Destroy Transportation system

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 21 Accepted Submission(s): 17

Problem 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.

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.

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

Sample Output

Case #1: happy
Case #2: unhappy

HintIn first sample, for any set S, X=2, Y=4. 
In second sample. S= {1}, T= {2, 3}, X=10, Y=4.

Author

UESTC

Source

2014 Multi-University Training Contest 7

官方题解：http://blog.sina.com.cn/s/blog_6bddecdc0102uzka.html

寻找是否存在能单独作为S集合的点！

代码例如以下：

//#pragma warning (disable:4786)
#include <cstdio>
#include <cstring>
typedef long long LL;
#define N 1017
LL X[N], Y[N];
void init()
{
    memset(X, 0, sizeof(X));
    memset(Y, 0, sizeof(Y));
}
int main()
{
    int n, m;
    int u, v, D, B;
    int t;
    int cas = 0;
    int flag = 0;
    int i;
    scanf("%d", &t);
    while(t--)
    {
        init();
        flag = 0;
        scanf("%d%d", &n,&m);
        for(i=1; i<=m; i++)
        {
            scanf("%d%d%d%d", &u, &v, &D, &B);
            X[u]+=D;
            Y[v]+=(D+B);

        }
        for(i = 1; i <= n; i++)
        {
            if(Y[i] < X[i])
            {
                 flag = 1;
                 break;
            }
        }
        if(flag)
            printf("Case #%d: unhappy
",++cas);
        else
            printf("Case #%d: happy
",++cas);
    }
    return 0;
}

贴一发官方题解：

#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 209
#define maxm 20000
#define INF 1e9
using namespace std;
struct Edge
{
    int v,cap,next;
}edge[maxm];
int n,tot,src,des;
int head[maxn],h[maxn],gap[maxn],B[maxn];
void addedge(int u,int v,int cap)
{
    edge[tot].v=v;
    edge[tot].cap=cap;
    edge[tot].next=head[u];
    head[u]=tot++;
    edge[tot].v=u;
    edge[tot].cap=0;
    edge[tot].next=head[v];
    head[v]=tot++;
}
int dfs(int u,int cap)
{
    if(u==des)return cap;
    int minh=n-1;
    int lv=cap,d;
    for(int e=head[u];e!=-1;e=edge[e].next)
    {
        int v=edge[e].v;
        if(edge[e].cap>0)
        {
            if(h[v]+1==h[u])
            {
                d=min(lv,edge[e].cap);
                d=dfs(v,d);
                edge[e].cap-=d;
                edge[e^1].cap+=d;
                lv-=d;
                if(h[src]>=n)return cap-lv;
                if(lv==0)
                    break;
            }
            minh=min(minh,h[v]);
        }
    }
    if(lv==cap)
    {
        --gap[h[u]];
        if(gap[h[u]]==0)
            h[src]=n;
        h[u]=minh+1;
        ++gap[h[u]];
    }
    return cap-lv;
}
int sap()
{
    int flow=0;
    memset(gap,0,sizeof(gap));
    memset(h,0,sizeof(h));
    gap[0]=n;
    while(h[src]<n)flow+=dfs(src,INF);
    return flow;
}
int main()
{
    int N,M;
    int cot=1;
    int CAS;
    scanf("%d", &CAS);
    while(CAS--)
    {
    	scanf("%d%d",&N,&M);
        memset(head,-1,sizeof(head));
        memset(B,0,sizeof(B));
        tot=0;src=0;des=N+1;n=des+1;
        for(int i=1;i<=M;i++)
        {
            int u,v,b,l;
            scanf("%d%d%d%d",&u,&v,&b,&l);
            addedge(u,v,l);
            B[v]+=b;B[u]-=b;
        }
        int sum=0;
        for(int i=1;i<=N;i++)
        {
            if(B[i]>0)
            {
                addedge(src,i,B[i]);
                sum+=B[i];
            }
            else if(B[i]<0)
            {
                addedge(i,des,-B[i]);
            }
        }
        int flow=sap();
        if(flow==sum)
        {
            printf("Case #%d: happy
",cot++);
        }
        else
        {
            printf("Case #%d: unhappy
",cot++);
        }
    }
    return 0;
}