题目链接：https://vjudge.net/contest/123674#problem/D
题意：一张个向图，求从点1开始到其他各点的最短路权值和加上从其他各点到点1的最短路权值和 首先注意的是这是一个有向图，既要求1到所有的点的距离又要求其他所有点到1的距离，由于只学过spfa算法，就又用了spfa算法，求1到其他点的距离比较好求，但其他点到1的距离就不太好求了,我用到了图的转置，就是把所有的边的方向都反向，给建立一个新的图，这样1以外的点到1的距离就是1到1以外的点的距离；
用习惯了Java还真不习惯c++，特别是看到指针就烦，但Java这时间也真吓人的，
AC代码： time: 21282ms 好可怕！
1 import java.util.*; 2 import java.io.*; 3 4 public class Main { 5 public static Ver[] array1; 6 public static Ver[] array2; 7 public static int[] head1; 8 public static int[] head2; 9 public static int n; 10 public static long[] dis1; 11 public static long[] dis2; 12 public static boolean[] judge; 13 public static long Max = 999999999; 14 public static long[] result; 15 public static int count; 16 17 /** 18 * @param args 19 */ 20 public static void main(String[] args) throws IOException { 21 // TODO Auto-generated method stub 22 Scanner s = new Scanner(new BufferedInputStream(System.in)); 23 int t = s.nextInt(); 24 25 while (t-- > 0) { 26 27 n = s.nextInt(); 28 int m = s.nextInt(); 29 array1 = new Ver[1000002]; 30 array2 = new Ver[1000002]; 31 head1 = new int[n + 1]; 32 head2 = new int[n + 1]; 33 count = 1; 34 for (int i = 0; i < m; i++) { //建图 35 int a = s.nextInt(); 36 int b = s.nextInt(); 37 long c = s.nextLong(); 38 array1[count] = new Ver(b, c); //正图 39 array1[count].next = head1[a]; 40 head1[a] = count; 41 42 43 array2[count] = new Ver(a, c); //反图 44 array2[count].next = head2[b]; 45 head2[b] = count; 46 count++; 47 } 48 result = new long[n + 1]; 49 spfa1(1); 50 spfa2(1); 51 long ans = 0; 52 for (int i = 2; i <= n; i++) { //求和 53 ans += dis1[i] + dis2[i]; 54 } 55 System.out.println(ans); 56 } 57 s.close(); 58 } 59 60 public static void spfa2(int a) { //反向图的spfa算法 61 dis2 = new long[n + 1]; 62 judge = new boolean[n + 1]; 63 for (int i = 1; i < n + 1; i++) { 64 dis2[i] = Max; 65 66 } 67 LinkedList<Integer> queue = new LinkedList<Integer>(); 68 judge[1] = true; 69 dis2[1] = 0; 70 queue.add(1); 71 while (!queue.isEmpty()) { 72 int vpoll = queue.poll(); 73 judge[vpoll] = false; 74 75 int temp = head2[vpoll]; 76 while (temp != 0) { 77 if (relax2(vpoll, array2[temp].value, array2[temp].distance)) { 78 if (!judge[array2[temp].value]) { 79 judge[array2[temp].value] = true; 80 queue.add(array2[temp].value); 81 } 82 } 83 temp = array2[temp].next; 84 } 85 } 86 87 } 88 89 public static void spfa1(int a) { //正向图的spfa算法 90 dis1 = new long[n + 1]; 91 judge = new boolean[n + 1]; 92 for (int i = 1; i < n + 1; i++) { 93 dis1[i] = Max; 94 95 } 96 LinkedList<Integer> queue1 = new LinkedList<Integer>(); 97 judge[a] = true; 98 dis1[a] = 0; 99 queue1.add(a); 100 while (!queue1.isEmpty()) { 101 int vpoll = queue1.poll(); 102 judge[vpoll] = false; 103 104 int temp = head1[vpoll]; 105 while (temp != 0) { 106 if (relax1(vpoll, array1[temp].value, array1[temp].distance)) { 107 if (!judge[array1[temp].value]) { 108 judge[array1[temp].value] = true; 109 queue1.add(array1[temp].value); 110 } 111 } 112 temp = array1[temp].next; 113 } 114 } 115 116 } 117 118 public static boolean relax1(int a, int b, long c) { 119 120 if (dis1[a] + c < dis1[b]) { 121 dis1[b] = dis1[a] + c; 122 return true; 123 } 124 return false; 125 } 126 127 public static boolean relax2(int a, int b, long c) { 128 129 if (dis2[a] + c < dis2[b]) { 130 dis2[b] = dis2[a] + c; 131 return true; 132 } 133 return false; 134 } 135 136 } 137 138 class Ver { //图类 139 long distance; 140 int value; 141 int next; 142 143 Ver(int value, long distance) { 144 this.value = value; 145 this.distance = distance; 146 } 147 }