POJ2983 Is the Information Reliable?

http://acm.sdut.edu.cn:8080/vjudge/contest/view.action?cid=267#problem/B

                       B - Is the Information Reliable?
Time Limit:3000MS     Memory Limit:131072KB     64bit IO Format:%I64d & %I64u

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

还是那句话:差分约束条件题目的难点是“怎么找到问题的约束条件”。
这题输入的边有两种格式:
      1. 边长确定,即xi - xj = b; 可以转化成 xi - xj <= b 和 xi - xj >=b (即 xj - xi <= -b).
      2. 边长不定,xi - xj >= 1; 可以转化成 xj - xi <= -1;

 

da-db>=x;

da-db<=x;

==>da>=db+x;

db>=da-x;

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
using namespace std;
int n,m,tt;
struct node
{
    int x,y,z;
} q[2000001];
int dis[100001];
void BF()
{
    int flag;
    memset(dis,0,sizeof(dis));
    for(int i=1; i<=n; i++)
    {
        flag=0;
        for(int j=0; j<tt; j++)
        {
            if(dis[q[j].x]<dis[q[j].y]+q[j].z)
            {
                dis[q[j].x]=dis[q[j].y]+q[j].z;
                flag=1;
            }
        }
        if(flag==0) break;
    }
    if(flag)
        cout<<"Unreliable"<<endl;
    else
        cout<<"Reliable"<<endl;
    return;
}
int main()
{
    char ch;
    int xx,yy,zz;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        tt=0;
        while(m--)
        {
            getchar();
            scanf("%c",&ch);
            if(ch=='P')
            {
                scanf("%d%d%d",&xx,&yy,&zz);
                q[tt].x=xx;
                q[tt].y=yy;
                q[tt++].z=zz;
                q[tt].x=yy;
                q[tt].y=xx;
                q[tt++].z=-zz;
            }
            else if(ch=='V')
            {
                scanf("%d%d",&xx,&yy);
                q[tt].x=xx;
                q[tt].y=yy;
                q[tt++].z=1;
            }
        }
        BF();
    }
    return 0;
}

 

posted @ 2014-08-23 15:45  人艰不拆_zmc  阅读(275)  评论(0编辑  收藏  举报