上了大二，又有新生來了，壓力也來了，不能像大一這麼鬆懈了，所以開始每天刷刷題，想根據專題一個知識點一個知識點過

　　今天是最短路專題，對於最短路，能想到的方法大概就是dijkstra演算法（求單源最短路不含負環）O（n^2）如果使用堆優化，就是O（mlogn）,還有就是floyd演算法（求圖上任意兩點的最短路）O(N^3），以及bellman演算法（可求解含負環的單源最短路問題）

　　對於a題 ，dijkstra演算法

A - Til the Cows Come Home

 Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants to get back as quickly as possible. 

Farmer John's field has N (2 <= N <= 1000) landmarks in it, uniquely numbered 1..N. Landmark 1 is the barn; the apple tree grove in which Bessie stands all day is landmark N. Cows travel in the field using T (1 <= T <= 2000) bidirectional cow-trails of various lengths between the landmarks. Bessie is not confident of her navigation ability, so she always stays on a trail from its start to its end once she starts it. 

Given the trails between the landmarks, determine the minimum distance Bessie must walk to get back to the barn. It is guaranteed that some such route exists.

　　求1-n的最短路，直接使用dij演算法，要注意邊的更新

#include<cstdio>
#include<cstring>
#include <iostream>
#define MAXN 1005
#define INF 1<<29
using namespace std;
    int n,m;
    int num[MAXN][MAXN];
    void dij(){
        int dp[MAXN];
        int vis[MAXN];
        for(int i=1;i<=n;i++){
            dp[i]
 
=num[1][i];
            vis[i]=0;
        }
        vis[1]=1;
        for(int i=1;i<=n;i++){
            int min=INF,sw=1;
            for(int j=1;j<=n;j++)
                if(vis[j]==0&&dp[j]<min){
                    sw=j;
                    min=dp[j];
                }
            vis[sw]
 
=1;
            for(int j=1;j<=n;j++)
                if(vis[j]==0&&dp[j]>dp[sw]+num[sw][j]) dp[j]=dp[sw]+num[sw][j];
        }
        printf("%d\n",dp[n]);
    }
int main()
{
    while(~scanf("%d %d",&m,&n)){
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                if(i==j) num[i][j]=0;
                else num[j][i]=num[i][j]=INF;
        while(m--){
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            if(c<num[a][b]) num[a][b]=num[b][a]=c;
        }
        dij();
    }
    return 0;
}

B - 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. 

　　wa了超級多次，結果告訴我不能用g++交，要用c++，哇

　　不過還是直接用dij演算法跑過去就行了

　　

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#define MAXN 305
#define INF 1e9
using namespace std;
double num[MAXN][MAXN];
int n,m;
    void dijkstra(int x){
        for(int k=1;k<=n;k++)
            for(int i=1;i<=n;i++)
                for(int j=1;j<=n;j++)
                    num[i][j]=min(num[i][j],max(num[i][k],num[k][j]));
        return ;
    }
int main()
{
    int x[MAXN],y[MAXN];
    int text=1;
    while(cin>>n&&n!=0){
    //printf("Scenario #%d\n",text++);
    memset(num,0,sizeof(num));
    for(int i=1; i<=n; i++)
            scanf("%d%d",&x[i],&y[i]);
        for(int i=1; i<=n; i++)
            for(int j=i+1; j<=n; j++)
                num[i][j]=num[j][i]=sqrt(double(x[i]-x[j])*(x[i]-x[j])+double(y[i]-y[j])*(y[i]-y[j]));

    dijkstra(1);
    printf("Scenario #%d\nFrog Distance = %.3lf\n\n",text++,num[1][2]);
    }
    return 0;
}

C - Heavy Transportation

Background 
Hugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand business. But he needs a clever man who tells him whether there really is a way from the place his customer has build his giant steel crane to the place where it is needed on which all streets can carry the weight. 
Fortunately he already has a plan of the city with all streets and bridges and all the allowed weights.Unfortunately he has no idea how to find the the maximum weight capacity in order to tell his customer how heavy the crane may become. But you surely know. 

Problem 
You are given the plan of the city, described by the streets (with weight limits) between the crossings, which are numbered from 1 to n. Your task is to find the maximum weight that can be transported from crossing 1 (Hugo's place) to crossing n (the customer's place). You may assume that there is at least one path. All streets can be travelled in both directions.

Input

The first line contains the number of scenarios (city plans). For each city the number n of street crossings (1 <= n <= 1000) and number m of streets are given on the first line. The following m lines contain triples of integers specifying start and end crossing of the street and the maximum allowed weight, which is positive and not larger than 1000000. There will be at most one street between each pair of crossings.

Output

The output for every scenario begins with a line containing "Scenario #i:", where i is the number of the scenario starting at 1. Then print a single line containing the maximum allowed weight that Hugo can transport to the customer. Terminate the output for the scenario with a blank line.

  思路：就是dij演算法跑呀，但是不知道為什麼一開始寫的dij一直tle，後面重新寫了一遍就過了

就把兩份程式碼都發一遍吧

第一份 TLE

#include <iostream>
#include <cstring>
#include <cmath>
#include <cstdio>
#define MAXN 1005
using namespace std;
int num[MAXN][MAXN];
int dp[MAXN];
int vis[MAXN];
int n;
 void dij(){
    for(int i=1;i<=n;i++){
            vis[i]=0;
            dp[i]=num[1][i];
    }
    vis[1]=1;
    for(int i=1;i<=n;i++){
        int maxx=0;
        int t;
        for(int j=1;j<=n;j++)
            if(!vis[j]&&dp[j]>maxx) maxx=dp[t=j];
        vis[t]=1;
        for(int k=1;k<=n;k++)
            if(!vis[k]) dp[k]=max(dp[k],min(dp[t],num[t][k]));
    }
    return ;
 }
int main(){
    int t;
    int m;
    cin>>t;
    for(int i=1;i<=t;i++){
        cin>>n>>m;
        memset(num,0,n*sizeof(num[0]));
        for(int i=0;i<m;i++){
            int a,b,c;
            cin>>a>>b>>c;
            num[a][b]=num[b][a]=c;
        }
        dij();
        printf("Scenario #%d:\n%d\n\n",i,dp[n]);
    }
    return 0;
}

第二份 AC

#include <iostream>
#include <cstring>
#include <cmath>
#include <cstdio>
#define MAXN 1005
using namespace std;
int num[MAXN][MAXN];
int dp[MAXN];
int vis[MAXN];
int n;
 void dij(){
    for(int i=1;i<=n;i++){
            vis[i]=0;
            dp[i]=num[1][i];
    }
    vis[1]=1;
    for(int i=1;i<=n;i++){
        int maxx=0;
        int t;
        for(int j=1;j<=n;j++)
            if(!vis[j]&&dp[j]>maxx) maxx=dp[t=j];
        vis[t]=1;
        for(int k=1;k<=n;k++)
            if(!vis[k]) dp[k]=max(dp[k],min(dp[t],num[t][k]));
    }
    return ;
 }
int main(){
    int t;
    int m;
    cin>>t;
    for(int i=1;i<=t;i++){
        cin>>n>>m;
        memset(num,0,sizeof(num));
        for(int i=0;i<m;i++){
            int a,b,c;
            cin>>a>>b>>c;
            num[a][b]=num[b][a]=c;
        }
        dij();
        printf("Scenario #%d:\n%d\n\n",i,dp[n]);
    }
    return 0;
}

D - Silver Cow Party

 

One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M≤ 100,000) unidirectional (one-way roads connects pairs of farms; road i requires T_i (1 ≤ T_i ≤ 100) units of time to traverse.

Each cow must walk to the party and, when the party is over, return to her farm. Each cow is lazy and thus picks an optimal route with the shortest time. A cow's return route might be different from her original route to the party since roads are one-way.

Of all the cows, what is the longest amount of time a cow must spend walking to the party and back?

Input

Line 1: Three space-separated integers, respectively:  N,  M, and  X 
Lines 2..  M+1: Line  i+1 describes road  i with three space-separated integers:  A_i, B_i, and  T_i. The described road runs from farm  A_i to farm  B_i, requiring  T_i time units to traverse.

Output

Line 1: One integer: the maximum of time any one cow must walk. 　思路：單向邊的圖，要從1到n然後又從n到1，就跑兩邊dij演算法唄

#include <iostream>
#include <cstring>
#include <cmath>
#include <cstdio>
#define MAXN 1005
#define INF 1<<30
using namespace std;
int num[MAXN][MAXN];
int dp[MAXN];
int vis[MAXN];
int dp1[MAXN];
int n;
    void dij(int x){
        for(int i=1;i<=n;i++){
            dp[i]=num[i][x];
            vis[i]=0;
        }
        vis[x]=1;
        for(int i=2;i<=n;i++){
            int min=INF,sw=1;
            for(int j=1;j<=n;j++)
                if(vis[j]==0&&dp[j]<min){
                    sw=j;
                    min=dp[j];
                }
            if(min==INF) break;
            vis[sw]=1;
            for(int j=1;j<=n;j++)
                if(vis[j]==0&&dp[j]>dp[sw]+num[j][sw]) dp[j]=dp[sw]+num[j][sw];
        }
    }
    void dij1(int x){
        for(int i=1;i<=n;i++){
            dp1[i]=num[x][i];
            vis[i]=0;
        }
        vis[x]=1;
        for(int i=2;i<=n;i++){
            int min=INF,sw=1;
            for(int j=1;j<=n;j++)
                if(vis[j]==0&&dp1[j]<min){
                    sw=j;
                    min=dp1[j];
                }
            if(min==INF) break;
            vis[sw]=1;
            for(int j=1;j<=n;j++)
                if(vis[j]==0&&dp1[j]>dp1[sw]+num[sw][j]) dp1[j]=dp1[sw]+num[sw][j];
        }
    }
int main()
{
    int m,x;
    while(cin>>n>>m>>x){
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            if(i==j) num[i][j]=0;
            num[i][j]=INF;
        }
    }
    for(int i=0;i<m;i++){
        int a,b,t;
        cin>>a>>b>>t;
        if(t<num[a][b]) num[a][b]=t;
    }
    dij(x);
    /*
    for(int i=1;i<=n;i++){
        for(int j=i;j<=n;j++){
        swap(num[i][j],num[j][i]);
        }
    }
    */
    dij1(x);
    int maxn=0;
    //for(int i=1;i<=n;i++) cout<<dp[i]<<' '<<dp1[i]<<endl;
    for(int i=1;i<=n;i++){
        if(dp[i]!=INF&&dp1[i]!=INF)
            maxn=max(maxn,dp[i]+dp1[i]);
    }
    cout<<maxn<<endl;
    }
    return 0;
}