【HDU 4940】Destroy Transportation system（数据水/无源无汇带上下界可行流）-布布扣-bubuko.com

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:
技术分享

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:
技术分享

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

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。

为什么错呢？输出happy时，即两个点单独作为S时，都有Y1>=X1,Y2>=X2，如果两个点之间没有边，它们一起作为S时，X=X1+X2,Y=Y1+Y2，则Y>=X；但是如果两个点有边相连，X=X1+X2-(S1到S2的D)-(S2到S1的D),Y=Y1+Y2-(S1到S2的D+B)-(S2到S1的D+B)，那么Y+B12+B21>=X,就有可能是Y<X了。当输入

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

$$\sum_{\substack{u\in S \\ v\in \overline {S}}} f_{uv} = \sum_{\substack{u\in S \\ v\in \overline {S}}}f_{uv}\\D_{uv} \leq f_{uv} \leq D_{uv}+B_{uv}$$

所以有

$$\sum_{\substack{u\in S \\ v\in \overline {S}}} D_{uv} \leq \sum_{\substack{u\in S \\ v\in \overline {S}}}D_{uv}+ B_{uv}$$

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 <cmath>
#include <queue>
#include <vector>
#include <map>
#define sf(a) scanf("%d",&a)
using namespace std;
#define N 500
#define M 100001
#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;
	sf(t);
	for(int cas=1;cas<=t;cas++){
		printf("Case #%d: ",cas);
		int n,m;
		sf(n);
		sf(m);
		int s=0,t=n+1,in[N];
		memset(in,0,sizeof in);
		init();
		for(int i=1;i<=m;i++){
			int u,v,d,b;
			sf(u);sf(v);sf(d);sf(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(s,i,in[i]);
				need+=in[i];
			}
			else add(i,t,-in[i]);
		}
		int ans=sap(s, t, t+1);
		if(need==ans)puts("happy");
		else puts("unhappy");
	}
}

【HDU 4940】Destroy Transportation system（数据水/无源无汇带上下界可行流）

原文：http://www.cnblogs.com/flipped/p/5792694.html

【HDU 4940】Destroy Transportation system（数据水/无源无汇带上下界可行流）