HDU - 4289 Control (Dinic)

　You, the head of Department of Security, recently received a top-secret information that a group of terrorists is planning to transport some WMD ¹ 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.
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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 10 ⁷.
　　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

Sample Output

3

题意：
在图中，删除点需要相应的花费，求最小的花费，使得s,t不连通。
思路：
最大流=最小割。
一开始忘记了，这个，想了半天费用流。。。
拆带限制点的流量即可。

#include<iostream>

#include<algorithm>

#include<vector>

#include<stack>

#include<queue>

#include<map>

#include<set>

#include<cstdio>

#include<cstring>

#include<cmath>

#include<ctime>

#define fuck(x) cerr<<#x<<" = "<<x<<endl;

#define debug(a, x) cerr<<#a<<"["<<x<<"] = "<<a[x]<<endl;

#define ls (t<<1)

#define rs ((t<<1)|1)

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

const int maxn = ;

const int maxm = ;

const int inf = 0x3f3f3f3f;

const ll Inf = ;

const int mod = ;

const double eps = 1e-;

const double pi = acos(-);

int Head[maxn],cnt;

struct edge{

    int Next,v;

    int w;

}e[maxm];

void add_edge(int u,int v,int w){

    e[cnt].Next=Head[u];

    e[cnt].v=v;

    e[cnt].w=w;

    Head[u]=cnt++;

}

int D_vis[maxn],D_num[maxn];

int source,meeting;

int n,m;

bool bfs()

{

    memset(D_vis,,sizeof(D_vis));

    for(int i=;i<=*n;i++){//注意要覆盖所有点

        D_num[i]=Head[i];

    }

    D_vis[source]=;

    queue<int>q;

    q.push(source);

    while(!q.empty()){

        int u=q.front();

        q.pop();

        int k=Head[u];

        while(k!=-){

            if(!D_vis[e[k].v]&&e[k].w){

                D_vis[e[k].v]=D_vis[u]+;

                q.push(e[k].v);

            }

            k=e[k].Next;

        }

    }

    return D_vis[meeting];

}

int dfs(int u,int f)

{

    if(u==meeting){return f;}

    int &k=D_num[u];

    while(k!=-){

        if(D_vis[e[k].v]==D_vis[u]+&&e[k].w){

            int d=dfs(e[k].v,min(f,e[k].w));

            if(d>){

                e[k].w-=d;

                e[k^].w+=d;

                return d;

            }

        }

        k=e[k].Next;

    }

    return ;

}

int Dinic()

{

    int ans=;

    while(bfs()){

        int f;

        while((f=dfs(source,inf))>){

            ans+=f;

        }

    }

    return ans;

}

int city1(int x){

    return x;

}

int city2(int x){

    return x+n;

}

int main() {

//    ios::sync_with_stdio(false);

//    freopen("in.txt", "r", stdin);

    while (scanf("%d%d",&n,&m)!=EOF){

        memset(Head,-,sizeof(Head));

        cnt=;

        scanf("%d%d",&source,&meeting);

        meeting+=n;

        for(int i=;i<=n;i++){

            int x;

            scanf("%d",&x);

            add_edge(city1(i),city2(i),x);

            add_edge(city2(i),city1(i),);

        }for(int i=;i<=m;i++){

            int x,y;

            scanf("%d%d",&x,&y);

            add_edge(city2(x),city1(y),inf);

            add_edge(city1(y),city2(x),);

            add_edge(city2(y),city1(x),inf);

            add_edge(city1(x),city2(y),);

        }

        printf("%d\n",Dinic());

    }

    return ;

}

巴特西

HDU - 4289 Control (Dinic)

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