POJ 2983 Is the Information Reliable?
2013-03-20 20:47
447 查看
Is the Information Reliable?
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
A is X light-years north of defense station B.
Vague tip is in the form of
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
Sample Output
Source
POJ Monthly--2006.08.27, Dagger
这是我做的第二道查分约束的题目,真是感觉自己太嫩了, a=z 我愣是没有想到可以转化为a>=z&&a<=z
很久不写spfa,本应该写> 而写成了>= 导致错了。 关键在于建图啊
Time Limit: 3000MS | Memory Limit: 131072K | |
Total Submissions: 9433 | Accepted: 2944 |
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
POJ Monthly--2006.08.27, Dagger
这是我做的第二道查分约束的题目,真是感觉自己太嫩了, a=z 我愣是没有想到可以转化为a>=z&&a<=z
很久不写spfa,本应该写> 而写成了>= 导致错了。 关键在于建图啊
#include <stdio.h> #include <string.h> #include <math.h> struct num { int end,next,val; }a[1000000]; int b[10010],sum[10010],d[10010],status[10010]; int queue[1000000]; int INF=0x7fffffff,n; int main() { int spfa(); int i,j,m,s,t,x,y,val,k; char c; while(scanf("%d %d%*c",&n,&m)!=EOF) { memset(b,-1,sizeof(b)); for(i=1,j=0;i<=n;i++) { a[j].end=i; a[j].val=0; a[j].next=b[0]; b[0]=j; j++; } for(i=0;i<=m-1;i++) { scanf("%c %d %d",&c,&x,&y); if(c=='P') { scanf("%d%*c",&val); a[j].end=y; a[j].val=-1*val; a[j].next=b[x]; b[x]=j; j++; a[j].end=x; a[j].val=val; a[j].next=b[y]; b[y]=j; j++; }else { getchar(); a[j].end=y; a[j].val=-1; a[j].next=b[x]; b[x]=j; j++; } } k=spfa(); if(k) { printf("Reliable\n"); }else { printf("Unreliable\n"); } } return 0; } int spfa() { int i,j,base,top,x,xend,k; k=1; memset(status,0,sizeof(status)); memset(sum,0,sizeof(sum)); for(i=1;i<=n;i++) { d[i]=INF; } base=top=0; d[0]=0; status[0]=1; sum[0]=1; queue[top++]=0; while(base<top) { x=queue[base++]; status[x]=0; for(j=b[x];j!=-1;j=a[j].next) { xend=a[j].end; if(d[xend]>(d[x]+a[j].val)) { d[xend]=d[x]+a[j].val; if(!status[xend]) { status[xend]=1; queue[top++]=xend; if(sum[xend]<=n-1) { sum[xend]+=1; }else { k=0; break; } } } } if(j!=-1) { break; } } return k; }
相关文章推荐
- POJ 2983-Is the Information Reliable?(差分约束系统)
- poj 2983——Is the Information Reliable?
- POJ 2983 Is the Information Reliable?(差分约束系统+BellmanFord)
- poj 2983 Is the Information Reliable?
- POJ 2983 Is the Information Reliable?(差分约束)
- POJ 2983 Is the Information Reliable?(差分约束第一发)
- poj 2983 差分约束Is the Information Reliable?
- POJ 2983 Is the Information Reliable?
- POJ 2983 Is the Information Reliable (差分约束 bell/spfa判断负环)
- poj 2983 Is the Information Reliable?(差分约束)
- POJ 2983 Is the Information Reliable?
- POJ训练计划2983_Is the Information Reliable?(差分约束)
- 学习笔记----差分约束系统 POJ 2983 Is the Information Reliable?
- Poj 2983 Is the Information Reliable?
- poj 2983 Is the Information Reliable? 差分约束判断回路
- POJ 2983 Is the Information Reliable?(差分约束系统)
- poj 2983 Is the Information Reliable?
- POJ 2983 Is the Information Reliable? (差分约束)
- POJ-2983-Is the Information Reliable?(差分约束)
- POJ 2983 Is the Information Reliable(差分约束系统)