題目描述 Like everyone else, FJ is always thinking up ways to increase his revenue. To this end, he has set up a series of tolls that the cows will pay wh ...
題目描述
Like everyone else, FJ is always thinking up ways to increase his revenue. To this end, he has set up a series of tolls that the cows will pay when they traverse the cowpaths throughout the farm.
The cows move from any of the N (1 <= N <= 250) pastures conveniently numbered 1..N to any other pasture over a set of M (1 <= M <= 10,000) bidirectional cowpaths that connect pairs of different pastures A_j and B_j (1 <= A_j <= N; 1 <= B_j <= N). FJ has assigned a toll L_j (1 <= L_j <= 100,000) to the path connecting pastures A_j and B_j.
While there may be multiple cowpaths connecting the same pair of pastures, a cowpath will never connect a pasture to itself. Best of all, a cow can always move from any one pasture to any other pasture by following some sequence of cowpaths.
In an act that can only be described as greedy, FJ has also assigned a toll C_i (1 <= C_i <= 100,000) to every pasture. The cost of moving from one pasture to some different pasture is the sum of the tolls for each of the cowpaths that were traversed plus a *single additional toll* that is the maximum of all the pasture tolls encountered along the way, including the initial and destination pastures.
The patient cows wish to investigate their options. They want you to write a program that accepts K (1 <= K <= 10,000) queries and outputs the minimum cost of trip specified by each query. Query i is a pair of numbers s_i and t_i (1 <= s_i <= N; 1 <= t_i <= N; s_i != t_i) specifying a starting and ending pasture.
Consider this example diagram with five pastures:
The 'edge toll' for the path from pasture 1 to pasture 2 is 3. Pasture 2's 'node toll' is 5.
To travel from pasture 1 to pasture 4, traverse pastures 1 to 3 to 5 to 4. This incurs an edge toll of 2+1+1=4 and a node toll of 4 (since pasture 5's toll is greatest), for a total cost of 4+4=8.
The best way to travel from pasture 2 to pasture 3 is to traverse pastures 2 to 5 to 3. This incurs an edge toll of 3+1=4 and a node toll of 5, for a total cost of 4+5=9.
跟所有人一樣，農夫約翰以著寧教我負天下牛，休叫天下牛負我的偉大精神，日日夜夜苦思生 財之道。為了發財，他設置了一系列的規章制度，使得任何一隻奶牛在農場中的道路行走，都 要向農夫約翰上交過路費。 農場中由N（1 <= N <= 250）片草地（標號為1到N），並且有M（1 <= M <= 10000）條 雙向道路連接草地A_j和B_j（1 <= A_j <= N; 1 <= B_j <= N）。
奶牛們從任意一片草 地出發可以抵達任意一片的草地。FJ已經在連接A_j和B_j的雙向道路上設置一個過路費L_j （1 <= L_j <= 100,000）。 可能有多條道路連接相同的兩片草地，但是不存在一條道路連接一片草地和這片草地本身。最 值得慶幸的是，奶牛從任意一篇草地出發，經過一系列的路徑，總是可以抵達其它的任意一片 草地。 除了貪得無厭，叫獸都不知道該說什麼好。
FJ竟然在每片草地上面也設置了一個過路費C_i （1 <= C_i <= 100000）。從一片草地到另外一片草地的費用，是經過的所有道路的過路 費之和，加上經過的所有的草地（包括起點和終點）的過路費的最大值。 任勞任怨的牛們希望去調查一下她們應該選擇那一條路徑。
她們要你寫一個程式，接受K（1 <= K <= 10,000）個問題並且輸出每個詢問對應的最小花費。第i個問題包含兩個數字s_i 和t_i（1 <= s_i <= N; 1 <= t_i <= N; s_i != t_i），表示起點和終點的草地。
輸入輸出格式
輸入格式：-
Line 1: Three space separated integers: N, M, and K
-
Lines 2..N+1: Line i+1 contains a single integer: C_i
- Lines N+2..N+M+1: Line j+N+1 contains three space separated
integers: A_j, B_j, and L_j
- Lines N+M+2..N+M+K+1: Line i+N+M+1 specifies query i using two space-separated integers: s_i and t_i
- Lines 1..K: Line i contains a single integer which is the lowest cost of any route from s_i to t_i
輸入輸出樣例
輸入樣例#1：5 7 2 2 5 3 3 4 1 2 3 1 3 2 2 5 3 5 3 1 5 4 1 2 4 3 3 4 4 1 4 2 3輸出樣例#1：
8 9
給大家介紹一種不用排序就能AC的方法
（排序的方法請看這位大神的博客：http://www.cnblogs.com/peter-le/p/6014643.html）
我們可以分別記錄下從點x到點y 的最短路徑的長度和在最短路上的花費
然後就是套Floyd的模板。
但是註意，因為你的長度和花費是分開記錄的
所以一遍Floyd跑出來的不一定是最小值
我們可以多跑幾遍試試
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 using namespace std; 6 const int MAXN=251; 7 const int maxn=0x7ffffff; 8 void read(int & n) 9 { 10 char c='+';int x=0; 11 while(c<'0'||c>'9')c=getchar(); 12 while(c>='0'&&c<='9') 13 { 14 x=x*10+c-48; 15 c=getchar(); 16 } 17 n=x; 18 19 } 20 int a[MAXN][MAXN]; 21 int spend[MAXN][MAXN]; 22 int b[MAXN]; 23 int n,m,q; 24 void floyd() 25 { 26 for(int k=1;k<=n;k++) 27 for(int i=1;i<=n;i++) 28 for(int j=1;j<=n;j++) 29 { 30 int p=a[i][k]+a[k][j]+max(spend[i][k],spend[k][j]); 31 if((p<a[i][j]+spend[i][j])&&a[i][k]<maxn&&a[k][j]<maxn) 32 { 33 //a[i][j]=a[i][j]-max(spend[i],spend[j]); 34 a[i][j]=a[i][k]+a[k][j]; 35 spend[i][j]=max(spend[i][k],spend[k][j]); 36 } 37 } 38 } 39 int main() 40 { 41 read(n);read(m);read(q); 42 for(int i=1;i<=n;i++) 43 read(b[i]); 44 for(int i=1;i<=n;i++) 45 for(int j=1;j<=n;j++) 46 { 47 if(i==j) 48 a[i][j]=0; 49 else 50 a[i][j]=maxn; 51 } 52 53 for(int i=1;i<=m;i++) 54 { 55 int x,y,z; 56 read(x);read(y);read(z); 57 if(a[x][y]>z) 58 { 59 a[x][y]=z; 60 a[y][x]=z; 61 spend[x][y]=max(b[x],b[y]); 62 spend[y][x]=max(b[x],b[y]); 63 } 64 65 } 66 floyd(); 67 floyd(); 68 floyd(); 69 for(int i=1;i<=q;i++) 70 { 71 int x,y; 72 read(x);read(y); 73 printf("%d\n",a[x][y]+spend[x][y]); 74 } 75 return 0; 76 }
