Gym 101485 E Elementary Math 网络流 或者 二分图
阅读原文时间:2023年07月08日阅读:4

题意:

输入一个n,后面输入n行,每一行两个数a、b。你可以对a、b进行三种操作:+、-、*

你需要保证对每一行a、b选取一个操作得到一个结果

你要保证这n行每一个式子选取的操作之后得到的结果都不一样。如果找不到就输出impossible

Sample Input 1

1 4

1 5

3 3

4 5

-1 -6

Sample Output

1 + 5 = 6

3 * 3 = 9

4 - 5 = -1

-1 - -6 = 5
Sample Input 2

2 4

-4 2

-4 2

-4 2

-4 2

Sample Output

impossible

题解:

网络流跑一边最大流就可以了

首先建一个汇点st,然后让st点和n个式子相连,把这n个式子看成一个点,然后一个式子对应三个结果,我们分别对着三个结果看成一个点,然后连线。之后把所有的结果点连接到一个尾点en

然后就是路径输出就行了,如果一个路径有流量,那么这个路径的流量信息肯定有改变,就判断这一点就可以输出路径

建图其实很简单,这里主要说一下不同网络流的复杂度

简单总结

FF算法: 利用dfs实现,时间复杂度O(V*E^2)

EK算法:利用bfs实现,时间复杂度O(V*E^2)

Dinic算法:递归实现,时间复杂度O(V^2*E)

SAP算法:时间复杂度O(V^2*E)但是加上弧优化和间隙优化之后时间复杂度非常可观

我以前认为点的数量肯定比边的数量小,所以dinic算法复杂度最小,所以就只记住了一个dinic算法

但是这一道题使用dinic算法就TLE了

1 #include
2 #include
3 #include
4 #include
5 #include
6 #define mem(a,b) memset(a,b,sizeof(a))
7 using namespace std;
8 typedef long long ll;
9 const int maxn=8000;
10 queue q;
11 int n,m,cnt,st,en,ans;
12 ll ref[10010],ra[2510],rb[2510];
13 int head[maxn],dis[maxn],cur[maxn];
14 map r;
15 struct edge
16 {
17 int v,next,c,flow;
18 } e[maxn<<2]; 19 void add_edge(int x,int y,int z) 20 { 21 e[cnt].v=y; 22 e[cnt].c=z; 23 e[cnt].flow=0; 24 e[cnt].next=head[x]; 25 head[x]=cnt++; 26 27 e[cnt].v=x; 28 e[cnt].c=0; 29 e[cnt].flow=0; 30 e[cnt].next=head[y]; 31 head[y]=cnt++; 32 } 33 int bfs() 34 { 35 mem(dis,0); 36 dis[st]=1; 37 queuer;
38 r.push(st);
39 while(!r.empty())
40 {
41 int x=r.front();
42 r.pop();
43 //printf("%d***\n",x);
44 for(int i=head[x]; i!=-1; i=e[i].next)
45 {
46 int v=e[i].v;
47 if(!dis[v] && e[i].c>e[i].flow)
48 {
49 //printf("%d***\n",v);
50 dis[v]=dis[x]+1;
51 r.push(v);
52 }
53 }
54 }
55 return dis[en];
56 }
57 int dinic(int s,int limit)
58 {
59 if(s==en || !limit) return limit;
60 int ans=0;
61 for(int &i=cur[s]; i!=-1; i=e[i].next)
62 {
63 int v=e[i].v,feed;
64 if(dis[v]!=dis[s]+1) continue;
65 feed=dinic(v,min(limit,e[i].c-e[i].flow));
66 if(feed)
67 {
68 e[i].flow+=feed;
69 e[i^1].flow-=feed;
70 limit-=feed;
71 ans+=feed;
72 if(limit==0) break;
73 }
74 }
75 if(!ans) dis[s]=-1;
76 return ans;
77 }
78 inline int rd()
79 {
80 int ret=0,f=1;
81 char gc=getchar();
82 while(gc<'0'||gc>'9')
83 {
84 if(gc=='-') f=-f;
85 gc=getchar();
86 }
87 while(gc>='0'&&gc<='9') ret=ret*10+(gc^'0'),gc=getchar(); 88 return ret*f; 89 } 90 int main() 91 { 92 n=m=rd(),st=0; 93 int i,j; 94 memset(head,-1,sizeof(head)); 95 for(i=1; i<=n; i++) 96 { 97 ra[i]=rd(),rb[i]=rd(); 98 if(!r[ra[i]+rb[i]]) r[ra[i]+rb[i]]=++m,ref[m]=ra[i]+rb[i]; 99 if(!r[ra[i]-rb[i]]) r[ra[i]-rb[i]]=++m,ref[m]=ra[i]-rb[i]; 100 if(!r[ra[i]*rb[i]]) r[ra[i]*rb[i]]=++m,ref[m]=ra[i]*rb[i]; 101 add_edge(i,r[ra[i]+rb[i]],1); 102 add_edge(i,r[ra[i]-rb[i]],1); 103 add_edge(i,r[ra[i]*rb[i]],1); 104 add_edge(st,i,1); 105 } 106 en=m+1; 107 for(i=n+1; i<=m; i++) add_edge(i,en,1); 108 int ans=0; 109 while(bfs()) 110 { 111 for(int i=0; i<=en+1; ++i) 112 { 113 cur[i]=head[i]; 114 } 115 ans+=dinic(st,1); 116 } 117 if(ans!=n) 118 { 119 printf("impossible"); 120 return 0; 121 } 122 for(i=1; i<=n; i++) 123 { 124 for(j=head[i]; j!=-1; j=e[j].next) 125 { 126 if(e[j].flow>0 && e[j].v!=st)
127 {
128 if(ref[e[j].v]==ra[i]+rb[i])
129 printf("%lld + %lld = %lld\n",ra[i],rb[i],ra[i]+rb[i]);
130 else if(ref[e[j].v]==ra[i]-rb[i])
131 printf("%lld - %lld = %lld\n",ra[i],rb[i],ra[i]-rb[i]);
132 else if(ref[e[j].v]==ra[i]*rb[i])
133 printf("%lld * %lld = %lld\n",ra[i],rb[i],ra[i]*rb[i]);
134 break;
135 }
136 }
137 }
138 return 0;
139 }

AC代码:

#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;
queue q;
int n,m,cnt,st,en,ans;
ll ref[10010],ra[2510],rb[2510];
int to[100000],nxt[100000],head[10010],val[100000],dis[10010];
map r;

inline void add(int a,int b,int c)
{
to[cnt]=b,val[cnt]=c,nxt[cnt]=head[a],head[a]=cnt++;
to[cnt]=a,val[cnt]=0,nxt[cnt]=head[b],head[b]=cnt++;
}
int dfs(int x,int mf)
{
if(x==en) return mf;
int i,k,temp=mf;
for(i=head[x]; i!=-1; i=nxt[i])
if(dis[to[i]]==dis[x]+1&&val[i])
{
k=dfs(to[i],min(temp,val[i]));
if(!k) dis[to[i]]=0;
temp-=k,val[i]-=k,val[i^1]+=k;
if(!temp) break;
}
return mf-temp;
}
int bfs()
{
while(!q.empty()) q.pop();
memset(dis,0,sizeof(dis));
q.push(st),dis[st]=1;
int i,u;
while(!q.empty())
{
u=q.front(),q.pop();
for(i=head[u]; i!=-1; i=nxt[i]) if(!dis[to[i]]&&val[i])
{
dis[to[i]]=dis[u]+1;
if(to[i]==en) return 1;
q.push(to[i]);
}
}
return 0;
}
inline int rd()
{
int ret=0,f=1;
char gc=getchar();
while(gc<'0'||gc>'9')
{
if(gc=='-') f=-f;
gc=getchar();
}
while(gc>='0'&&gc<='9') ret=ret*10+(gc^'0'),gc=getchar();
return ret*f;
}
int main()
{
n=m=rd(),st=0;
int i,j;
memset(head,-1,sizeof(head));
for(i=1; i<=n; i++)
{
ra[i]=rd(),rb[i]=rd();
if(!r[ra[i]+rb[i]])
r[ra[i]+rb[i]]=++m,ref[m]=ra[i]+rb[i];
if(!r[ra[i]-rb[i]])
r[ra[i]-rb[i]]=++m,ref[m]=ra[i]-rb[i];
if(!r[ra[i]*rb[i]])
r[ra[i]*rb[i]]=++m,ref[m]=ra[i]*rb[i];
add(i,r[ra[i]+rb[i]],1);
add(i,r[ra[i]-rb[i]],1);
add(i,r[ra[i]*rb[i]],1);
add(st,i,1);
}
en=m+1;
for(i=n+1; i<=m; i++) add(i,en,1);
while(bfs())
ans+=dfs(st,1<<30);
if(ans!=n)
{
printf("impossible");
return 0;
}
for(i=1; i<=n; i++)
{
for(j=head[i]; j!=-1; j=nxt[j])
{
if(!val[j])
{
if(ref[to[j]]==ra[i]+rb[i])
printf("%lld + %lld = %lld\n",ra[i],rb[i],ra[i]+rb[i]);
else if(ref[to[j]]==ra[i]-rb[i])
printf("%lld - %lld = %lld\n",ra[i],rb[i],ra[i]-rb[i]);
else if(ref[to[j]]==ra[i]*rb[i])
printf("%lld * %lld = %lld\n",ra[i],rb[i],ra[i]*rb[i]);
}
}
}
return 0;
}

二分图题解:

就是把n个式子看成n个点,把n个式子对应的三个结果看成3*n个点

二分图匹配就行

代码:

#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;
const int maxn=7500+10;
const int INF=0x3f3f3f3f;
const long long ll_INF=0x3f3f3f3f3f3f3f3fll;
ll match[maxn],visit[maxn],n,m,grap[2505][maxn],link[maxn];
mapr;
ll dfs_solve(ll x)
{
for(ll i=1; i<=m; ++i)
{
//printf("%d***\n",grap[x][i]);
if(grap[x][i] && !visit[i])
{
visit[i]=1;
//printf("%d %d %d****\n",x,i,match[i]);
if(match[i]==0 || dfs_solve(match[i]))
{
match[i]=x;
link[x]=i;
return 1;
}
}
}
return 0;
}
ll hungran()
{
memset(match,0,sizeof(match));
ll sum=0;
for(ll i=1; i<=n; ++i)
{
memset(visit,0,sizeof(visit));
sum+=dfs_solve(i);
}
return sum;
}
ll que[maxn];
struct shudui
{
ll a,b;
} w[2505];
int main()
{
scanf("%lld",&n);
m=0;
for(ll i=1; i<=n; ++i)
{
ll a,b;
scanf("%lld%lld",&a,&b);
w[i].a=a;
w[i].b=b;
if(r[a*b]==0)
{
r[a*b]=++m;
que[m]=a*b;
}
grap[i][r[a*b]]=1;
if(r[a+b]==0)
{
r[a+b]=++m;
que[m]=a+b;
}
grap[i][r[a+b]]=1;
if(r[a-b]==0)
{
r[a-b]=++m;
que[m]=a-b;
}
grap[i][r[a-b]]=1;
}
ll ans=hungran();
//printf("%lld***%lld %lld\n",ans,n,m);
if(ans!=n)
{
printf("impossible\n");
return 0;
}
for(ll i=1; i<=n; ++i)
{
ll x=que[link[i]];
ll a=w[i].a;
ll b=w[i].b;
//printf("%lld %lld %lld\n",a,b,x);
if(x==a+b)
printf("%lld + %lld = %lld\n",a,b,x);
else if(x==a-b)
printf("%lld - %lld = %lld\n",a,b,x);
else if(x==a*b)
printf("%lld * %lld = %lld\n",a,b,x);
}
return 0;
}

/*
4
1 5
3 3
4 5
-1 -6

*/