POJ 3308 Paratroopers 最小割、最大流

這道題目的建圖是：從源點到行，在從行到列，最後從列到匯點。然後注意權值，因為這裡讓求的是乘積的最大值，所以要是用EK求最大時，用到的是加法，所以需要用對數轉化一下。log(a)+log(b) = log(a*b),然後再轉化回來就行了啊。

這道題目陣列開小了，但是poj一直返回wa，然後就提交了好多遍才找到原因啊，以後的注意一下。

Description

Input

Output

Sample Input

1
4 4 5
2.0 7.0 5.0 2.0
1.5 2.0 2.0 8.0
1 1
2 2
3 3
4 4
1 4

Sample Output

16.0000

EK演算法：

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-7
//#define M 1000100
//#define LL __int64
#define LL long long
#define INF 0x3f3f3f3f
#define PI 3.1415926535898

const int maxn = 110;
using namespace std;

double c[110][110];
int pre[maxn];
double f[maxn];
int n, m;
double sum;

double spfa(int s, int t)
{
    memset(pre, -1 , sizeof(pre));
    memset(f , 0 , sizeof(f));
    queue<int>q;
    while(!q.empty())
        q.pop();
    q.push(s);
    f[s] = INF;
    pre[s] = 0;
    while(!q.empty())
    {
        int temp = q.front();
        q.pop();
        if(temp == t)
            break;
        for(int i = 0; i <= n+m+1; i++)
        {
            if(c[temp][i] > 0 && pre[i] == -1)
            {
                pre[i] = temp;
                f[i] = min(c[temp][i], f[temp]);
                q.push(i);
            }
        }
    }
    return f[t];
}

void EK(int s, int t)
{
    while(1)
    {
        double temp = spfa(s, t);
        if(temp == 0)
            break;
        for(int i = t; i != s; i = pre[i])
        {
            c[pre[i]][i] -= temp;
            c[i][pre[i]] += temp;
        }
        sum += temp;
    }
}

int main()
{
    int t, l;
    cin >>t;
    while(t--)
    {
        cin >>m>>n>>l;
        memset(c, 0 , sizeof(c));
        for(int i = 1; i <= m; i++)
        {
            double a;
            cin >>a;
            c[0][i] = log(a);
        }
        for(int i = 1; i <= n; i++)
        {
            double a;
            cin >>a;
            c[i+m][n+m+1] = log(a);
        }
        for(int i = 1; i <= l; i++)
        {
            int x, y;
            cin >>x>>y;
            c[x][y+m] = INF;
        }
        sum = 0;
        EK(0, m+n+1);
        cout<<fixed<<setprecision(4)<<exp(sum)<<endl;
    }
    return 0;
}

dinic演算法：

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-7
//#define M 1000100
//#define LL __int64
#define LL long long
#define INF 0x3f3f3f3f
#define PI 3.1415926535898

const int maxn = 1010;
using namespace std;

int pre[maxn];
int head[maxn];
int cnt;
int n, m;
int deep[maxn];

struct node
{
    int v,next;
    double w;
} edge[maxn*5];

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

void add(int u, int v, double w)
{
    edge[cnt].v = v;
    edge[cnt].w = w;
    edge[cnt].next = head[u];
    head[u] = cnt++;

    edge[cnt].v = u;
    edge[cnt].w = 0;
    edge[cnt].next = head[v];
    head[v] = cnt++;
}
bool bfs()//bfs一次找到所有的增廣路
{
    memset(deep, -1 , sizeof(deep));
    queue<int>q;
    while(!q.empty())
        q.pop();
    q.push(0);
    deep[0] = 0;
    while(!q.empty())
    {
        int t = q.front();
        q.pop();
        int p = head[t];
        while(p != -1)
        {
            int v = edge[p].v;
            if(deep[v] == -1 && edge[p].w > 0)
            {
                deep[v] = deep[t]+1;
                q.push(v);
            }
            p = edge[p].next;
        }
    }
    if(deep[n+m+1] == -1)
        return false;
    return true;
}

void fid(int u, int v, double w)//更新反向邊
{
    int p = head[u];
    while(edge[p].v != v)
    {
        p = edge[p].next;
    }
    edge[p].w += w;
}

double dfs(int s, double f)
{
    if(s == n+m+1)
        return f;
    double sum = 0;
    int p = head[s];
    while(p != -1)
    {
        int v = edge[p].v;
        if(deep[v] == deep[s]+1 && edge[p].w > 0)
        {
            double temp = dfs(v, min(f-sum, edge[p].w));//這裡要等於最小的流量，保證所有的可以通過
            sum += temp;
            edge[p].w -= temp;//正向邊流量減少
            fid(v, s, temp);//反向邊流量增加
        }
        p = edge[p].next;
    }
    return sum;
}

double dinic()
{
    double ans=0;
    while(bfs())
    {
        ans += dfs(0,INF);
    }
    return ans;
}

int main()
{
    int t, l;
    cin >>t;
    while(t--)
    {
        cin >>m>>n>>l;
        init();
        for(int i = 1; i <= m; i++)
        {
            double a;
            cin >>a;
            add(0, i, log(a));
        }
        for(int i = 1; i <= n; i++)
        {
            double a;
            cin >>a;
            add(i+m, n+m+1, log(a));
        }
        for(int i = 1; i <= l; i++)
        {
            int x, y;
            cin >>x>>y;
            add(x, y+m, INF);
        }
        double sum;
        sum = dinic();
        cout<<fixed<<setprecision(4)<<exp(sum)<<endl;
    }
    return 0;
}