Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 12681 | Accepted: 4597 |
Description
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
Output
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
/*http://blog.csdn.Net/wangjian8006
題目大意:現在假設有一個這樣的序列,S={a1,a2,a3,a4...ai...at}
其中ai=a*si,其實這句可以忽略不看
現在給出一個不等式,使得ai+a(i+1)+a(i+2)+...+a(i+n)<ki或者是ai+a(i+1)+a(i+2)+...+a(i+n)>ki
首先給出兩個數分別代表S序列有多少個,有多少個不等式
不等式可以這樣描述
給出四個參數第一個數i可以代表序列的第幾項,然後給出n,這樣前面兩個數就可以描述爲ai+a(i+1)+...a(i+n),即從i到n的連續和,再給出一個符號和一個ki
當符號爲gt代表‘>’,符號爲lt代表‘<'
那麼樣例可以表示
1 2 gt 0
a1+a2+a3>0
2 2 lt 2
a2+a3+a4<2
最後問你所有不等式是否都滿足條件,若滿足輸出lamentable kingdom,不滿足輸出successful conspiracy,這裏要注意了,不要搞反了
解題思路:一個典型的差分約束,很容易推出約束不等式
首先設Si=a1+a2+a3+...+ai
那麼根據樣例可以得出
S3-S0>0---->S0-S3<=-1
S4-S1<2---->S4-S1<=1
因爲差分約束的條件是小於等於,所以我們將ki-1可以得到一個等於號
那麼通式可以表示爲
a b gt c
S[a-1]-s[a+b]<=-ki-1
a b lt c
S[a+b]-S[a-1]<=ki-1
那麼根據差分約束建圖,加入這些有向邊
gt: <a+b,a-1>=-ki-1
lt: <a-1,a+b>=ki-1
再根據bellman_ford判斷是否有無負環即可
若出現負環了則這個序列不滿足所有的不等式
我寫的代碼:
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <iostream>
#define M 301
using namespace std;
const int inf = (1<<25);
struct node
{
int to, dist;
int next;
}E[4*M];
int head[M], dis[M];
bool vis[M];
int sum[M];
int n, m, top;
void init()
{
memset ( head, -1, sizeof(head) );
memset ( vis, false, sizeof(vis) );
memset ( dis, 0, sizeof(dis) );
memset ( sum, 0, sizeof(sum) );
top = 0;
}
void AddEdge(int u, int v, int dist)
{
E[top].to = v;
E[top].dist = dist;
E[top].next = head[u];
head[u] = top++;
}
bool SPFA()
{
int i;
queue<int> q;
for ( i = 0;i <= n+1; i++ )
q.push(i);
vis[0] = true;
while ( !q.empty() )
{
int u = q.front();
q.pop();
vis[u] = false;
for ( i = head[u];i != -1; i = E[i].next )
{
int v = E[i].to;
if ( dis[v] > dis[u]+E[i].dist )
{
dis[v] = dis[u]+E[i].dist;
if ( !vis[v] )
{
vis[v] = true;
q.push(v);
sum[v]++;
if ( sum[v] > n+1 )
return false;
}
}
}
}
return true;
}
int main()
{
int i;
char ch[5];
int u, v, dist;
while ( ~scanf ( "%d", &n ) && n != 0 )
{
init();
scanf ( "%d", &m );
for ( i = 0;i < m; i++ )
{
scanf ( "%d %d %s %d", &u, &v, ch, &dist );
if ( ch[0] == 'g' )
AddEdge(u+v+1, u, -dist-1);
else
AddEdge(u, u+v+1, dist-1);
}
printf ( !SPFA() ? "successful conspiracy\n" : "lamentable kingdom\n" );
}
}
代碼菜鳥,如有錯誤,請多包涵!!!
如有幫助記得支持我一下,謝謝!!!