hdu 1364(差分约束)

King

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 12056		Accepted: 4397

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

题意(恶心的题意):
现在假设有一个这样的序列，S={a1，a2，a3，a4...ai...at}
其中ai=a*si,其实这句可以忽略不看
现在给出一个不等式，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
首先给出两个数分别代表S序列有多少个，有多少个不等式
不等式可以这样描述
给出四个参数第一个数i可以代表序列的第几项，然后给出n，这样前面两个数就可以描述为ai+a(i+1)+...a(i+n)，即从i到n的连续和，再给出一个符号和一个ki
当符号为gt代表‘>’,符号为lt代表‘<'
那么样例可以表示
1 2 gt 0
a1+a2+a3>0
2 2 lt 2
a2+a3+a4<2
最后问你所有不等式是否都满足条件，若满足输出lamentable kingdom，不满足输出successful conspiracy，这里要注意了，不要搞反了

题解:
每一个序列之和都可以用前缀和表示出来，比如说第一个测试用例,sum[3]-sum[0]>0,sum[4]-sum[1]<2，又因为差分约束系统必须是>=或者<=，这题又全是整数,所以我们可以通过加减
一来得到标准差分约束式,这些点不一定是联通的,所以加个超级源点,然后判断负环.

#include <iostream>
#include <cstdio>
#include <string.h>
#include <queue>
#include <algorithm>
#include <math.h>
using namespace std;
typedef long long LL;
const int INF = ;
const int N = ;
struct Edge{
    int v,w,next;
}edge[*N];
int head[N];
int tot ;
void init(){
    memset(head,-,sizeof(head));
    tot = ;
}
void addEdge(int u,int v,int w,int &k){
    edge[k].v = v,edge[k].w = w,edge[k].next = head[u],head[u]=k++;
}
int n,m;
int low[N],time[N];
bool vis[N];
bool spfa(int s){
    for(int i=;i<=;i++){
        vis[i] = false;
        low[i] = INF;
        time[i] = ;
    }
    low[s] = ;
    time[s]++;
    queue<int> q;
    q.push(s);
    while(!q.empty()){
        int u = q.front();
        q.pop();
        vis[u] = false;
        for(int k=head[u];k!=-;k=edge[k].next){
            int v = edge[k].v,w=edge[k].w;
            if(low[v]>low[u]+w){
                low[v] = low[u]+w;
                if(!vis[v]){
                    vis[v] = true;
                    q.push(v);
                    if(time[v]++>n+) return false;
                }
            }
        }
    }
    return true;
}
int main(){
    while(scanf("%d",&n)!=EOF,n){
        init();
        scanf("%d",&m);
        int super = ;
        for(int i=;i<=m;i++){
            int u,v,w;
            char op[];
            scanf("%d%d%s%d",&u,&v,op,&w);
            v = u+v;
            if(op[]=='g'){
                w+=;
                addEdge(v,u-,-w,tot);
            }else{
                w-=;
                addEdge(u-,v,w,tot);
            }
            addEdge(super,u-,,tot);
            addEdge(super,v,,tot);
        }
        if(spfa(super)){
            printf("lamentable kingdom\n");
        }else{
            printf("successful conspiracy\n");
        }
    }
    return ;
}