hdu 4940

Destroy Transportation system

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 900    Accepted Submission(s): 465

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

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

 

析: 有一个n个顶点m条边的有向图, 每条边有两个权值d, b表示拆掉该边花费以及将该边重建为无向边的花费, 要求从该图中任意选取一个子图S, 其补图为T,

X为拆掉边uv(u属于S, v属于T)的总花费, Y为拆掉边uv(u属于T, v属于S)以及重建该边的花费,  问是否任意都有X <= Y

看 X <= Y 的情况:

　　由于改图为强连通图, 每个点都有出边和入边, 出边加X的值, 入边加Y的值, 要是满足 X == Y 则肯定满足 X <= Y, 现在要使 X == Y , 即要求每个点的入流等于出流, 即求无源汇有上下界的网络流是否存在可行流

 

　　1.构建一个虚拟源点与汇点

　　2.计算每一个点的D[i] = 该点所有入流 - 观点所有出流

　　3.判断:

　　　　如果D[i] > 0, 累加D[i], 并在源点与该点之间建有向边

　　　　如果D[i] <= 0, 在该点与汇点之间建有向边

　　4.计算最大流

 

#include <queue>
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <functional>
#define ll long long
#define ull unsigned ll
#define Inf 0x7fffffff
#define maxn 50005
#define maxm 100005
#define P pair<int, int>
#define eps 1e-6
#define mod 10000
using namespace std;
const int N = 205;
int t, n, m, len, ans;
int u, v, d, b, S, T;
int D[N], h[N], dis[N];
struct Node{
    int v, w, nxt;
}g[maxn];
void init(){
    ans = 0;
    S = 0, T = n+1;
    fill(h, h+N, -1);
    fill(D, D+N, 0);
}
void add(int u, int v, int w){
    g[ans].v = v;
    g[ans].w = w;
    g[ans].nxt = h[u];
    h[u] = ans++;
}
bool bfs(){
    fill(dis, dis+N, -1);
    queue<int>Q;
    Q.push(S);
    dis[S] = 0;
    while(!Q.empty()){
        int u = Q.front();
        Q.pop();
        for(int i = h[u]; i+1; i = g[i].nxt){
            int v = g[i].v;
            int w = g[i].w;
            if(w > 0 && !(dis[v]+1)){
                dis[v] = dis[u]+1;
                Q.push(v);
            }
        }
    }
    return dis[T]+1;
}
int dfs(int u, int flow){
    int cost = 0;
    if(u == T || !flow)
        return flow;
    for(int i = h[u]; i+1; i = g[i].nxt){
        int v = g[i].v;
        int w = g[i].w;
        if(w > 0 && dis[v] == dis[u]+1){
            int f = dfs(v, min(w, flow-cost));
            if(f > 0){
                g[i].w -= f;
                g[i^1].w += f;
                cost += f;
                if(cost == flow) //剪枝
                    break;
            }
        }
    }
    return cost;
}
int dinic(){
    int f, flow = 0;
    while(bfs()){
        while((f = dfs(S, Inf)) > 0){
            flow += f;
        }
    }
    return flow;
}
int main(){
    int tot = 1;
    scanf("%d", &t);
    while(t--){
        scanf("%d%d", &n, &m);
        init();
        for(int i = 0; i < m; i ++){
            scanf("%d%d%d%d", &u, &v, &d, &b);
            add(u, v, b);
            add(v, u, 0);
            D[u] -= d;
            D[v] += d;
        }
        int flow = 0;
        for(int i = 1; i <= n; i ++){
            if(D[i] > 0){
                flow += D[i];
                add(S, i, D[i]);
                add(i, S, 0);
            }else{
                add(i, T, -D[i]);
                add(T, i, 0);
            }
        }
        if(flow == dinic())
            printf("Case #%d: happy
", tot++);
        else
            printf("Case #%d: unhappy
", tot++);
    }
    return 0;
}