1;i<=w;i++)
{
ll tmp1,tmp2,tmp3;
cin>>tmp1>>tmp2>>tmp3;
cons(tmp1,tmp2,-tmp3);
}
if(spfa(1,n))
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
}
return 0;
}POJ1860
本题是一个关于货币兑换的问题，这类问题一般利用最长路松弛式判断图内是否存在正环（即走过这一个环之后收益会变大）。
在本题中这个式子是这个样子:dis[t]<(dis[current]-nodes[i].cost)*nodes[i].value
struct node
{
ll next,to;
double value,cost;
}nodes[523333];
ll heads[523333],cnt1=0;
double dis[523332];
void construction(ll from,ll to,double value,double cost)
{
nodes[++cnt1]=node{heads[from],to,value,cost};
heads[from]=cnt1;
}
bool vis[352322];
ll cnt[341222];
bool spfa(ll st,ll limit,double v)
{
for(int i=1;i<=limit+87;i++)
{
dis[i]=(double)0;
cnt[i]=0;
vis[i]=false;
}
queue<ll>q;
q.push(st);
dis[st]=v;
while(!q.empty())
{
ll current=q.front();
q.pop();
vis[current]=false;
for(int i=heads[current];i!=-1;i=nodes[i].next)
{
ll t=nodes[i].to;
if(dis[t]<(dis[current]-nodes[i].cost)*nodes[i].value)
{
dis[t]=(dis[current]-nodes[i].cost)*nodes[i].value;
//cout<<dis[t]<<endl;
if(cnt[t]>=limit)
return true;
if(!vis[t])
{
vis[t]=true;
cnt[t]++;
q.push(t);
}
}
}
}
return false;
}
int DETERMINATION()
{
ios::sync_with_stdio(false);
cin.tie(0),cout.tie(0);
ll n,m,s;
ld v;
cin>>n>>m>>s>>v;
reset(heads,-1);
for(int i=1;i<=m;i++)
{
ll a,b;
double tmp1,tmp2,tmp3,tmp4;
cin>>a>>b>>tmp1>>tmp2>>tmp3>>tmp4;
construction(a,b,tmp1,tmp2);
construction(b,a,tmp3,tmp4);
}
if(spfa(s,n,v))
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
//cout<<dis[s]<<endl;
return 0;
}POJ3159
这是差分约束系统的一个题。
最短路表达式总是要满足dis[y]<=dis[x]+c，也即是dis[y]-dis[x]<=c，将题意中的不等式变换成差分约束方程组即可
struct node
{
ll next,to,value;
}nodes[250000];
ll heads[30055];
ll cnt=0;
void cons(ll from,ll to,ll value)
{
nodes[++cnt]=node{heads[from],to,value};
heads[from]=cnt;
}
struct status
{
ll pos,dis;
};
bool operator<(status a,status b)
{
return a.dis>b.dis;
}
ll dis[57666];
void dijkstra(ll st,ll limit)
{
for(int i=1;i<=limit;i++)
dis[i]=INF;
dis[st]=0;
priority_queue<status>pq;
pq.push(status{st,0});
while(!pq.empty())
{
status current=pq.top();
pq.pop();
if(current.dis>dis[current.pos])
continue;
else
{
for(int i=heads[current.pos];i!=-1;i=nodes[i].next)
{
ll t=nodes[i].to;
if(dis[t]>dis[current.pos]+nodes[i].value)
{
dis[t]=dis[current.pos]+nodes[i].value;
pq.push(status{t,dis[t]});
}
}
}
}
}
int DETERMINATION()
{
//ios::sync_with_stdio(false);
//cin.tie(0),cout.tie(0);
ll n,m;
lldin(n),lldin(m);
reset(heads,-1);
for(int i=1;i<=m;i++)
{
ll tmp1,tmp2,tmp3;
lldin(tmp1),lldin(tmp2),lldin(tmp3);
cons(tmp1,tmp2,tmp3);
}
dijkstra(1,n);
ll ans=0;
println(dis[n]);
return 0;
}
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