差分约束系统

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如果一个系统由n个变量和m个约束条件组成,其中每个约束条件形如xj-xi<=bk(i,j∈[1,n],k∈[1,m]),则称其为差分约束系统(system of difference constraints)。亦即,差分约束系统是求解关于一组变量的特殊不等式组的方法。
求解差分约束系统,可以转化成图论的单源最短路径(或最长路径)问题。
观察xj-xi<=bk,会发现它类似最短路中的三角不等式d[v]<=d[u]+w[u,v],即d[v]-d[u]<=w[u,v]。因此,以每个变量xi为结点,对于约束条件xj-xi<=bk,连接一条边(i,j),边权为bk。我们再增加一个源点s,s与所有定点相连,边权均为0。对这个图,以s为源点运行Bellman-ford算法(或SPFA算法),最终{d[ i]}即为一组可行解。
引理:设x=(x1,x2,…,xn)是差分约束系统Ax≤b的一个解,d为任意常数。则x+d=(x1+d,x2+d,…,xn+d)也是该系统Ax≤b的一个解。
bellman-ford算法伪代码:
for each v V do d[v] <-- 无限大; d[s] <-- 0
Relaxation
for i =1,...,|V|-1 do
for each edge (u,v) 属于 E do
d[v] <-- min{d[v], d[u]+w(u,v)}
Negative cycle checking
for each v 属于V do if d[v]> d[u] + w(u,v) then no solution
在实际的应用中,一般使用SPFA(Shortest Path Fast Algorithm)算法来实现。
差分约束系统中源点到每个点的距离确定
关于Dist[]的初始化化
1.如果将源点到各点的距离初始化为0,最终求出的最短路满足 它们之间相互最接近了
2.如果将源点到各点的距离初始化为INF(无穷大),其中之1为0,最终求出的最短路满足 它们与该点之间相互差值最大。
3.差分约束系统的确立要根据自己确定的约束条件,从约束点走向被约束点
连边一般有两种方法,第一种是连边后求最长路的方法,第二种是连边后求最短路的方法。
例:d[x]-d[y]>=Z
如果想连边后求最长路 那么将不等式变形为这种形式 d[x]>=d[y]+z y---x连一条权值为z的边
求最短路则变形成d[y]<=d[x]-z x---y连一条权值为-z的边。
如果是别的不等式,也可以根据情况变形。但是要保证的是 两个变量(x,y)的系数一定要是正的。而常量则不一定。

  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 

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$S_i = \{a_{s_i}, a_{s_i+1}, \dots, a_{s_i+n_i}\}$ of a sequence$S = \{a_1, a_2, \dots, a_n\}$. The king thought a minute and then decided, i.e. he set for the sum$a_{s_i} + a_{s_i+1} + \dots + a_{s_i+n_i}$ of each subsequenceSi an integer constraint ki (i.e.$a_{s_i} + a_{s_i+1} + \dots + a_{s_i+n_i} < k_i$ or$a_{s_i} + a_{s_i+1} + \dots + a_{s_i+n_i} > k_i$ 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 sequenceS 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 integersn, and m where $0 < n \leŸ 100$ is length of the sequenceS and $0 < m \leŸ 100$ is the number of subsequencesSi. Next m lines contain particular decisions coded in the form of quadruplessi, ni, oi,ki, where oi represents operator > (coded asgt) 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 containing0.

Output 

The output file contains the lines corresponding to the blocks in the input file. A line contains textsuccessful conspiracy when such a sequence does not exist. Otherwise it contains textlamentable kingdom. There is no line in the output file corresponding to the last ``null'' block of the input file.

Sample Input 

4 21 2 gt 02 2 lt 21 21 0 gt 01 0 lt 00

Sample Output 

lamentable kingdomsuccessful 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;}


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