| Time Limit: 3000MS | Memory Limit: 131072K | |
| Total Submissions: 10303 | Accepted: 3204 |
Description
The galaxy war between the Empire Draco and the Commonwealth of Zibu broke out 3 years ago. Draco established a line of defense called Grot. Grot is a straight line with N defense stations. Because of the cooperation of the stations, Zibu’s Marine Glory cannot march any further but stay outside the line.
A mystery Information Group X benefits form selling information to both sides of the war. Today you the administrator of Zibu’s Intelligence Department got a piece of information about Grot’s defense stations’ arrangement from Information Group X. Your task is to determine whether the information is reliable.
The information consists of M tips. Each tip is either precise or vague.
Precise tip is in the form of P A B X, means defense station A is X light-years north of defense station B.
Vague tip is in the form of V A B, means defense station A is in the north of defense station B, at least 1 light-year, but the precise distance is unknown.
Input
There are several test cases in the input. Each test case starts with two integers N (0 < N ≤ 1000) and M (1 ≤ M ≤ 100000).The next M line each describe a tip, either in precise form or vague form.
Output
Output one line for each test case in the input. Output “Reliable” if It is possible to arrange N defense stations satisfying all the M tips, otherwise output “Unreliable”.
Sample Input
3 4 P 1 2 1 P 2 3 1 V 1 3 P 1 3 1 5 5 V 1 2 V 2 3 V 3 4 V 4 5 V 3 5
Sample Output
Unreliable Reliable
Source
题意：
给定n个变量，m组不等式，求解。
其中P操作表示权值为X的双向变，V操作表示权值为1的单向边。
差分约束：
其实看到这类题就应该想到是差分约束。因为做得还不多，了解还不是很深，构图时出现麻烦，WA了很多次。
这题也是要用到求最短路的算法，但是不用求最短路，只要求是否存在负权环，建图模型为：
bellman_ford:
P: g[a][b]=-x,
g[b][a]=x;
V:g[a][b]=1;
spfa:
此方法要在bellman的基础上加上
for i=0 ... n do
g[0][i]=0;
照旧给出两种方法的代码：
bellman_ford:
1 //Accepted 2488K 485MS C++ 1315B 2013-11-29 09:33:23 2 #include<stdio.h> 3 #include<string.h> 4 #define N 1005 5 #define inf 0x7ffffff 6 struct node{ 7 int u,v,w; 8 }edge[400005]; 9 int d[N]; 10 int n,m,edgenum; 11 void addedge(int u,int v,int w) 12 { 13 edge[edgenum].u=u; 14 edge[edgenum].v=v; 15 edge[edgenum++].w=w; 16 } 17 bool bellman_ford() 18 { 19 memset(d,0,sizeof(d)); 20 for(int i=0;i<n;i++){ 21 int flag=1; 22 for(int j=0;j<edgenum;j++){ 23 if(d[edge[j].u]!=inf && d[edge[j].v]>d[edge[j].u]+edge[j].w){ 24 d[edge[j].v]=d[edge[j].u]+edge[j].w; 25 flag=0; 26 } 27 } 28 if(flag) break; 29 } 30 for(int j=0;j<edgenum;j++) 31 if(d[edge[j].u]!=inf && d[edge[j].v]>d[edge[j].u]+edge[j].w) 32 return 0; 33 return 1; 34 } 35 int main(void) 36 { 37 char op; 38 int a,b,c; 39 while(scanf("%d%d",&n,&m)!=EOF) 40 { 41 edgenum=0; 42 for(int i=0;i<m;i++){ 43 getchar(); 44 scanf("%c",&op); 45 if(op=='P'){ 46 scanf("%d%d%d",&a,&b,&c); 47 addedge(a,b,-c); 48 addedge(b,a,c); 49 }else{ 50 scanf("%d%d",&a,&b); 51 addedge(a,b,-1); 52 } 53 } 54 if(bellman_ford()) puts("Reliable"); 55 else puts("Unreliable"); 56 } 57 return 0; 58 }
spfa
1 //Accepted 2704K 704MS C++ 1641B 2013-11-29 09:18:36 2 #include<iostream> 3 #include<queue> 4 #include<vector> 5 #include<stdio.h> 6 #include<string.h> 7 #define inf 0x7ffffff 8 #define N 1005 9 using namespace std; 10 struct node{ 11 int v,w; 12 node(int a,int b){ 13 v=a;w=b; 14 } 15 }; 16 vector<node>V[N]; 17 int vis[N]; 18 int d[N]; 19 int n,m; 20 bool spfa() 21 { 22 int in[N]={0}; 23 memset(vis,0,sizeof(vis)); 24 for(int i=0;i<=n;i++) d[i]=inf; 25 queue<int>Q; 26 d[0]=0; 27 Q.push(0); 28 while(!Q.empty()){ 29 int u=Q.front(); 30 Q.pop(); 31 if(in[u]>=n) return false; 32 vis[u]=0; 33 int n0=V[u].size(); 34 for(int i=0;i<n0;i++){ 35 int v=V[u][i].v; 36 int w=V[u][i].w; 37 if(d[v]>d[u]+w){ 38 d[v]=d[u]+w; 39 in[v]++; 40 if(!vis[v]){ 41 Q.push(v); 42 vis[v]=1; 43 } 44 } 45 } 46 } 47 return true; 48 } 49 int main(void) 50 { 51 char op; 52 int a,b,c; 53 while(scanf("%d%d%*c",&n,&m)!=EOF) 54 { 55 for(int i=0;i<=n;i++) V[i].clear(); 56 for(int i=0;i<=n;i++) 57 V[0].push_back(node(i,0)); 58 for(int i=0;i<m;i++){ 59 scanf("%c",&op); 60 if(op=='P'){ 61 scanf("%d%d%d",&a,&b,&c); 62 V[a].push_back(node(b,-c)); 63 V[b].push_back(node(a,c)); 64 }else{ 65 scanf("%d%d",&a,&b); 66 V[a].push_back(node(b,-1)); 67 } 68 getchar(); 69 } 70 if(spfa()) puts("Reliable"); 71 else puts("Unreliable"); 72 } 73 return 0; 74 }