POJ 1364-King（差分约束系统）

女爷i 2022-09-25 04:18 181阅读 0赞

King

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;10000K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;12246</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;4452</td> 
  </tr> 
 </tbody> 
</table>

Description

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

[Central Europe 1997][]

## 题目意思： ##

N表示序列S中元素的个数，M表示从S中取出的子序列的个数。

从序列S=\{A1，A2，…An\}中取出子序列Si=\{A\[si\]，A\[si+1\],…A\[si+ni\]\}。

对于每个子序列Si中的所有元素求和，对这个和设定一个约束，使之大于Ki或者小于Ki。

请问能否找出这样一个长度为N的序列S，使得设定的约束成立。

## 解题思路： ##

差分约束系统。

使用前N项和的概念，将子序列\[a，b\]内所有元素求和表示成：Sum\[b\]-Sum\[a-1\]>K\[i\]（或小于）。

那么，Sum\[b\]-Sum\[a-1\]>K\[i\]可以改写成Sum\[a-1\]-Sum\[b\]≤K\[i\]，由此构建一个有向图的边b->(a-1),边权为K\[i\]。

根据输入可以构建多组这样的边，从源点0开始，如果存在负权值回路，说明可以找到；反之不行。

要注意另设的这一个源点0，到其他所有点路径都为0。在我的程序中，N在判断时变成N+2。

#include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <vector>
    #include <queue>
    #include <iomanip>
    #include <algorithm>
    #define MAXN 10010
    #define INF 0xfffffff
    using namespace std;
    
    struct ArcNode
    {
        int to;
        int weight;
        ArcNode *next;
    };
    queue<int> Q;//队列中的节点为顶点序号
    int n;//顶点个数
    int m,W;
    ArcNode * List[MAXN];//每个顶点的边链表表头指针
    int inq[MAXN];//每个顶点是否在队列中的标志
    int dist[MAXN],path[MAXN];
    int cnt[MAXN];
    bool SPFA(int src)
    {
        memset(cnt,0,sizeof(cnt));
        memset(inq,0,sizeof(inq));
        int i,u;//u为队列头顶点序号
        ArcNode * temp;
        for(i=0; i<=n+2; ++i)//初始化
        {
            dist[i]=INF;
            path[i]=src;
            inq[i]=0;
        }
        dist[src]=0;
        path[src]=src;
        ++inq[src];
        Q.push(src);
        ++cnt[src];
        while(!Q.empty())
        {
            u=Q.front();
            Q.pop();
            --inq[u];
            if(cnt[u]>n+2) return true;//存在负权回路
            temp=List[u];
            while(temp!=NULL)
            {
                int v=temp->to;
                if(dist[v]>dist[u]+temp->weight)
                {
                    dist[v]=dist[u]+temp->weight;
                    path[v]=u;
                    if(!inq[v])
                    {
                        Q.push(v);
                        ++inq[v];
                        ++cnt[v];
                    }
    
                }
                temp=temp->next;
            }
        }
        return false;
    }
    int main()
    {
        /*#ifdef ONLINE_JUDGE
        #else
            freopen("G:/x/read.txt","r",stdin);
            freopen("G:/x/out.txt","w",stdout);
        #endif*/
        while(cin>>n&&n)
        {
            cin>>m;
            int i;
            memset(List,0,sizeof(List));
            ArcNode *temp;
            while(!Q.empty()) Q.pop();
            while(m--)
            {
                int u,v;
                cin>>u>>v;
                char g,t;
                cin>>g>>t;
                int k;
                cin>>k;
                if(g=='g')//大于
                {
                    temp=new ArcNode;
                    temp->to=u;//构造邻接表
                    temp->weight=-k-1;
                    temp->next=NULL;
                    if(List[u+v+1]==NULL) List[u+v+1]=temp;
                    else
                    {
                        temp->next=List[u+v+1];
                        List[u+v+1]=temp;
                    }
                }
                else if(g=='l')//小于
                {
                    temp=new ArcNode;
                    temp->to=u+v+1;//构造邻接表
                    temp->weight=k-1;
                    temp->next=NULL;
                    if(List[u]==NULL) List[u]=temp;
                    else
                    {
                        temp->next=List[u];
                        List[u]=temp;
                    }
                }
            }
            for(i=0; i<=n; ++i)
            {
                temp=new ArcNode;
                temp->to=i;//构造邻接表
                temp->weight=0;
                temp->next=NULL;
                if(List[0]==NULL) List[0]=temp;
                else
                {
                    temp->next=List[0];
                    List[0]=temp;
                }
            }
            if(SPFA(0)) puts("successful conspiracy");//求源点到其他顶点的最短路径
            else puts("lamentable kingdom");
            for(i=0; i<=n; ++i)
            {
                temp=List[i];
                while(temp!=NULL)
                {
                    List[i]=temp->next;
                    delete temp;
                    temp=List[i];
                }
            }
        }
        return 0;
    }
    /*
    4 2
    1 2 gt 0
    2 2 lt 2
    1 2
    1 0 gt 0
    1 0 lt 0
    0
    */

[Central Europe 1997]: http://poj.org/searchproblem?field=source&key=Central+Europe+1997