pku 1459 Power Network EK&&Dinic算法解决。

http://poj.org/problem?id=1459

才开始套自己的那个模板就是过不了一直TLE,最后看了别人过得修改了一下自己的模板最后1391s险过。。。

本题是一个多源点，多汇点问题，建立超级源点与汇点，然后套用模板求解。。

EK O(V*E^2)

#include<iostream>
#include<queue>
#include<cstdio>
#include<cstring>
using namespace std;
const int inf=100000000;
const int maxn=105;
int map[maxn][maxn],flow[maxn];
bool vt[maxn];
int n,np,nc,m;
int pre[maxn];

int bfs(int s,int e)
{
    int i;
    for (i = 0; i <= n + 1; ++i)
    flow[i] = inf;
    memset(pre,-1,sizeof(pre));
    queue<int>q;
    q.push(s);
    while (!q.empty())
    {
        int p = q.front(); q.pop();
        for (i = 0; i <= n + 1; ++i)
        {
            if (i == s || map[p][i] <= 0 || pre[i] != -1)
            continue;
            vt[i] = true;
            pre[i] = p;
            flow[i] = min(flow[p],map[p][i]);//就是这里的优化，这里直接就找出了最小流量
            //省去了我在EK中再循环一遍了。。。
            q.push(i);
        }
    }
    if (flow[e] == inf) return -1;
    return flow[e];
}
int EK(int s,int e)
{
    int i,d;
    int ans = 0;
    while (1)
    {
          d = bfs(s,e);
          if (d == -1) break;
        for (i = e; i != s; i = pre[i])
        {
            map[pre[i]][i] -= d;
            map[i][pre[i]] += d;
        }
        ans += d;
    }
    return ans;
}
int main()
{
    int i,x,y,v;
    while (~scanf("%d%d%d%d",&n,&np,&nc,&m))
    {
        memset(map,0,sizeof(map));
        for (i = 0; i < m; ++i)
        {
            scanf(" (%d,%d)%d",&x,&y,&v);
            map[x][y] = v;
        }
        for (i = 0; i < np; ++i)
        {
            scanf(" (%d)%d",&x,&v);
            map[n][x] = v;
        }
        for (i = 0; i < nc; ++i)
        {
            scanf(" (%d)%d",&x,&v);
            map[x][n + 1] = v;
        }
        printf("%d\n",EK(n,n + 1));
    }
    return 0;
}

Dinic算法79过。。好快。。。O(V^2*E)

#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#define maxn 107
#define inf 99999999
using namespace std;
int g[maxn][maxn],level[maxn];
bool vt[maxn];
//分层
bool layer(int s,int e)
{
    queue<int>q;
    memset(level,-1,sizeof(level));
    q.push(s);
    level[s] = 1;
    int i,p;
    while (!q.empty())
    {
        p = q.front();
        q.pop();
        for (i = 0; i <= e; ++i)
        {
            if (g[p][i] > 0 && level[i] == -1)//满足还没有分层，且p与i有流量
            {
                level[i] = level[p] + 1;
                if (i == e) { return true;}
                else q.push(i);
            }
        }
    }
    return false;
}
int dinic(int s,int e)
{
    deque<int>q;//利用双端队列
    int i,p,vs,ve,min,pos,sum;
    sum = 0;
    while (layer(s,e))
    {
        memset(vt,false,sizeof(vt));
        q.push_back(s);//从后端插入
        vt[s] = true;
        while (!q.empty())
        {
            p = q.back();//从后端取值
            if (p == e)//如果得到了终点
            {
                min = inf;
                for (i = 1; i < q.size(); ++i)
                {
                    vs = q[i - 1]; ve = q[i];
                    if (g[vs][ve] > 0 && g[vs][ve] < min)
                    {
                        min = g[vs][ve];
                        pos = vs;
                    }
                }
                sum += min;
                for (i = 1; i < q.size(); ++i)
                {
                    vs = q[i - 1]; ve = q[i];
                    if (g[vs][ve] > 0)
                    {
                        g[vs][ve] -= min;
                        g[ve][vs] += min;
                    }
                }
                while (!q.empty() && q.back() != pos)//返回最小流量的边的定点，重新找增广路
                {
                    vt[q.back()] = false;
                    q.pop_back();
                }
            }
            else//继续找下一层
            {
                for (i = 0; i <= e; ++i)
                {
                    if (g[p][i] > 0 && !vt[i] && level[i] == level[p] + 1)
                    {
                        vt[i] = true;
                        q.push_back(i);
                        break;
                    }
                }
                if (i > e) q.pop_back();
            }
        }
    }
    return sum;
}
int main()
{
    int i,n,m,np,nc;
    int x,y,v;
    while (~scanf("%d%d%d%d",&n,&np,&nc,&m))
    {
        memset(g,0,sizeof(g));
        for (i = 0; i < m; ++i)
        {
            scanf(" (%d,%d)%d",&x,&y,&v);
            g[x][y] = v;
        }
        for (i = 0; i < np; ++i)
        {
            scanf(" (%d)%d",&x,&v);
            g[n][x] = v;
        }
        for (i = 0; i < nc; ++i)
        {
            scanf(" (%d)%d",&x,&v);
            g[x][n + 1] = v;
        }
        printf("%d\n",dinic(n,n+1));
    }
    return 0;
}