Description
Tom is a commander, his task is destroying his enemy’s transportation system.
Let’s represent his enemy’s transportation system as a simple directed graph G with n nodes and m edges. Each node is a city and each directed edge is a directed road. Each edge from node u to node v is associated with two values D and B, D is the cost to destroy/remove such edge, B is the cost to build an undirected edge between u and v.
His enemy can deliver supplies from city u to city v if and only if there is a directed path from u to v. At first they can deliver supplies from any city to any other cities. So the graph is a strongly-connected graph.
He will choose a non-empty proper subset of cities, let’s denote this set as S. Let’s denote the complement set of S as T. He will command his soldiers to destroy all the edges (u, v) that u belongs to set S and v belongs to set T.
To destroy an edge, he must pay the related cost D. The total cost he will pay is X. You can use this formula to calculate X:
After that, all the edges from S to T are destroyed. In order to deliver huge number of supplies from S to T, his enemy will change all the remained directed edges (u, v) that u belongs to set T and v belongs to set S into undirected edges. (Surely, those edges exist because the original graph is strongly-connected)
To change an edge, they must remove the original directed edge at first, whose cost is D, then they have to build a new undirected edge, whose cost is B. The total cost they will pay is Y. You can use this formula to calculate Y:
At last, if Y>=X, Tom will achieve his goal. But Tom is so lazy that he is unwilling to take a cup of time to choose a set S to make Y>=X, he hope to choose set S randomly! So he asks you if there is a set S, such that Y<X. If such set exists, he will feel unhappy, because he must choose set S carefully, otherwise he will become very happy.
Let’s represent his enemy’s transportation system as a simple directed graph G with n nodes and m edges. Each node is a city and each directed edge is a directed road. Each edge from node u to node v is associated with two values D and B, D is the cost to destroy/remove such edge, B is the cost to build an undirected edge between u and v.
His enemy can deliver supplies from city u to city v if and only if there is a directed path from u to v. At first they can deliver supplies from any city to any other cities. So the graph is a strongly-connected graph.
He will choose a non-empty proper subset of cities, let’s denote this set as S. Let’s denote the complement set of S as T. He will command his soldiers to destroy all the edges (u, v) that u belongs to set S and v belongs to set T.
To destroy an edge, he must pay the related cost D. The total cost he will pay is X. You can use this formula to calculate X:
After that, all the edges from S to T are destroyed. In order to deliver huge number of supplies from S to T, his enemy will change all the remained directed edges (u, v) that u belongs to set T and v belongs to set S into undirected edges. (Surely, those edges exist because the original graph is strongly-connected)
To change an edge, they must remove the original directed edge at first, whose cost is D, then they have to build a new undirected edge, whose cost is B. The total cost they will pay is Y. You can use this formula to calculate Y:
At last, if Y>=X, Tom will achieve his goal. But Tom is so lazy that he is unwilling to take a cup of time to choose a set S to make Y>=X, he hope to choose set S randomly! So he asks you if there is a set S, such that Y<X. If such set exists, he will feel unhappy, because he must choose set S carefully, otherwise he will become very happy.
Input
There are multiply test cases.
The first line contains an integer T(T<=200), indicates the number of cases.
For each test case, the first line has two numbers n and m.
Next m lines describe each edge. Each line has four numbers u, v, D, B.
(2=<n<=200, 2=<m<=5000, 1=<u, v<=n, 0=<D, B<=100000)
The meaning of all characters are described above. It is guaranteed that the input graph is strongly-connected.
The first line contains an integer T(T<=200), indicates the number of cases.
For each test case, the first line has two numbers n and m.
Next m lines describe each edge. Each line has four numbers u, v, D, B.
(2=<n<=200, 2=<m<=5000, 1=<u, v<=n, 0=<D, B<=100000)
The meaning of all characters are described above. It is guaranteed that the input graph is strongly-connected.
Output
For each case, output "Case #X: " first, X is the case number starting from 1.If such set doesn’t exist, print “happy”, else print “unhappy”.
Sample Input
2 3 3 1 2 2 2 2 3 2 2 3 1 2 2 3 3 1 2 10 2 2 3 2 2 3 1 2 2
Sample Output
Case #1: happy Case #2: unhappy
Hint
In first sample, for any set S, X=2, Y=4. In second sample. S= {1}, T= {2, 3}, X=10, Y=4.Source
2014 Multi-University Training Contest 7
先来一个错误但是AC了(数据水)的算法：
每个点单独作为S集合时，如果存在满足Y<X，就输出unhappy。否则输出happy。1
4 5
1 4 10 2
4 1 6 2
2 1 1 1
4 3 6 2
3 2 4 2
时，输出的是happy。可是实际上，选择1、4节点，Y=2，X=8，显然是unhappy。
附上非正解代码：（AC了只能说明题目数据太水了）
#include <cstdio> #include <cstring> #define N 205 #define sf(x) scanf("%d",&x) int x[N],y[N]; int main(){ int t; sf(t); for(int cas=1;cas<=t;cas++){ memset(x,0,sizeof x); memset(y,0,sizeof y); printf("Case #%d: ",cas); int n,m; sf(n); sf(m); for(int i=1;i<=m;i++){ int u,v,d,b; sf(u);sf(v);sf(d);sf(b); x[u]+=d; y[v]+=d+b; } int ok=1; for(int i=1;i<=n;i++) if(x[i]>y[i])ok=0; if(ok)puts("happy"); else puts("unhappy"); } }
正解是无源无汇带上下界判断是否有可行流。
将问题转化为网络流问题：
每条边下界为D，上界为D+B，如果存在可行流，那么
$$sum_{substack{uin S \ vin overline {S}}} f_{uv} = sum_{substack{uin S \ vin overline {S}}}f_{uv}\D_{uv} leq f_{uv} leq D_{uv}+B_{uv}$$
所以有
$$sum_{substack{uin S \ vin overline {S}}} D_{uv} leq sum_{substack{uin S \ vin overline {S}}}D_{uv}+ B_{uv}$$
因此只要求无源无汇上下界网络流是否存在可行流，如果不存在就是unhappy。
而无源汇有上下界的网络流，是否有可行流可以这样求：
人为加上源点s，汇点t，
边权改为上界-下界（这样转化为下界为0），
流入i点的下界和为in，流出的下界和为out，
in>out则s 到 i 连边，流量为in-out；
in<out则 i 到 t 连边，流量为out-in。
求s到t的最大流，如果源点汇点连接的边全部满流则有可行解。
参考国家集训队论文《一种简易的方法求解流量有上下界的网络中网络流问题》
#include <iostream> #include <algorithm> #include <cstdio> #include <cstring> #include <queue> using namespace std; #define N 500 #define M 200001 #define inf 0x3f3f3f3f struct edge{ int to,next,cap,flow; }e[M]; int head[N],cnt; int gap[N],dep[N],cur[N]; void init(){ cnt=0; memset(head, -1, sizeof head); } void add(int u,int v,int w,int rw=0){ e[cnt]=(edge){v,head[u],w,0}; head[u]=cnt++; e[cnt]=(edge){u,head[v],rw,0}; head[v]=cnt++; } int q[N]; void bfs(int st,int ed){ memset(dep,-1,sizeof dep); memset(gap,0,sizeof gap); gap[0]=1; int front=0,rear=0; dep[ed]=0; q[rear++]=ed; while(front!=rear){ int u=q[front++]; for(int i=head[u];~i;i=e[i].next){ int v=e[i].to; if(dep[v]!=-1)continue; q[rear++]=v; dep[v]=dep[u]+1; gap[dep[v]]++; } } } int s[N]; int sap(int st,int ed,int n){ bfs(st,ed); memcpy(cur,head,sizeof head); int top=0; int u=st; int ans=0; while(dep[st]<n){ if(u==ed){ int Min=inf; int inser; for(int i=0;i<top;i++) if(Min>e[s[i]].cap-e[s[i]].flow){ Min=e[s[i]].cap-e[s[i]].flow; inser=i; } for(int i=0;i<top;i++){ e[s[i]].flow+=Min; e[s[i]^1].flow-=Min; } ans+=Min; top=inser; u=e[s[top]^1].to; continue; } bool flag=false; int v; for(int i=cur[u];~i;i=e[i].next){ v=e[i].to; if(e[i].cap-e[i].flow&&dep[v]+1==dep[u]){ flag=true; cur[u]=i; break; } } if(flag){ s[top++]=cur[u]; u=v; continue; } int Min=n; for(int i=head[u];~i;i=e[i].next) if(e[i].cap-e[i].flow &&dep[e[i].to]<Min){ Min=dep[e[i].to]; cur[u]=i; } gap[dep[u]]--; if(!gap[dep[u]])return ans; gap[dep[u]=Min+1]++; if(u!=st)u=e[s[--top]^1].to; } return ans; } int main(){ int t; scanf("%d",&t); for(int cas=1;cas<=t;cas++){ printf("Case #%d: ",cas); int n,m; scanf("%d%d",&n,&m); int in[N]; int st=0,ed=n+1; memset(in,0,sizeof in); init(); for(int i=1;i<=m;i++){ int u,v,d,b; scanf("%d%d%d%d",&u,&v,&d,&b); add(u,v,b); in[v]+=d; in[u]-=d; } int need=0; for(int i=1;i<=n;i++){ if(in[i]>0){ add(st,i,in[i]); need+=in[i]; } else add(i,ed,-in[i]); } int ans=sap(st, ed, ed+1); if(need==ans)puts("happy"); else puts("unhappy"); } }