标签：stream using lang 地方 pac cee 通过 ring cow
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 Ti (1 ≤ Ti ≤ 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
Output
Sample Input
4 8 2 1 2 4 1 3 2 1 4 7 2 1 1 2 3 5 3 1 2 3 4 4 4 2 3
Sample Output
10
Hint
1 #include <cstdio> 2 #include <fstream> 3 #include <algorithm> 4 #include <cmath> 5 #include <deque> 6 #include <vector> 7 #include <queue> 8 #include <string> 9 #include <cstring> 10 #include <map> 11 #include <stack> 12 #include <set> 13 #include <sstream> 14 #include <iostream> 15 #define mod 998244353 16 #define eps 1e-6 17 #define ll long long 18 #define INF 0x3f3f3f3f 19 using namespace std; 20 21 //ma用来存放两个点之间的距离 22 int ma[1010][1010]; 23 //dis用来存放到1之间的距离 24 int dis1[1010]; 25 int dis2[1010]; 26 //vis数组标记已经找过的最大起重 27 bool vis[1010]; 28 29 //从x到其他地方的正向图 30 void dijkstra1(int m,int q) 31 { 32 memset(vis,0,sizeof(vis)); 33 memset(dis1,INF,sizeof(dis1)); 34 vis[q]=0; 35 //初始时间为0； 36 dis1[q]=0; 37 for(int i=1;i<=m;i++) 38 { 39 int mi=INF; 40 int k=-1; 41 //寻找x到其他地方的最小值 42 for(int j=1;j<=m;j++) 43 { 44 if(!vis[j]&&mi>dis1[j]) 45 { 46 mi=dis1[j]; 47 k=j; 48 } 49 } 50 //如果没有最小值，则退出循环 51 if(k==-1) 52 { 53 break; 54 } 55 vis[k]=1; 56 for(int j=1;j<=m;j++) 57 { 58 //从x到k的时间与从k到其他地方的时间之和与从x到其他地方的时间取最小值 59 dis1[j]=min(dis1[j],dis1[k]+ma[k][j]); 60 } 61 } 62 } 63 //从x到其他地方的反向图 64 void dijkstra2(int m,int q) 65 { 66 memset(vis,0,sizeof(vis)); 67 memset(dis2,INF,sizeof(dis2)); 68 vis[q]=0; 69 //初始时间为0； 70 dis2[q]=0; 71 for(int i=1;i<=m;i++) 72 { 73 int mi=INF; 74 int k=-1; 75 //寻找x到其他地方的最小值 76 for(int j=1;j<=m;j++) 77 { 78 if(!vis[j]&&mi>dis2[j]) 79 { 80 mi=dis2[j]; 81 k=j; 82 } 83 } 84 //如果没有最小值，则退出循环 85 if(k==-1) 86 { 87 break; 88 } 89 vis[k]=1; 90 for(int j=1;j<=m;j++) 91 { 92 //从x到k的时间与从其他地方到k的时间之和与从x到其他地方的时间取最小值 93 dis2[j]=min(dis2[j],dis2[k]+ma[j][k]); 94 } 95 } 96 } 97 98 int main() 99 { 100 int n,m,x; 101 scanf("%d %d %d",&n,&m,&x); 102 int a,b,c; 103 //因为要求的是最小时间，所以初始化为无穷 104 memset(ma,INF,sizeof(ma)); 105 for(int i=1;i<=m;i++) 106 { 107 scanf("%d %d %d",&a,&b,&c); 108 ma[a][b]=c; 109 } 110 //回来的 111 dijkstra1(n,x); 112 //去的 113 dijkstra2(n,x); 114 int ans=0; 115 for(int i=1;i<=n;i++) 116 { 117 //所有牛花费时间中最大的一个 118 ans=max(ans,dis1[i]+dis2[i]); 119 } 120 printf("%d\n",ans); 121 }
标签：stream using lang 地方 pac cee 通过 ring cow
原文地址：https://www.cnblogs.com/mzchuan/p/11477897.html
