Guessing Game

题目连接：

Description

Jaehyun has two lists of integers, namely a1,...,aN and b1,...,bM.Je!rey wants to know what these

numbers are, but Jaehyun won’t tell him the numbers directly. So, Je!rey asks Jaehyun a series of questions

of the form “How big is ai + bj ?” Jaehyun won’t even tell him that, though; instead, he answers either

“It’s at least c,” or “It’s at most c.” (Right, Jaehyun simply doesn’t want to give his numbers for whatever

reason.) After getting Jaehyun’s responses, Je!rey tries to guess the numbers, but he cannot figure them out

no matter how hard he tries. He starts to wonder if Jaehyun has lied while answering some of the questions.

Write a program to help Je!rey.

Input

The input consists of multiple test cases. Each test case begins with a line containing three positive integers

N, M,and Q, which denote the lengths of the Jaehyun’s lists and the number of questions that Je!rey

asked. These numbers satisfy 2 ! N + M ! 1,000 and 1 ! Q ! 10,000. Each of the next Q lines is of the

form ij<=c or ij>=c.Theformerrepresents ai + bj ! c, and the latter represents ai + bj " c. It is

guaranteed that #1,000 ! c ! 1,000. The input terminates with a line with N = M = Q = 0. For example:

Output

For each test case, print a single line that contains “Possible” if there exist integers a1,...,aN and b1,...,bM

that are consistent with Jaehyun’s answers, or “Impossible” if it can be proven that Jaehyun has definitely

lied (quotes added for clarity). The correct output for the sample input above would be:

Sample Input

2 1 3

1 1 <= 3

2 1 <= 5

1 1 >= 4

2 2 4

1 1 <= 3

2 1 <= 4

1 2 >= 5

2 2 >= 7

0 0 0

Sample Output

Impossible

Possible

Hint

题意

a数组有n个数，b数组有m个数

然后告诉你一些不等式表示a[i]+b[j]<=C之类的

问你能否找到一组解

题解：

差分约束的裸题

我们建边之后，跑最短路，看是否有负环，如果存在负环的话，就说明这个不等式显然是不成立的

就好了

代码

#include
    
     using namespace std;const int inf=0x3f3f3f3f;struct node{    int x,y;};vector
     
       E[2005];int n,m,q;int inq[2005],dis[2005];int flag=0;void solve(int x){    if(flag)        return;    inq[x]=1;    for(int i=0;i
      
       dis[x]+v.y)        {            dis[v.x]=dis[x]+v.y;            if(inq[v.x])            {                flag=1;                return;            }            if(!inq[v.x])            {                dis[v.x]=dis[x]+v.y;                solve(v.x);            }        }    }    inq[x]=0;}int main(){    //freopen("1.in","r",stdin);    while(scanf("%d%d%d",&n,&m,&q)!=EOF)    {        if(n==0&&m==0&&q==0)            break;        flag = 0;        memset(inq,0,sizeof(inq));        memset(dis,0,sizeof(dis));        for(int i=0;i<=n+m;i++)            E[i].clear();        for(int i=0;i<=n;i++)            dis[i]=inf;        for(int i=0;i
       
        >s;            scanf("%d",&z);            if(s==">=")                E[x].push_back((node){n+y,-z});            if(s=="<=")                E[y+n].push_back((node){x,z});        }        for(int i=1;i<=n+m;i++)            dis[i]=0,solve(i);        if(flag)printf("Impossible\n");        else printf("Possible\n");    }}