HDU 3879 最大權閉合圖

Base Station

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 65768/32768 K (Java/Others)
Total Submission(s): 1788    Accepted Submission(s): 758

Problem Description

A famous mobile communication company is planning to build a new set of base stations. According to the previous investigation, n places are chosen as the possible new locations to build those new stations. However, the condition of each position varies much, so the costs to built a station at different places are different. The cost to build a new station at the ith place is P_i (1<=i<=n).

When complete building, two places which both have stations can communicate with each other.

Besides, according to the marketing department, the company has received m requirements. The ith requirement is represented by three integers A_i, B_i and C_i, which means if place A_i and B_i can communicate with each other, the company will get C_i profit.

Now, the company wants to maximize the profits, so maybe just part of the possible locations will be chosen to build new stations. The boss wants to know the maximum profits.

 

Input

Multiple test cases (no more than 20), for each test case:
The first line has two integers n (0<n<=5000) and m (0<m<=50000).
The second line has n integers, P1 through Pn, describes the cost of each location.
Next m line, each line contains three integers, A_i, B_i and C_i, describes the ith requirement.

 

Output

One integer each case, the maximum profit of the company.

 

Sample Input

5 5 1 2 3 4 5 1 2 3 2 3 4 1 3 3 1 4 2 4 5 3

 

Sample Output

4

 

Author

liulibo

 

Source

2011 Multi-University Training Contest 5 - Host by BNU

 

Recommend

lcy   |   We have carefully selected several similar problems for you:  3657 3338 3277 3887 3889 

分析：最大權閉合圖模型。

代碼：

//Isap算法，複雜度O(n^2m)
#pragma comment(linker,"/STACK:102400000,102400000")
#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <string>
#include <math.h>
#include <queue>
#include <stack>
#include <map>
#include <set>
using namespace std;
typedef long long ll;   //記得必要的時候改成無符號
const int maxn=55005;
const int maxm=1000005;
const int INF=1000000000;
struct EdgeNode
{
    int from;
    int to;
    int cost;
    int next;
}edge[maxm];
int head[maxn],cnt;
void add(int x,int y,int z)
{
    edge[cnt].from=x;edge[cnt].to=y;edge[cnt].cost=z;edge[cnt].next=head[x];head[x]=cnt++;
    edge[cnt].from=y;edge[cnt].to=x;edge[cnt].cost=0;edge[cnt].next=head[y];head[y]=cnt++;
}

void init()
{
    cnt=0;
    memset(head,-1,sizeof(head));
}

int S,T,n,m;
int d[maxn],gap[maxn],curedge[maxn],pre[maxn];
//curedge[]爲當前弧數組，pre爲前驅數組

int sap(int S,int T,int n)  //n爲點數
{
    int cur_flow,flow_ans=0,u,tmp,neck,i;
    memset(d,0,sizeof(d));
    memset(gap,0,sizeof(gap));
    memset(pre,-1,sizeof(pre));
    for(i=0;i<=n;i++)curedge[i]=head[i]; //初始化當前弧爲第一條鄰接表
    gap[0]=n;
    u=S;
    while(d[S]<n)             //當d[S]>=n時，網絡中肯定出現了斷層
    {
        if(u==T)
        {
            cur_flow=INF;
            for(i=S;i!=T;i=edge[curedge[i]].to)
            {                           //增廣成功，尋找瓶頸邊
                if(cur_flow>edge[curedge[i]].cost)
                {
                    neck=i;
                    cur_flow=edge[curedge[i]].cost;
                }
            }
            for(i=S;i!=T;i=edge[curedge[i]].to)
            {                             //修改路徑上的邊容量
                tmp=curedge[i];
                edge[tmp].cost-=cur_flow;
                edge[tmp^1].cost+=cur_flow;
            }
            flow_ans+=cur_flow;
            u=neck;                     //下次增廣從瓶頸邊開始
        }
        for(i=curedge[u];i!=-1;i=edge[i].next)
            if(edge[i].cost&&d[u]==d[edge[i].to]+1)
               break;
        if(i!=-1)
        {
            curedge[u]=i;
            pre[edge[i].to]=u;
            u=edge[i].to;
        }
        else
        {
            if(0==--gap[d[u]])break;    //gap優化
            curedge[u]=head[u];
            for(tmp=n,i=head[u];i!=-1;i=edge[i].next)
                if(edge[i].cost)
                   tmp=min(tmp,d[edge[i].to]);
            d[u]=tmp+1;
            ++gap[d[u]];
            if(u!=S)u=pre[u];           //重標號並且從當前點前驅重新增廣
        }
    }
    return flow_ans;
}

int main()
{
    int ans,i,x,y,z,sum;
    while(~scanf("%d%d",&n,&m))
    {
        sum=0;
        init(); S=0; T=n+m+1; ans=n;
        for(i=1;i<=n;i++){
            scanf("%d",&x);
            add(i,T,x);
        }
        for(i=1;i<=m;i++){
            scanf("%d%d%d",&x,&y,&z);
            ans++;
            sum+=z;
            add(S,ans,z);
            add(ans,x,INF);
            add(ans,y,INF);
        }
        n=T+1;
        printf("%d\n",sum-sap(S,T,n));
    }
    return 0;
}