[Atcoder SoundHound Contest 2018]E.+ Graph

题面

Time limit : 2sec / Memory limit : 1024MB

Score : 600 points

Problem Statement-题目描述

Kenkoooo found a simple connected graph. The vertices are numbered 1" role="presentation">11 through n" role="presentation">nn. The i" role="presentation">ii-th edge connects Vertex ui" role="presentation">uiui and vi" role="presentation">vivi, and has a fixed integer si" role="presentation">sisi.

Kenkoooo is trying to write a positive integer in each vertex so that the following condition is satisfied:

For every edge i" role="presentation">ii, the sum of the positive integers written in Vertex ui" role="presentation">uiui and vi" role="presentation">vivi is equal to si" role="presentation">sisi.

Find the number of such ways to write positive integers in the vertices.

Constraints-数据范围

2≤n≤105" role="presentation" style="position: relative;">2≤n≤1052≤n≤105

1≤m≤105" role="presentation" style="position: relative;">1≤m≤1051≤m≤105

1≤ui&lt;vi≤n" role="presentation" style="position: relative;">1≤ui<vi≤n1≤ui<vi≤n

2≤si≤109" role="presentation" style="position: relative;">2≤si≤1092≤si≤109

If i≠j" role="presentation" style="position: relative;">i≠ji≠j, then ui≠uj" role="presentation" style="position: relative;">ui≠ujui≠uj or vi≠vj." role="presentation" style="position: relative;">vi≠vj.vi≠vj.

The graph is connected.

All values in input are integers.

题目大意

给你一个无向图G=(V,E)" role="presentation" style="position: relative;">G=(V,E)G=(V,E),定义任意一条边e=(i,j)" role="presentation" style="position: relative;">e=(i,j)e=(i,j) (i≠j,i∈V,j∈V)" role="presentation" style="position: relative;">(i≠j,i∈V,j∈V)(i≠j,i∈V,j∈V)的权值为wi,j" role="presentation" style="position: relative;">wi,jwi,j,任意一个点ci(c∈V)" role="presentation" style="position: relative;">ci(c∈V)ci(c∈V)的权值为wi" role="presentation" style="position: relative;">wiwi,且wi" role="presentation" style="position: relative;">wiwi为正整数,现在给出所有边e" role="presentation" style="position: relative;">ee和边的权值wi,j" role="presentation" style="position: relative;">wi,jwi,j,求所有边与点都满足wi,j=wi+wj" role="presentation" style="position: relative;">wi,j=wi+wjwi,j=wi+wj时,1号结点有多少种不同的取值。

2≤n≤105" role="presentation" style="position: relative;">2≤n≤1052≤n≤105

1≤m≤105" role="presentation" style="position: relative;">1≤m≤1051≤m≤105

1≤ui&lt;vi≤n" role="presentation" style="position: relative;">1≤ui<vi≤n1≤ui<vi≤n

2≤si≤109" role="presentation" style="position: relative;">2≤si≤1092≤si≤109

题解

1.确定&方向

学过差分约束的同学可能会感觉这题与差分约束十分类似,是的。差分约束可以维护ai−aj≤X" role="presentation" style="position: relative;">ai−aj≤Xai−aj≤X的情况,同时,ai−aj=X" role="presentation" style="position: relative;">ai−aj=Xai−aj=X也可以转成ai−aj≤X" role="presentation" style="position: relative;">ai−aj≤Xai−aj≤X进行差分约束。但题目的要求是ai+aj=X" role="presentation" style="position: relative;">ai+aj=Xai+aj=X，至少在蒟蒻的知识范围内，是无法转化(或者很难转化)的。

这时候我们就又需要从暴力开始思考了。首先考虑枚举的变量应该是什么。

我们以样例2作为例子

4 3

1 2 6

2 3 7

3 4 5

显然我们可以将这些条件化成方程组

{a1+a2=6a2+a3=7a3+a4=5" role="presentation">⎧⎩⎨a1+a2=6a2+a3=7a3+a4=5{a1+a2=6a2+a3=7a3+a4=5

我们很快可以发现只要我们知道其中的任意一个值，我们就可以求出所有的a" role="presentation" style="position: relative;">aa。

我们不妨直接枚举a1" role="presentation" style="position: relative;">a1a1,判断它是否成立即可。

然而这样做一定是超时的，因此我们需要一个更好的方法来节省时间。

对方程比较敏感的同学很快会忍不住将1,2" role="presentation" style="position: relative;">1,21,2式相减得出a1−a3=−1" role="presentation" style="position: relative;">a1−a3=−1a1−a3=−1,再将其与3" role="presentation" style="position: relative;">33式相加得出a1+a4=4" role="presentation" style="position: relative;">a1+a4=4a1+a4=4。

这样一来，我们就求出a1" role="presentation" style="position: relative;">a1a1与其他a" role="presentation" style="position: relative;">aa的关系。

但是这有何用呢。

再看看题目,题目要求ai&gt;0" role="presentation" style="position: relative;">ai>0ai>0。显然我们需要往求范围的方向想问题。

因为a2=6−a1&gt;0" role="presentation" style="position: relative;">a2=6−a1>0a2=6−a1>0,a3=a1+1&gt;0" role="presentation" style="position: relative;">a3=a1+1>0a3=a1+1>0,a4=4−a1&gt;0" role="presentation" style="position: relative;">a4=4−a1>0a4=4−a1>0。

我们就可以得到a1&lt;6" role="presentation" style="position: relative;">a1<6a1<6,a1&gt;−1" role="presentation" style="position: relative;">a1>−1a1>−1,a1&lt;4" role="presentation" style="position: relative;">a1<4a1<4,和a1&gt;0" role="presentation" style="position: relative;">a1>0a1>0结合得0&lt;a1&lt;4" role="presentation" style="position: relative;">0<a1<40<a1<4即a1" role="presentation" style="position: relative;">a1a1有3" role="presentation" style="position: relative;">33个取值。没错，这就是正确答案了。

根据上面的操作我们可以发现几个有用的规律

1.当n≡0(mod2)" role="presentation" style="position: relative;">n≡0(mod2)n≡0(mod2)(即n%2=0)时,总有a1+an=A" role="presentation" style="position: relative;">a1+an=Aa1+an=A,换成不等式即a1&lt;A" role="presentation" style="position: relative;">a1<Aa1<A。

2.当n≡1(mod2)" role="presentation" style="position: relative;">n≡1(mod2)n≡1(mod2)(即n%2=1)时,总有a1−an=B" role="presentation" style="position: relative;">a1−an=Ba1−an=B,换成不等式即a1&gt;B" role="presentation" style="position: relative;">a1>Ba1>B。

显然，答案a1" role="presentation" style="position: relative;">a1a1的范围即为max(B1,B2,...,Bk1,0)&lt;a1&lt;min(A1,A2,...,Ak2)" role="presentation" style="position: relative;">max(B1,B2,...,Bk1,0)<a1<min(A1,A2,...,Ak2)max(B1,B2,...,Bk1,0)<a1<min(A1,A2,...,Ak2)

直接用DFS/BFS" role="presentation" style="position: relative;">DFS/BFSDFS/BFS遍历整张图很容易求出范围与a1" role="presentation" style="position: relative;">a1a1的合法个数。

难道这就是这道题的正解？

2.尝试&完善

作为这场比赛的最后一题，绝对不可能这么简单。

上面的样例显然遗漏了我们的老朋友(233)——环。

这个关系在环上也成立吗?

让我们来看看第一组样例吧。

3 3

1 2 3

2 3 5

1 3 4

(莫名画成直角三角形…)

我们再写出它的对应方程组:

{a1+a2=3a2+a3=5a1+a3=4" role="presentation">⎧⎩⎨a1+a2=3a2+a3=5a1+a3=4{a1+a2=3a2+a3=5a1+a3=4

求出它的对应式子

a1+a2=3" role="presentation" style="position: relative;">a1+a2=3a1+a2=3

a1−a2=−1" role="presentation" style="position: relative;">a1−a2=−1a1−a2=−1

a1−a3=−2" role="presentation" style="position: relative;">a1−a3=−2a1−a3=−2

a1+a3=4" role="presentation" style="position: relative;">a1+a3=4a1+a3=4

咦?…怎么每个a" role="presentation" style="position: relative;">aa都有两个关系式？

实质上,是因为每个点都可以通过节点1" role="presentation" style="position: relative;">11,通过两条不同的路径到达节点n" role="presentation" style="position: relative;">nn所造成的。

可以发现,每个a" role="presentation" style="position: relative;">aa所拥有的两个关系式都可以相加直接求出唯一的a1" role="presentation" style="position: relative;">a1a1

即设a1+an=A" role="presentation" style="position: relative;">a1+an=Aa1+an=A,a1−an=B" role="presentation" style="position: relative;">a1−an=Ba1−an=B,那么a1=A+B2" role="presentation" style="position: relative;">a1=A+B2a1=A+B2

那么环都能直接能确定a1" role="presentation" style="position: relative;">a1a1吗？

实际上，是不行的。



例如上图中的例子，a1" role="presentation" style="position: relative;">a1a1就有三个合法解1,2,3" role="presentation" style="position: relative;">1,2,31,2,3(具体求法可以参考第1" role="presentation" style="position: relative;">11部分的方程组)。

实质上是因为从节点1" role="presentation" style="position: relative;">11节点n" role="presentation" style="position: relative;">nn的两条路径都是奇数,以至于两者的关系式不仅符号相同，连数值也相同。(使用1" role="presentation" style="position: relative;">11部分中的规律)

也就相当于一个式子了。

经过一系列的归纳与整理，我们发现:只要是奇环(即环的节点数有奇数个)，就能求出唯一的a1" role="presentation" style="position: relative;">a1a1,反之则不行(当然,在本题中,这并不是一个重要的结论)

这就是这道题的正解？

好像还漏了些什么…

3.排查&漏洞

没错…还有无解的情况

让我们返回样例1" role="presentation" style="position: relative;">11看看。

我们说过对于每一个a" role="presentation" style="position: relative;">aa,都能确定唯一一个a1" role="presentation" style="position: relative;">a1a1。

但在无解的情况下它仍然成立吗?

肯定不一定。

因此，要判断无解，首先要确定通过其他a" role="presentation" style="position: relative;">aa求出的每个a1" role="presentation" style="position: relative;">a1a1都相同。这是第一。

还有,前文说道环上的每个结点对于a1" role="presentation" style="position: relative;">a1a1都有两条路径。但如果有多个环，那就有多个关系式，那就有许多关系式相同的情况。如果当他们的关系式符号相同时(即an+a1=A" role="presentation" style="position: relative;">an+a1=Aan+a1=A,an+a1=B" role="presentation" style="position: relative;">an+a1=Ban+a1=B或者an−a1=A" role="presentation" style="position: relative;">an−a1=Aan−a1=A,an−a1=B" role="presentation" style="position: relative;">an−a1=Ban−a1=B),我们一定能得出A=B" role="presentation" style="position: relative;">A=BA=B的结论,但如果A≠B" role="presentation" style="position: relative;">A≠BA≠B,显然无解。这是第二。

第三是最简单的，即1" role="presentation" style="position: relative;">11部分中说提到的范围max(B1,B2,...,Bk1,0)&lt;a1&lt;min(A1,A2,...,Ak2)" role="presentation" style="position: relative;">max(B1,B2,...,Bk1,0)<a1<min(A1,A2,...,Ak2)max(B1,B2,...,Bk1,0)<a1<min(A1,A2,...,Ak2)无解。

设A=min(A1,A2,...,Ak2)" role="presentation" style="position: relative;">A=min(A1,A2,...,Ak2)A=min(A1,A2,...,Ak2),B=max(B1,B2,...,Bk1,0)" role="presentation" style="position: relative;">B=max(B1,B2,...,Bk1,0)B=max(B1,B2,...,Bk1,0),那么当A−B−1≤0" role="presentation" style="position: relative;">A−B−1≤0A−B−1≤0时，a1" role="presentation" style="position: relative;">a1a1无解。

这就是这道题的正解？

4.汇集&实现

没错，这就是这题的正解(QWQ" role="presentation" style="position: relative;">QWQQWQ)

至于如何BFS/DFS" role="presentation" style="position: relative;">BFS/DFSBFS/DFS,一般只有遍历顺序的问题，这里直接将记录关系式(不等式)的问题与遍历问题结合，即vis[N][2]" role="presentation" style="position: relative;">vis[N][2]vis[N][2]

,这里vis[N][0]" role="presentation" style="position: relative;">vis[N][0]vis[N][0]中存的是a1&gt;B" role="presentation" style="position: relative;">a1>Ba1>B中的−B" role="presentation" style="position: relative;">−B−B(看后面代码的时候一定要注意这一点),目的是更简单直观，vis[N][1]" role="presentation" style="position: relative;">vis[N][1]vis[N][1]中存的是a1&lt;A" role="presentation" style="position: relative;">a1<Aa1<A中的A" role="presentation" style="position: relative;">AA;

然而还有一些细节需要提一下

1.图是双向的(蒟蒻没看清楚以至于调了2天的代码..)

2.2" role="presentation" style="position: relative;">22部分中，用奇环求a1" role="presentation" style="position: relative;">a1a1时最好放在DFS/BFS" role="presentation" style="position: relative;">DFS/BFSDFS/BFS后单独求解

3.2" role="presentation" style="position: relative;">22部分中，用奇环求出的a1" role="presentation" style="position: relative;">a1a1一定要满足1" role="presentation" style="position: relative;">11部分中说提到的范围max(B1,B2,...,Bk1,0)&lt;a1&lt;min(A1,A2,...,Ak2)" role="presentation" style="position: relative;">max(B1,B2,...,Bk1,0)<a1<min(A1,A2,...,Ak2)max(B1,B2,...,Bk1,0)<a1<min(A1,A2,...,Ak2),否则无解。

4.1" role="presentation" style="position: relative;">11部分中，一定要记得a1&lt;A" role="presentation" style="position: relative;">a1<Aa1<A,a1&gt;B" role="presentation" style="position: relative;">a1>Ba1>B,不是a1≤A" role="presentation" style="position: relative;">a1≤Aa1≤A,a1≥B" role="presentation" style="position: relative;">a1≥Ba1≥B。

对了….BFS/DFS" role="presentation" style="position: relative;">BFS/DFSBFS/DFS的时间复杂度是接近O(n)" role="presentation" style="position: relative;">O(n)O(n)的。可以非常轻松地通过本题。

现在就是你们最喜欢的代码了233…

有点丑..

DFS版

#include<iostream>
#include<cstdio>
#include<cstring>
#include<climits>
#include<algorithm>
#include<cmath>
#include<queue>
using namespace std;
#define MAXN 100000
#define MAXM 100000
#define INF 1000000000000000
struct node
{
    int next,to;
    long long w;
}e[MAXM*2+5];
long long T;
int cnt,m,n;
int x,y,z,flag;
long long ri=INF,le=1;
long long vis[2][MAXN+5];
int b[MAXN+5];
void fpush(int u,int v,long long w)
{
    e[++cnt].next=b[u];
    e[cnt].w=w;
    e[cnt].to=v;
    b[u]=cnt;
}
int Ch(int x){if(x==-1)x=0;return x;}
void dfs(int xr,int lim,int dec)
{
    for(int i=b[xr];i;i=e[i].next)
    {
        int xnext=e[i].to;
        int FD=Ch(dec);
        if(vis[FD][xnext]!=INF)
        {
            if(vis[FD][xnext]!=lim+e[i].w*dec){flag=1;return ;}
            else continue;
        }
        vis[FD][xnext]=lim+e[i].w*dec;
        if(!FD)le=max(le,-vis[FD][xnext]);
        else ri=min(ri,vis[FD][xnext]);
        dfs(xnext,lim+e[i].w*dec,-dec);
    }
}
int main()
{
    fill(vis[0],vis[0]+MAXN+1,INF);
    fill(vis[1],vis[1]+MAXN+1,INF);
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d%d",&x,&y,&z);
        fpush(x,y,z);fpush(y,x,z);
    }
    dfs(1,0,1);
    long long T=0;
    for(int i=1;i<=n;i++)
        if(vis[0][i]!=INF&&vis[1][i]!=INF)
        {
            long long NT=(vis[1][i]-vis[0][i])/2;
            if(NT<=le&&NT>=ri){flag=1;break;}
            if(NT<=0){flag=1;break;}
            if(T==0)T=NT;
            else if(T!=NT){flag=1;break;}
        }
    if(flag){printf("0\n");;}
    if(T)printf("1\n");
    else printf("%lld",max((long long)0,ri-le-1));
}

BFS版

#include<iostream>
#include<cstdio>
#include<cstring>
#include<climits>
#include<algorithm>
#include<cmath>
#include<queue>
using namespace std;
#define MAXN 100000
#define MAXM 100000
#define INF 1000000000000000
struct node
{
    int next,to;
    long long w;
}e[MAXM*2+5];
struct Qr
{
    int xr,dec;
    long long lim;
};
long long T;
int cnt,m,n;
int x,y,z,flag;
long long ri=INF,le=-INF;
long long vis[2][MAXN+5];
int b[MAXN+5];
void fpush(int u,int v,long long w)
{
    e[++cnt].next=b[u];
    e[cnt].w=w;
    e[cnt].to=v;
    b[u]=cnt;
}
int Ch(int x){if(x==-1)x=0;return x;}
int main()
{
    fill(vis[0],vis[0]+MAXN+1,INF);
    fill(vis[1],vis[1]+MAXN+1,INF);
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d%d",&x,&y,&z);
        fpush(x,y,z);fpush(y,x,z);
    }
    queue<Qr> Q;
    Q.push((Qr){1,1,0});
    while(!Q.empty())
    {
        Qr A=Q.front();
        Q.pop();
        int xr=A.xr,dec=A.dec;
        long long lim=A.lim;
        for(int i=b[xr];i;i=e[i].next)
        {
            int xnext=e[i].to;
            long long nlim=e[i].w-lim;
            int FD=Ch(dec);
            if(vis[FD][xnext]!=INF)
            {
                if(vis[FD][xnext]!=nlim){printf("0\n");return 0;}
                else continue;
            }
            vis[FD][xnext]=nlim;
            if(!FD)le=max(le,-vis[FD][xnext]);
            else ri=min(ri,vis[FD][xnext]);
            Q.push((Qr){xnext,-dec,nlim});
        }
    }
    long long T=-1;
    for(int i=1;i<=n;i++)
        if(vis[0][i]!=INF&&vis[1][i]!=INF)
        {
            long long NT=(vis[1][i]-vis[0][i])/2;
            if(NT<0){flag=1;break;}
            if(T==-1)T=NT;
            else if(T!=NT){flag=1;break;}
        }
    if(flag){printf("0\n");}
    if(T!=-1)
    {
        if(T>le&&T<ri)printf("1\n");
        else printf("0\n");
    }
    else printf("%lld",max((long long)0,ri-le-1));
}

END