首页 > 代码库 > hdu 4289 Control（网络流 最大流+拆点）（模板）

hdu 4289 Control（网络流 最大流+拆点）（模板）

2024-07-16 17:02:45   223人阅读

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4289


Control

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1545    Accepted Submission(s): 677


Problem Description
　　You, the head of Department of Security, recently received a top-secret information that a group of terrorists is planning to transport some WMD1 from one city (the source) to another one (the destination). You know their date, source and destination, and they are using the highway network.
　　The highway network consists of bidirectional highways, connecting two distinct city. A vehicle can only enter/exit the highway network at cities only.
　　You may locate some SA (special agents) in some selected cities, so that when the terrorists enter a city under observation (that is, SA is in this city), they would be caught immediately.
　　It is possible to locate SA in all cities, but since controlling a city with SA may cost your department a certain amount of money, which might vary from city to city, and your budget might not be able to bear the full cost of controlling all cities, you must identify a set of cities, that:
　　* all traffic of the terrorists must pass at least one city of the set.
　　* sum of cost of controlling all cities in the set is minimal.
　　You may assume that it is always possible to get from source of the terrorists to their destination.
------------------------------------------------------------
1 Weapon of Mass Destruction
 

Input
　　There are several test cases.
　　The first line of a single test case contains two integer N and M ( 2 <= N <= 200; 1 <= M <= 20000), the number of cities and the number of highways. Cities are numbered from 1 to N.
　　The second line contains two integer S,D ( 1 <= S,D <= N), the number of the source and the number of the destination.
　　The following N lines contains costs. Of these lines the ith one contains exactly one integer, the cost of locating SA in the ith city to put it under observation. You may assume that the cost is positive and not exceeding 107.
　　The followingM lines tells you about highway network. Each of these lines contains two integers A and B, indicating a bidirectional highway between A and B.
　　Please process until EOF (End Of File).
 
Output
　　For each test case you should output exactly one line, containing one integer, the sum of cost of your selected set.
　　See samples for detailed information.
 
Sample Input
5 6
5 3
5
2
3
4
12
1 5
5 4
2 3
2 4
4 3
2 1
 
Sample Output
3
 
Source
2012 ACM/ICPC Asia Regional Chengdu Online 

（此题如用超级源点，汇点解 去掉注释即可）；
代码如下：
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int MAXN = 3017;//点数的最大值
const int MAXM = 100017;//边数的最大值
const int INF = 0x3f3f3f3f;
struct Edge
{
    int to, cap, flow;
	int next;
}edge[MAXM];//注意是MAXM
int tol;
int head[MAXN];
int dep[MAXN],pre[MAXN],cur[MAXN];
int gap[MAXN];//gap[x]=y :说明残留网络中dep[i]==x的个数为y
void init()
{
    tol = 0;
    memset(head,-1,sizeof (head));
}
//加边，单向图三个参数，双向图四个参数
void addedge (int u,int v,int w,int rw=0)
{
    edge[tol].to = v;edge[tol].cap = w;
	edge[tol].next = head[u];
    edge[tol].flow = 0;
	head[u] = tol++;
    edge[tol].to = u;edge[tol].cap = rw;
	edge[tol]. next = head[v];
    edge[tol].flow = 0;head[v]=tol++;
}
//输入参数：起点、终点、点的总数
//点的编号没有影响，只要输入点的总数
int sap(int start,int end, int N)
{
    memset(gap,0,sizeof(gap));
    memset(dep,0,sizeof(dep));
    memcpy(cur,head,sizeof(head));
    int u = start;
    pre[u] = -1;
    gap[0] = N;
    int ans = 0;
    int i;
    while(dep[start] < N)
    {
        if(u == end)
        {
            int Min = INF;
            for( i = pre[u];i != -1; i = pre[edge[i^1]. to])
            {
                if(Min > edge[i].cap - edge[i]. flow)
                    Min = edge[i].cap - edge[i].flow;
            }
            for( i = pre[u];i != -1; i = pre[edge[i^1]. to])
            {
                edge[i].flow += Min;
                edge[i^1].flow -= Min;
            }
            u = start;
            ans += Min;
            continue;
        }
        bool flag =  false;
        int v;
        for( i = cur[u]; i != -1;i = edge[i].next)
        {
            v = edge[i]. to;
            if(edge[i].cap - edge[i].flow && dep[v]+1 == dep[u])
            {
                flag =  true;
                cur[u] = pre[v] = i;
                break;
            }
        }
        if(flag)
        {
            u = v;
            continue;
        }
        int Min = N;
        for( i = head[u]; i !=  -1; i = edge[i]. next)
        {
            if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min)
            {
                Min = dep[edge[i].to];
                cur[u] = i;
            }
        }
        gap[dep[u]]--;
        if(!gap[dep[u]]) 
			return ans;
        dep[u] = Min+1;
        gap[dep[u]]++;
        if(u != start) 
			u = edge[pre[u]^1].to;
    }
    return ans;
}
int main()
{
    int n,m;
	int a,b;
	int c,s,t;
	int i;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        init();//初始化
        scanf("%d%d",&s,&t);//源点，汇点
		int w;
	//	int ds = 0, dt = 2*n+1;//超级源点，超级汇点，初始化
        for(i = 1; i <= n; i++)
        {
            scanf("%d",&w);//每个顶点的值
            addedge(i,i+n,w,0);//对于每个节点id，拆成 id 和 id+n 两个点
        }
        for(i = 1; i <= m; i++)//边数
        {
            scanf("%d%d",&a,&b);
            addedge(a+n,b,INF,0);
            addedge(b+n,a,INF,0);
        }
	//	addedge(ds,s,INF,0);//添加超级源点
	//	addedge(t+n,dt,INF,0);//添加超级汇点
	//	int ans = sap(ds, dt, 2*(n+1));// 总节点数的变化
		int ans = sap(s, t+n, 2*n);
        printf("%d\n",ans);
    }
    return 0;
}


联系
我们