1 #include<stdio.h> 2 #include<stdlib.h> 3 #include<string.h> 4 #include<math.h> 5  6 #define Infp 0x7fffffff 7 #define Infn 0x80000000 8 #define DisCon 1000000000 9 #define MAX  99999999910 #define MIN -99999999911 12 #define Z(a) memset(a, 0, sizeof(a))13 #define C(a, b) memcpy(a, b, sizeof(b))14 #define Pi acos(-1)15 #define FMAX (1E300)16 #define FMIN (-1E300)17 #define feq(a, b) (fabs((a)-(b))<1E-6)18 #define flq(a, b) ((a)<(b)||feq(a, b))19 #define RG 100520 21 struct Edge22 {23     int b;24     int e;25     int dis;26 }e[200010];//简单的记录边的信息27 28 int bell(int en, int pn)29 {30     int dis[pn+1];31     int chg;32 33     dis[0]=0;34     for(int i=1; i<=pn; i++)35         dis[i]=DisCon;//这是为了防止溢出才不用最大值36     for(int i=0; i<pn; i++)37     {38         chg=0;39         for(int j=0; j<en; j++)40             if(dis[e[j].e]>dis[e[j].b]+e[j].dis)//松弛操作41             {42                 dis[e[j].e]=dis[e[j].b]+e[j].dis;43                 chg=1;//如果最终趋于稳定，就是说肯定没有负权回路了44             }45         if(!chg)return 1;46 /*************解释*****************************47 结合《算法导论》的解释，如果从单源出发，而且图是无负权回路的，48 那么到任何一个短点的最短路径最多只包含|V|-1条边，而上述代码最49 外层的循环次数是|V|次，就是最多将构造含有|V|条边的最短路集合。50 假设循环不断进行，一直有最短路被发现（也即一直有松弛的路径更新51 ），直到i>|V|而退出，这说明这时构造出了一条最短路是有多于|V|条52 边的，与初衷矛盾，所以这时说明了图是有负权回路的！53 *********************************************/54     }55     return 0;56 }57 58 int main()59 {60     int N, M;61 62     while(scanf("%d%d", &N, &M)!=EOF)63     {64         int en=0;65         while(M--)66         {67             getchar();68 69             char c;70             scanf("%c", &c);71             if(c=='P')//具体的信息，双向的边都要构造72             {73                 int a, b, dis;74 75                 scanf("%d%d%d", &a, &b, &dis);76                 e[en].b=a;77                 e[en].e=b;78                 e[en++].dis=dis;79                 e[en].b=b;80                 e[en].e=a;81                 e[en++].dis=-dis;82             }83             else//模糊的信息，据说边长一定大于1，只构造单向的边84             {85                 int a, b;86                 scanf("%d%d", &a, &b);87                 e[en].b=b;88                 e[en].e=a;89                 e[en++].dis=-1;90             }91         }92 93         if(bell(en, N))printf("Reliable\n");94         else printf("Unreliable\n");95     }96 97     return 0;98 }

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <algorithm>
 4 #include <cstring>
 5
 6 using namespace std;
 7
 8 const int N = 1111;
 9 const int M = 111111;
10 const int INF = 0x11111111;
11 struct Edge {
12     int u, v, c;
13     Edge() {}
14     Edge(int u, int v, int c) : u(u), v(v), c(c) {}
15 } edge[M << 1];
16 int dis[N];
17
18 bool bellman(int m) {
19     bool ok;
20     for (int i = 0; i < N; i++) dis[i] = INF;
21     for (int i = 0; i < N; i++) {
22         ok = true;
23         for (int j = 0; j < m; j++) {
24             if (dis[edge[j].v] > dis[edge[j].u] + edge[j].c) {
25                 dis[edge[j].v] = dis[edge[j].u] + edge[j].c;
26                 ok = false;
27             }
28         }
29         if (ok) return true;
30     }
31     return false;
32 }
33
34 int main() {
35 //    freopen("in", "r", stdin);
36     int n, m;
37     int x, y, c;
38     char buf[3];
39     while (~scanf("%d%d", &n, &m)) {
40         int cnt = 0;
41         for (int i = 0; i < m; i++) {
42             scanf("%s%d%d", buf, &x, &y);
43             if (buf[0] == 'P') {
44                 scanf("%d", &c);
45                 edge[cnt++] = Edge(x, y, c);
46                 edge[cnt++] = Edge(y, x, -c);
47             } else edge[cnt++] = Edge(y, x, -1);
48         }
49         if (bellman(cnt)) puts("Reliable");
50         else puts("Unreliable");
51     }
52     return 0;
53 }