POJ——3169Layout（差分约束）

POJ——3169Layout

Layout

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14702		Accepted: 7071

Description

Like everyone else, cows like to stand close to their friends when queuing for feed. FJ has N (2 <= N <= 1,000) cows numbered 1..N standing along a straight line waiting for feed. The cows are standing in the same order as they are numbered, and since they can be rather pushy, it is possible that two or more cows can line up at exactly the same location (that is, if we think of each cow as being located at some coordinate on a number line, then it is possible for two or more cows to share the same coordinate).

Some cows like each other and want to be within a certain distance of each other in line. Some really dislike each other and want to be separated by at least a certain distance. A list of ML (1 <= ML <= 10,000) constraints describes which cows like each other and the maximum distance by which they may be separated; a subsequent list of MD constraints (1 <= MD <= 10,000) tells which cows dislike each other and the minimum distance by which they must be separated.

Your job is to compute, if possible, the maximum possible distance between cow 1 and cow N that satisfies the distance constraints.

Input

Line 1: Three space-separated integers: N, ML, and MD.

Lines 2..ML+1: Each line contains three space-separated positive integers: A, B, and D, with 1 <= A < B <= N. Cows A and B must be at most D (1 <= D <= 1,000,000) apart.

Lines ML+2..ML+MD+1: Each line contains three space-separated positive integers: A, B, and D, with 1 <= A < B <= N. Cows A and B must be at least D (1 <= D <= 1,000,000) apart.

Output

Line 1: A single integer. If no line-up is possible, output -1. If cows 1 and N can be arbitrarily far apart, output -2. Otherwise output the greatest possible distance between cows 1 and N.

Sample Input

4 2 1
1 3 10
2 4 20
2 3 3

Sample Output

27

Hint

Explanation of the sample:

There are 4 cows. Cows #1 and #3 must be no more than 10 units apart, cows #2 and #4 must be no more than 20 units apart, and cows #2 and #3 dislike each other and must be no fewer than 3 units apart.

The best layout, in terms of coordinates on a number line, is to put cow #1 at 0, cow #2 at 7, cow #3 at 10, and cow #4 at 27.

Source

USACO 2005 December Gold

 

 

题目大意：

n头牛编号为1到n，按照编号的顺序排成一列，每两头牛的之间的距离 >= 0。这些牛的距离存在着一些约束关系：

1.有ml组（u, v, w）的约束关系，表示牛[u]和牛[v]之间的距离必须 <= w。

2.有md组（u, v, w）的约束关系，表示牛[u]和牛[v]之间的距离必须 >= w。

问如果这n头无法排成队伍，则输出-1，如果牛[1]和牛[n]的距离可以无限远，则输出-2，否则则输出牛[1]和牛[n]之间的最大距离。

 

差分约束入门题

首先记$d[i]$为第$i$头奶牛到第1头奶牛的距离，由于奶牛们按顺序排列，$d[i]<=d[i+1]+0$

第一种约束：$d[u]-d[v]<=w$,可作如下转化$d[u]<=d[v]+w$

观察发现这个式子很像求解最短路时spfa的松弛操作，的确与他有关，建立的模型便是$差分约束系统$。

第二种约束：$d[u]-d[v]>=w$,可转化为$d[v]<=d[u]-w$

 

根据红色标记的三种约束（如上）建图：

1.由i+1向i建立一条权值为0的边

记u=max(u,v),v=min(u,v),因为奶牛们是按顺序排列的，根据$d[i]$的意义，$d[u]>=d[v]$

2.由$u->v$连一条权值为w的边

3.由$u->v$连一条权值为-w的边

 

定理1：问题是否有解等价于图G是否没有负权回路。

证明：若G中无负权回路，我们可以求出v₁其他顶点u的最短路长，设为d(u)。由于是最短路，因此对于任意边eE，e=uv，有d(u)+w(e)>=d(v)，从而所有的约束条件都被满足，问题一定有解。若G中有负权回路，说明在任何时刻，G中至少有一个点v的最短路长可以更新，因此必须存在一条边e=uv，使得d(u)+w(e)<d(v)。所以无论何时，都会有某个约束条件不被满足，问题无解。（证毕）

定理2：若运行Bellman-Ford后，标号为N的顶点的最短路估计值仍为充分大，那么N和1的距离可以任意大。

证明：从刚才的操作可以看出，到了这一步，已经把含有负权回路的情况排除掉了。在图G中，该充分大的值比可能得到的最大距离大，因此，它和任意大的值对于G的效果都是一样的（同样大于合法的最大距离）。由于充分大的值在G中满足约束，所以任意大的值亦满足约束，从而距离可以任意大。（证毕）

定理3：若运行Bellman-Ford后，标号为N的顶点的最短路估计值比充分大小，那么它是N和1可能的最大距离。

证明：设D[i]是顶点i和1的最短路径估计值，d[i]是顶点i和1可能的最大距离。

我们首先证明，d[n]<=D[n]，运用反证法。

假如d[n]>D[n]，那么在Bellman-Ford运行之前，将赋予每个顶点i的充分大的值换成对应的d[i]。由于d本身满足所有约束条件，所以运行后，得出D'=d。由于充分大的值比所有d[i]都大，而求最短路运用的是逐步松弛操作，我们设立一个更大的初值不可能导致我们的终值反而更小。所以对于任意i，必定有D[i]>=D'[i]，即有D[n]>=d[n]，这与我们的假设矛盾。

然后我们证明，d[n]>=D[n]。

根据d[i]的定义，它是i和1的可能最大距离。由于D[i]是满足题目的所有约束的，所以D[i]是顶点i和1可能的距离。如果D[i]>d[i]，那么与d[i]的定义矛盾。

综合上述，有D[i]=d[i]。从而D[n]是n和1可能的最大距离。（证毕）

 

#pragma GCC optimize(2)
#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
#include<cmath>

#define inf 0x7fffffff
#define N 100010
using namespace std;

int n,ml,md,head[N],tot;
struct node{
    int to,next,w;
}e[N];

void add(int u,int v,int w){
    e[++tot].to=v,e[tot].next=head[u],head[u]=tot,e[tot].w=w;
}

int d[N],in[N];
bool vis[N];
queue<int>Q;
int spfa(){
    memset(vis,,sizeof(vis));
    fill(d+,d++n,inf);
    d[]=;Q.push();vis[]=;
    while(!Q.empty()){
        int u=Q.front();Q.pop();vis[u]=;
        for(int i=head[u];i;i=e[i].next){
            int v=e[i].to;
            if(d[v]>d[u]+e[i].w){
                d[v]=d[u]+e[i].w;
                if(!vis[v]){
                    vis[v]=;
                    in[v]++;
                    if(in[v]>n) return -;
                    Q.push(v);
                }
            }
        }
    }
    if(d[n]==inf) return-;
    else return d[n];
}

int main()
{
    scanf("%d%d%d",&n,&ml,&md);
    for(int u,v,w,i=;i<=ml;i++){
        scanf("%d%d%d",&u,&v,&w);
        if(v>u) swap(v,u);
        add(v,u,w);
    }
    for(int u,v,w,i=;i<=md;i++){
        scanf("%d%d%d",&u,&v,&w);
        if(u<v) swap(u,v);
        add(u,v,-w);
    }
    for(int i=;i<n;i++) add(i+,i,);
    printf("%d",spfa());
    return ;
}

 

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