[分类讨论 思路题] BZOJ 3135 [Baltic2013] pipes

官方题解：http://boi2013.informatik-olympiade.de/wp-content/uploads/2013/05/pipes-spoiler.pdf

输入构成一个n个m元方程组 当m<=n是有唯一解，而m>=n-1

所以只需考虑 m==n-1 m==n

m==n-1 

这是一棵树 从叶子往上推就好了 在根节点判一下是否矛盾

m==n

这是一个环套树 先处理到只剩一个环

然后如果环是奇数 可以解方程

如果是偶数 就有无数解

#include<cstdio>#include<cstdlib>#include<algorithm>#define V G[p].vusing namespace std;typedef long long ll;inline char nc(){static char buf[100000],*p1=buf,*p2=buf;if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }return *p1++;}inline void read(int &x){char c=nc(),b=1;for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;}inline void read(ll &x){char c=nc(),b=1;for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;}struct edge{int u,v;int idx;int next;};edge G[1000005];int head[100005],inum;inline void add(int u,int v,int idx,int p){G[p].u=u; G[p].v=v; G[p].idx=idx; G[p].next=head[u]; head[u]=p;}int n,m;ll s[100005],w[100005];int deg[100005];namespace Work1{inline ll dfs(int u,int fa){ll sum=0;for (int p=head[u];p;p=G[p].next)if (V!=fa)sum+=(w[G[p].idx]=dfs(V,u));return s[u]-sum;}inline void Solve(){if (dfs(1,0))printf("0\n");else{for (int i=1;i<=m;i++)printf("%lld\n",2*w[i]);}}}namespace Work2{int Q[100005],l,r;int cnt;int a[100005],e[100005];int del[100005];inline void Solve(){l=r=-1;for (int i=1;i<=n;i++)if (deg[i]==1)Q[++r]=i;while (l<r){int u=Q[++l]; del[u]=1;for (int p=head[u];p;p=G[p].next)if (!del[V]){w[G[p].idx]=s[u];s[V]-=s[u];deg[V]--;if (deg[V]==1)Q[++r]=V;}}for (int i=1,flag=0;i<=n && !flag;i++)if (!del[i])a[++cnt]=i,flag=1;while (1){int u=a[cnt]; del[u]=1;int flag=0;for (int p=head[u];p && !flag;p=G[p].next)if (!del[V]) a[++cnt]=V,e[cnt-1]=G[p].idx,flag=1;if (!flag)break;}for (int p=head[a[cnt]],flag=0;p && !flag;p=G[p].next)if (V==a[1])e[cnt]=G[p].idx,flag=1;if (cnt&1){ll itmp=0;for (int i=1,t=1;i<=cnt;i++,t=-t)itmp+=s[a[i]]*t;w[e[cnt]]=itmp/2;for (int i=cnt-1;i>=1;i--)w[e[i]]=s[a[i+1]]-w[e[i+1]];for (int i=1;i<=m;i++)printf("%lld\n",2*w[i]);}elseprintf("0\n");}}int main(){int iu,iv;freopen("t.in","r",stdin);freopen("t.out","w",stdout);read(n); read(m);if (m>n) return printf("0\n"),0;for (int i=1;i<=n;i++) read(s[i]);for (int i=1;i<=m;i++)read(iu),read(iv),add(iu,iv,i,++inum),add(iv,iu,i,++inum),deg[iu]++,deg[iv]++;if (m==n-1)Work1::Solve();else if (m==n)Work2::Solve();return 0;}