最短路 summary

有四种类型；

单源：dij，spfa，bellman-ford

多源：floyd

dij有两种：

一个复杂度为n^2，一个复杂度是m*logn

畅通工程续

某省自从实行了很多年的畅通工程计划后，终于修建了很多路。不过路多了也不好，每次要从一个城镇到另一个城镇时，都有许多种道路方案可以选择，而某些方案要比另一些方案行走的距离要短很多。这让行人很困扰。

现在，已知起点和终点，请你计算出要从起点到终点，最短需要行走多少距离。

Input本题目包含多组数据，请处理到文件结束。

每组数据第一行包含两个正整数N和M(0<N<200,0<M<1000)，分别代表现有城镇的数目和已修建的道路的数目。城镇分别以0～N-1编号。

接下来是M行道路信息。每一行有三个整数A,B,X(0<=A,B<N,A!=B,0<X<10000),表示城镇A和城镇B之间有一条长度为X的双向道路。

再接下一行有两个整数S,T(0<=S,T<N)，分别代表起点和终点。Output对于每组数据，请在一行里输出最短需要行走的距离。如果不存在从S到T的路线，就输出-1.

Sample Input

3 3
0 1 1
0 2 3
1 2 1
0 2
3 1
0 1 1
1 2

Sample Output

2
-1

#include <stdio.h>//n^2
#include <stdlib.h>
#include <string.h>
#define inf 0x3f3f3f3f
using namespace std;
const int maxn=220;
int w[maxn][maxn],d[maxn];
bool vis[maxn];
int n,m;

void dijkstra(int s)
{
	memset(vis,0,sizeof(vis));
	for(int i=0;i<n;i++) d[i]=inf;
	d[s]=0;

	for(int i=0;i<n;i++)
	{
		int x=-1,y=inf;
		for(int j=0;j<n;j++)
		{
			if(vis[j]==0&&d[j]<=y) y=d[x=j];
		}
		if(x==-1) continue;
		vis[x]=1;
		for(int j=0;j<n;j++)
		{
			if(!vis[j]&&d[j]>d[x]+w[x][j]) d[j]=d[x]+w[x][j];
		}
	}
}

int main()
{
	while(scanf("%d %d",&n,&m)!=EOF)
	{
		for(int i=0;i<=n;i++)
		{
			for(int j=0;j<=n;j++)
			{
				if(i==j) w[i][j]=0;
				else w[i][j]=inf;
			}
		}
		for(int i=0;i<m;i++)
		{
			int a,b,x;
			scanf("%d%d%d",&a,&b,&x);
			if(w[a][b]>x) w[a][b]=w[b][a]=x;
		}
		int s,t;
		scanf("%d%d",&s,&t);
		dijkstra(s);
		if(d[t]>=inf) printf("-1\n");
		else printf("%d\n",d[t]);
	}
	return 0;
}

　　

#include <stdio.h>//mlogn
#include <stdlib.h>
#include <string.h>
#include <queue>
#include <vector>
#define inf 0x3f3f3f3f
using namespace std;
const int maxn=1100;
int d[maxn],n,m;
bool vis[maxn];
struct edge
{
	int from,to,dist;
	edge(int from,int to,int dist):from(from),to(to),dist(dist){}
};
struct heapnode
{
	int d,u;
	heapnode(int d,int u) : d(d),u(u) {}
	bool operator<(const heapnode &a) const{
	    return a.d<d;
	}
};

queue<edge>q[maxn];

void dijkstra(int s)
{
	priority_queue<heapnode>que;
	for(int i=0;i<=n;i++) d[i]=inf;
	d[s]=0;
	memset(vis,0,sizeof(vis));
	que.push(heapnode(0,s));
	while(!que.empty())
	{
		heapnode x=que.top();que.pop();
		int u=x.u;
		if(vis[u]) continue;
		vis[u]=1;
		while(!q[u].empty())
		{
			edge e=q[u].front();
			q[u].pop();
			if(d[e.to]>d[u]+e.dist)
			{
				d[e.to]=d[u]+e.dist;
				que.push(heapnode(d[e.to],e.to));
			}
		}
	}
}

int main()
{
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(int i=0;i<n;i++) while(!q[i].empty()) q[i].pop();
		for(int i=0;i<m;i++)
		{
		    int a,b,x;
		    scanf("%d%d%d",&a,&b,&x);
		    q[a].push(edge(a,b,x));
		    q[b].push(edge(b,a,x));
		}
		int s,t;
		scanf("%d%d",&s,&t);
		dijkstra(s);
		if(d[t]>=inf) printf("-1\n");
		else printf("%d\n",d[t]);
	}
	return 0;
}

　　

#include <stdio.h>//mlogn
#include <stdlib.h>
#include <string.h>
#include <queue>
#include <vector>
#define inf 0x3f3f3f3f
using namespace std;
const int maxn=1100;
int d[maxn],n,m;
bool vis[maxn];
struct edge
{
	int from,to,dist;
	edge(int from,int to,int dist):from(from),to(to),dist(dist){}
};
struct heapnode
{
	int d,u;
	heapnode(int d,int u) : d(d),u(u) {}
	bool operator<(const heapnode &a) const{
	    return a.d<d;
	}
};

vector<edge> vec;
vector<int> g[maxn];

void dijkstra(int s)
{
	priority_queue<heapnode>que;
	for(int i=0;i<=n;i++) d[i]=inf;
	d[s]=0;
	memset(vis,0,sizeof(vis));
	que.push(heapnode(0,s));
	while(!que.empty())
	{
		heapnode x=que.top();que.pop();
		int u=x.u;
		if(vis[u]) continue;
		vis[u]=1;
		for(int i=0;i<g[u].size();i++)
		{
			edge &e=vec[g[u][i]];
			if(d[e.to]>d[u]+e.dist)
			{
				d[e.to]=d[u]+e.dist;
				que.push(heapnode(d[e.to],e.to));
			}
		}
	}
}

int main()
{
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(int i=0;i<=n;i++) g[i].clear();
		vec.clear();
		for(int i=0;i<m;i++)
		{
		    int a,b,x;
		    scanf("%d%d%d",&a,&b,&x);
		    vec.push_back(edge(a,b,x));
		    int c=vec.size();
		    g[a].push_back(c-1);
		    vec.push_back(edge(b,a,x));
		    c=vec.size();
		    g[b].push_back(c-1);
		}
		int s,t;
		scanf("%d%d",&s,&t);
		dijkstra(s);
		if(d[t]>=inf) printf("-1\n");
		else printf("%d\n",d[t]);
	}
	return 0;
}

　　

C - find the longest of the shortest

Marica is very angry with Mirko because he found a new girlfriend and she seeks revenge.Since she doesn't live in the same city, she started preparing for the long journey.We know for every road how many minutes it takes to come from one city to another.
Mirko overheard in the car that one of the roads is under repairs, and that it is blocked, but didn't konw exactly which road. It is possible to come from Marica's city to Mirko's no matter which road is closed.

Marica will travel only by non-blocked roads, and she will travel by shortest route. Mirko wants to know how long will it take for her to get to his city in the worst case, so that he could make sure that his girlfriend is out of town for long enough.Write a program that helps Mirko in finding out what is the longest time in minutes it could take for Marica to come by shortest route by non-blocked roads to his city.

InputEach case there are two numbers in the first row, N and M, separated by a single space, the number of towns,and the number of roads between the towns. 1 ≤ N ≤ 1000, 1 ≤ M ≤ N*(N-1)/2. The cities are markedwith numbers from 1 to N, Mirko is located in city 1, and Marica in city N.

In the next M lines are three numbers A, B and V, separated by commas. 1 ≤ A,B ≤ N, 1 ≤ V ≤ 1000.Those numbers mean that there is a two-way road between cities A and B, and that it is crossable in V minutes.OutputIn the first line of the output file write the maximum time in minutes, it could take Marica to come to Mirko.Sample Input

5 6
1 2 4
1 3 3
2 3 1
2 4 4
2 5 7
4 5 1

6 7
1 2 1
2 3 4
3 4 4
4 6 4
1 5 5
2 5 2
5 6 5

5 7
1 2 8
1 4 10
2 3 9
2 4 10
2 5 1
3 4 7
3 5 10

Sample Output

11
13
27

这个和之前差不多，有一点点区别，就是让你求出最短路，然后在最短路里面枚举每一边断了，这个最坏的情况用的时间。

这个用邻接矩阵写的，用队列的太麻烦

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#define inf 0x3f3f3f3f
using namespace std;
const int maxn=1100;
int w[maxn][maxn];
int d[maxn];
int path[maxn];
bool flag,vis[maxn];
int n,m;

void dijkstra(int s)
{
	for(int i=0;i<=n;i++)
	{
		d[i]=inf;
		vis[i]=0;
	}
	d[s]=0;

	for(int i=1;i<=n;i++)
	{
		int x,m=inf;
	    for(int j=1;j<=n;j++) if(!vis[j]&&d[j]<m) m=d[x=j];
	    vis[x]=1;
	    for(int j=1;j<=n;j++)
	    {
	    	if(d[j]>d[x]+w[x][j])
			{
				d[j]=d[x]+w[x][j];
	    	    if(!flag) path[j]=x;
			}
		}
	}
}

int main()
{
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(int i=1;i<=n;i++)
		{
			for(int j=1;j<=n;j++)
			{
				if(i==j) w[i][j]=0;
				else w[i][j]=inf;
			}
		}
		for(int i=1;i<=m;i++)
		{
			int a,b,x;
			scanf("%d%d%d",&a,&b,&x);
			w[a][b]=x;
			w[b][a]=x;
		}
		flag=0;
		dijkstra(1);
	//	printf("%d\n",d[n]);
		flag=1;
		int tmp,ans=d[n];
		for(int i=n;i>1;i=path[i])
		{
			tmp=w[path[i]][i];
			w[path[i]][i]=inf;
			dijkstra(1);
			w[path[i]][i]=tmp;
			if(d[n]>=inf) continue;
			ans=max(ans,d[n]);
		//	printf("%d\n",i);
		}
		printf("%d\n",ans);
	}
	return 0;
}

　　

D - Frogger

Freddy Frog is sitting on a stone in the middle of a lake. Suddenly he notices Fiona Frog who is sitting on another stone. He plans to visit her, but since the water is dirty and full of tourists' sunscreen, he wants to avoid swimming and instead reach her by jumping.
Unfortunately Fiona's stone is out of his jump range. Therefore Freddy considers to use other stones as intermediate stops and reach her by a sequence of several small jumps.

To execute a given sequence of jumps, a frog's jump range obviously must be at least as long as the longest jump occuring in the sequence.

The frog distance (humans also call it minimax distance) between two stones therefore is defined as the minimum necessary jump range over all possible paths between the two stones.

You are given the coordinates of Freddy's stone, Fiona's stone and all other stones in the lake. Your job is to compute the frog distance between Freddy's and Fiona's stone.

Input

The input will contain one or more test cases. The first line of each test case will contain the number of stones n (2<=n<=200). The next n lines each contain two integers xi,yi (0 <= xi,yi <= 1000) representing the coordinates of stone #i. Stone #1 is Freddy's stone, stone #2 is Fiona's stone, the other n-2 stones are unoccupied. There's a blank line following each test case. Input is terminated by a value of zero (0) for n.

Output

For each test case, print a line saying "Scenario #x" and a line saying "Frog Distance = y" where x is replaced by the test case number (they are numbered from 1) and y is replaced by the appropriate real number, printed to three decimals. Put a blank line after each test case, even after the last one.

Sample Input

2
0 0
3 4

3
17 4
19 4
18 5

0

Sample Output

Scenario #1
Frog Distance = 5.000

Scenario #2
Frog Distance = 1.414

这个题目有点特别，要结合贪心，在d[i]的i的取值的时候要注意不是直接取求出到达i的最短路径长度，而是，从1号位置到这个位置的最长边的最小值。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <queue>
#include <algorithm>
#include <math.h>
#define inf 0x3f3f3f3f
using namespace std;
const int maxn=250;
int n;
double d[maxn];
bool vis[maxn];
double w[maxn][maxn];

struct node
{
	double x,y;
	node(double x=0,double y=0):x(x),y(y){}
}exa[maxn];

void dijkstra()
{
	memset(vis,0,sizeof(vis));
	for(int i=1;i<=n;i++) d[i]=w[1][i];
	for(int i=1;i<=n;i++)
	{
		int x=-1;
		double m=inf;
		for(int j=1;j<=n;j++)
		{
			if(!vis[j]&&d[j]<m) m=d[x=j];
		}
		if(x!=-1)
		{
			vis[x]=1;
		    for(int j=1;j<=n;j++)
		    {
		    	if(!vis[j]&&d[j]>max(d[x],w[x][j])) d[j]=max(d[x],w[x][j]);
			}
		}
	}
} 

int main()
{
	int cnt=0;
	while(scanf("%d\n",&n)!=EOF&&n)
	{
		for(int i=1;i<=n;i++)
		{
			scanf("%lf%lf",&exa[i].x,&exa[i].y);
		}
		for(int i=1;i<=n;i++)
		{
			for(int j=i;j<=n;j++)
			{
				w[i][j]=w[j][i]=sqrt((exa[i].x-exa[j].x)*(exa[i].x-exa[j].x)+(exa[i].y-exa[j].y)*(exa[i].y-exa[j].y));
			}
		}
		dijkstra();
		printf("Scenario #%d\nFrog Distance = %.3lf\n\n",++cnt,d[2]);
	}
	return 0;
}