又是一道差分约束的题目。最短路的算法我用了spfa
题目中给出了一个区间的和，这个和<或者>一个给定的值，这里注意，差分约束只能处理<=,>=因为题目中序列的每个数都是整数，所以我们可以把<,>处理成<=,>=
约束：
sum[si+ni]-sum[si-1]>=a+1
sum[si+ni]-sum[si-1]<=a-1
将约束处理之后，在添加一个源点 n+1
约束关系
sum[i]-sum[n+1]<=0;
这样一共有了n+2个点，判断负权回路的时候入队超过n+2次就返回false
【图论|POJ1364 King 差分约束】。对自己的英语水平越来越没自信了。。。。。。
King
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 7596 | Accepted: 2913 |
Once, in one kingdom, there was a queen and that queen was expecting a baby. The queen prayed: ``If my child was a son and if only he was a sound king.'' After nine months her child was born, and indeed, she gave birth to a nice son.
Unfortunately, as it used to happen in royal families, the son was a little retarded. After many years of study he was able just to add integer numbers and to compare whether the result is greater or less than a given integer number. In addition, the numbers had to be written in a sequence and he was able to sum just continuous subsequences of the sequence.
The old king was very unhappy of his son. But he was ready to make everything to enable his son to govern the kingdom after his death. With regards to his son's skills he decided that every problem the king had to decide about had to be presented in a form of a finite sequence of integer numbers and the decision about it would be done by stating an integer constraint (i.e. an upper or lower limit) for the sum of that sequence. In this way there was at least some hope that his son would be able to make some decisions.
After the old king died, the young king began to reign. But very soon, a lot of people became very unsatisfied with his decisions and decided to dethrone him. They tried to do it by proving that his decisions were wrong.
Therefore some conspirators presented to the young king a set of problems that he had to decide about. The set of problems was in the form of subsequences Si = {aSi, aSi+1, ..., aSi+ni} of a sequence S = {a1, a2, ..., an}. The king thought a minute and then decided, i.e. he set for the sum aSi + aSi+1 + ... + aSi+ni of each subsequence Si an integer constraint ki (i.e. aSi + aSi+1 + ... + aSi+ni < ki or aSi + aSi+1 + ... + aSi+ni > ki resp.) and declared these constraints as his decisions.
After a while he realized that some of his decisions were wrong. He could not revoke the declared constraints but trying to save himself he decided to fake the sequence that he was given. He ordered to his advisors to find such a sequence S that would satisfy the constraints he set. Help the advisors of the king and write a program that decides whether such a sequence exists or not.
Input
The input consists of blocks of lines. Each block except the last corresponds to one set of problems and king's decisions about them. In the first line of the block there are integers n, and m where 0 < n <= 100 is length of the sequence S and 0 < m <= 100 is the number of subsequences Si. Next m lines contain particular decisions coded in the form of quadruples si, ni, oi, ki, where oi represents operator > (coded as gt) or operator < (coded as lt) respectively. The symbols si, ni and ki have the meaning described above. The last block consists of just one line containing 0. Output
The output contains the lines corresponding to the blocks in the input. A line contains text successful conspiracy when such a sequence does not exist. Otherwise it contains text lamentable kingdom. There is no line in the output corresponding to the last ``null'' block of the input. Sample Input
4 2 1 2 gt 0 2 2 lt 2 1 2 1 0 gt 0 1 0 lt 0 0
Sample Output
lamentable kingdom successful conspiracy
Source
#include
#include
#include
#include
#include
#include
#includeusing namespace std;
#define MAXN 110
#define INF 0xFFFFFFstruct node
{
int to;
int dis;
};
vector g[MAXN];
bool in[MAXN];
int dis[MAXN],counts[MAXN];
int n,m;
void init()
{
int a,b,c;
string s;
node temp;
cin>>m;
for(int i=0;
i<=n+1;
i++)
{
g[i].clear();
}
for(int i=1;
i<=m;
i++)
{
cin>>a>>b>>s>>c;
b=a+b;
if(s[0]=='l')
{
temp.to=b;
temp.dis=c-1;
g[a-1].push_back(temp);
}
else
{
temp.to=a-1;
temp.dis=-(c+1);
g[b].push_back(temp);
}
}
for(int i=0;
iq;
for(int i=0;
i<=n+1;
i++)
{
in[i]=false;
dis[i]=INF;
counts[i]=0;
}
dis[s]=0;
in[s]=true;
counts[s]++;
q.push(s);
while(!q.empty())
{
int tag=q.front();
q.pop();
in[tag]=false;
for(int i=0;
idis[tag]+g[tag][i].dis)
{
dis[j]=dis[tag]+g[tag][i].dis;
if(!in[j])
{
in[j]=true;
q.push(j);
counts[j]++;
if(counts[j]>=n+2)
return false;
}
}
}
}
return true;
}
int main()
{
while(cin>>n && n)
{
init();
if(spfa(n+1))
cout<<"lamentable kingdom"< 推荐阅读
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