上面是来自百度百科,其实就是由xj-xi<=k(注意有等号)建立图后判断有没有负环。因为图可能不连通,所以要再加一个原点到每个点的权值都是0(不影响原来的负环判断),这样确保图是连通的。经过bellman-ford或SPFA算法后,求出的最短路正好是满足要求的一组解。
uva515 - King
King
King |
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 of a sequence
. The king thought a minute and then decided, i.e. he set for the sum
of each subsequence Si an integer constraint ki (i.e.
or
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 file 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

Output
The output file contains the lines corresponding to the blocks in the input file. 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 file corresponding to the last ``null'' block of the input file.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
Si=a(si)+a(si+1)...+a(si+ni),Si大于或小于ki。设xi=a0+a1+...+a[i],那么Si=x(ni)-x(si-1)。这样构成一个差分约束系统,只要xi都存在,Si就存在。注意共有N+2个点(x0..xn是N+1个,再建立一个原点,共N+2个),有M+N+1条边。还要注意的是如果x(ni)-x(si-1)<ki,则x(ni)-x(si-1)<=ki-1。如果x(ni)-x(si-1)>ki,则x(ni)-x(si-1)>=ki+1,也就是x(si-1)-x(ni)<=-ki-1。
#include<cstring>
#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#define INF 0x3f3f3f3f
#define MAXN 220
#define pii pair<int,int>
using namespace std;
int N,M,K,d[MAXN];
struct edge{
int u,v,d;
}e[MAXN];
void add_edge(int a,int b,int t){
e[K].u=a;
e[K].v=b;
e[K++].d=t;
}
int Bellman(){
int i,j;
memset(d,INF,sizeof(d));
d[N+1]=0;
for(i=0;i<N+1;i++){
for(j=0;j<K;j++){
int u=e[j].u,v=e[j].v;
if(d[u]!=INF&&d[u]+e[j].d<d[v]) d[v]=d[u]+e[j].d;
}
}
for(j=0;j<K;j++){
int u=e[j].u,v=e[j].v;
if(d[u]!=INF&&d[u]+e[j].d<d[v]) return 1;
}
return 0;
}
int main(){
freopen("in.txt","r",stdin);
while(scanf("%d",&N),N){
scanf("%d",&M);
int i,a,b,t;
char s[10];
K=0;
for(i=0;i<M;i++){
scanf("%d%d%s%d",&a,&b,s,&t);
if(s[0]=='l') add_edge(a+b,a-1,t-1);
else add_edge(a-1,a+b,-t-1);
}
for(i=0;i<=N;i++) add_edge(N+1,i,0);
if(Bellman()) printf("successful conspiracy\n");
else printf("lamentable kingdom\n");
}
return 0;
}