Title Description
Once upon a kingdom, in n between cities has m path is connected, and at most any road directly connected between the two cities. After a serious war, there are d roads have been destroyed. The king wants to repair the road system in the country, there are now two major cities A and B traffic between the interrupt, the king hoped to restore the connection between the two cities as soon as possible. Your task is to repair some of the roads A and B minimum connection between the restoration and repair of road length requirements.
Input Format
A first behavior file input integer n- ( 2 <n≤100 ), represents the number of cities. These cities are numbered from 1 to the n- .
Conduct a second integer m ( n--1≤m≤2n (n--. 1) ), the number of road representation.
The next m lines, each line . 3 integers I, J, K ( 1≤i, j ≦ N, I ≠ J, 0 <k≤100 ), represents city II I and jj have a length between J K connected to the road.
Next, a behavior of an integer d ( 1≤d≤m ), representing the number of post-war damaged roads. In the next d rows, each row two integers i and j , represent city II i and jj road directly connected between j is destroyed.
The last row of two integers A and B, on behalf of two major cities need to restore traffic.
Output Format
The output file a single integer representing restore A and the minimum total length of road traffic between A and B in need of repair.
Destroyed some paths, fix his price is the length of the path, not the path of destruction, fix it at the cost of 0 (* Front chain to build the star map)
SPFA for the most short-circuit:
1 #include <cstdio> 2 #include <iostream> 3 #include <algorithm> 4 #include <cmath> 5 #include <queue> 6 using namespace std; 7 int n,m,d; 8 int a,b; 9 int x,y; 10 int dis[10000]; 11 bool vis[10000]; 12 bool mark[1000][1000]; 13 struct node 14 { 15 int x,y,w; 16 }edge[10000]; 17 struct ask 18 { 19 int w; 20 int to; 21 int next; 22 }ed[10000]; 23 int head [10000]; 24 int tot =0; 25 void add (int u,int v,int w) 26 { 27 ed[++tot].w=w; 28 ed[tot].to=V; 29 ED [TOT] = .next head [U]; 30 head [U] = TOT; 31 is } 32 void the SPFA ( int S) 33 is { 34 is Queue < int > Q; 35 for ( int I = . 1 ; I <= n-; I ++) { // initialization, 36 DIS [I] = 0x7FFFFFFF ; 37 [ VIS [I] = 0 ; 38 is } 39 q.push (S); // put starting 40 DIS [S] =0 ; // themselves to themselves as 0; 41 is VIS [S] = . 1 ; 42 is the while (! Q.empty ()) { 43 is int U = q.front (); 44 is q.pop (); VIS [U] = 0 ; // pop-up head of the queue 45 for ( int I = head [U]; I; I = ED [I] .next) { 46 is int V = ED [I] .to; 47 IF (VIS [V]) Continue ; 48 IF (DIS [V]> DIS [U] + ED [I] .W) { 49 DIS [V] = DIS [U] + ED [I] .W; 50 if(!vis[v]){ 51 vis[v]=1; 52 q.push(v);//重新放入； 53 } 54 } 55 } 56 } 57 } 58 int main() 59 { 60 scanf ("%d",&n); 61 scanf ("%d",&m); 62 for (int i = 1;i <= m;i++) 63 { 64 scanf ("%d%d%d",&edge[i].x,&edge[i].y,&edge[i].w); 65 } 66 scanf ("%d",&d); 67 for (int i = 1;i <= d;i++) 68 { 69 scanf ("%d%d",&x,&y); 70 mark[x][y]=1; 71 mark[y][x]=1; 72 } 73 scanf ("%d%d",&a,&b); 74 for (int i = 1;i <= m;i++) 75 { 76 if (mark[edge[i].x][edge[i].y]==1) 77 { 78 add(edge[i].x,edge[i].y,edge[i].w); 79 add(edge[i].y,edge[i].x,edge[i].w); 80 } 81 else 82 { 83 add(edge[i].x,edge[i].y,0); 84 add(edge[i].y,edge[i].x,0); 85 } 86 } 87 SPFA(a); 88 cout<<dis[b]<<endl; 89 return 0; 90 }