問題 -m memory ret wid bsp man 最大 中轉

Power Network

Time Limit: 2000MS		Memory Limit: 32768K
Total Submissions: 25108		Accepted: 13077

Description

A power network consists of nodes (power stations, consumers and dispatchers) connected by power transport lines. A node u may be supplied with an amount s(u) >= 0 of power, may produce an amount 0 <= p(u) <= p_max(u) of power, may consume an amount 0 <= c(u) <= min(s(u),c_max(u)) of power, and may deliver an amount d(u)=s(u)+p(u)-c(u) of power. The following restrictions apply: c(u)=0 for any power station, p(u)=0 for any consumer, and p(u)=c(u)=0 for any dispatcher. There is at most one power transport line (u,v) from a node u to a node v in the net; it transports an amount 0 <= l(u,v) <= l_max(u,v) of power delivered by u to v. Let Con=Σ_uc(u) be the power consumed in the net. The problem is to compute the maximum value of Con.

An example is in figure 1. The label x/y of power station u shows that p(u)=x and p_max(u)=y. The label x/y of consumer u shows that c(u)=x and c_max(u)=y. The label x/y of power transport line (u,v) shows that l(u,v)=x and l_max(u,v)=y. The power consumed is Con=6. Notice that there are other possible states of the network but the value of Con cannot exceed 6.

Input

There are several data sets in the input. Each data set encodes a power network. It starts with four integers: 0 <= n <= 100 (nodes), 0 <= np <= n (power stations), 0 <= nc <= n (consumers), and 0 <= m <= n^2 (power transport lines). Follow m data triplets (u,v)z, where u and v are node identifiers (starting from 0) and 0 <= z <= 1000 is the value of l_max(u,v). Follow np doublets (u)z, where u is the identifier of a power station and 0 <= z <= 10000 is the value of p_max(u). The data set ends with nc doublets (u)z, where u is the identifier of a consumer and 0 <= z <= 10000 is the value of c_max(u). All input numbers are integers. Except the (u,v)z triplets and the (u)z doublets, which do not contain white spaces, white spaces can occur freely in input. Input data terminate with an end of file and are correct.

Output

For each data set from the input, the program prints on the standard output the maximum amount of power that can be consumed in the corresponding network. Each result has an integral value and is printed from the beginning of a separate line.

Sample Input

2 1 1 2 (0,1)20 (1,0)10 (0)15 (1)20
7 2 3 13 (0,0)1 (0,1)2 (0,2)5 (1,0)1 (1,2)8 (2,3)1 (2,4)7
         (3,5)2 (3,6)5 (4,2)7 (4,3)5 (4,5)1 (6,0)5
         (0)5 (1)2 (3)2 (4)1 (5)4

Sample Output

15
6

Hint

The sample input contains two data sets. The first data set encodes a network with 2 nodes, power station 0 with pmax(0)=15 and consumer 1 with cmax(1)=20, and 2 power transport lines with lmax(0,1)=20 and lmax(1,0)=10. The maximum value of Con is 15. The second data set encodes the network from figure 1.

題目給出非常多都是廢話，特別是符號s(u)，d(u)，Con還有那條公式都別管。混淆視聽

難點在於構圖

電站p(u)均為源點。用戶c(u)均為匯點，中轉站當普通點處理

結點和邊都有 x/y（流量和容量），這個非常easy使人產生矛盾（由於學習最大流問題是。僅僅有 邊 才有流量和容量。可是不難發現。題目所給的例圖中有多個源點，多個匯點，多個普通點。僅僅有源點和匯點才標有 x/y，普通點沒有標x/y，並且所給出的全部邊都有x/y。 這無疑在促使我們對圖做一個變形： 建議一個超級源 s，一個超級匯 t。使 s 指向全部源點，並把源點的 容量y 分別作為這些邊的 容量，使全部匯點指向 t。並把匯點的容量y分別作為這些邊的 容量，然後本來是源點和匯點的點，全部變為普通點。這樣就把“多源多匯最大流”變形為“單源單匯最大流”問題。

學習最大流問題時。會發現邊上的流量值是給定初始值的，可是這題的輸入僅僅有容量。沒有流量，非常多人立即感覺到無從入手。事實上邊上的流量初始值為多少都沒有所謂，解最大流須要用到的僅僅有容量。

可是一般為了方便起見， 會把全部邊的流量初始化為0。

這樣做有一個最大的優點，就是能夠回避 反向弧 的存在。

以上解析來自http://www.cnblogs.com/lyy289065406/archive/2011/07/30/2122116.html

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
#define maxn 300
#define maxm 100000
#define INF 0x3f3f3f3f
using namespace std;

int head[maxn], cur[maxn], cnt;
int dist[maxn], vis[maxn];
int n, np, nc, m;

struct node{
    int u, v, cap, flow, next;
};

node edge[maxm];

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

void add(int u, int v, int w){
    edge[cnt] = {u, v, w, 0, head[u]};
    head[u] = cnt++;
    edge[cnt] = {v, u, 0, 0, head[v]};
    head[v] = cnt++;
}

void getmap(){
    int u, v, w;
    while(m--){
        scanf(" (%d,%d)%d", &u, &v, &w);//註意有空格
        add(u, v, w);
    }
    while(np--){
        scanf(" (%d)%d", &u, &w);
        add(n, u, w);// n 為源點, 源點和電站連接
    }
    while(nc--){
        scanf(" (%d)%d", &u, &w);
        add(u, n + 1, w); // n + 1 為匯點 ，消費者和匯點連接
    }
}

bool BFS(int st ,int ed){
    queue<int>q;
    memset(vis, 0 ,sizeof(vis));
    memset(dist, -1, sizeof(dist));
    vis[st] = 1;
    dist[st] = 0;
    q.push(st);
    while(!q.empty()){
        int u = q.front();
        q.pop();
        for(int i = head[u]; i != -1; i = edge[i].next){
            node E = edge[i];
            if(!vis[E.v] && E.cap > E.flow){
                vis[E.v] = 1;
                dist[E.v] = dist[u] + 1;
                if(E.v == ed)
                    return true;
                q.push(E.v);
            }
        }
    }
    return false;
}

int DFS(int x, int ed, int a){
    if(x == ed || a == 0)
        return a;
    int flow = 0, f;
    for(int &i = cur[x]; i != -1; i = edge[i].next){
        node &E = edge[i];
        if(dist[E.v] == dist[x] + 1 && (f = DFS(E.v, ed, min(a, E.cap - E.flow))) > 0){
            E.flow += f;
            edge[i ^ 1].flow -= f;
            a -= f;
            flow += f;
            if(a == 0)
                break;
        }
    }
    return flow;
}

int maxflow(int st, int ed){
    int flowsum = 0;
    while(BFS(st,ed)){
        memcpy(cur, head, sizeof(head));
        flowsum += DFS(st, ed, INF);
    }
    return flowsum;
}

int main (){
    while(scanf("%d%d%d%d", &n, &np, &nc, &m) != EOF){
        init();
        getmap();
        printf("%d\n", maxflow(n, n + 1));
    }
    return 0;
}

POJ 1459 &amp;&amp; ZOJ 1734--Power Network【最大流dinic】