• <ins id="pjuwb"></ins>
    <blockquote id="pjuwb"><pre id="pjuwb"></pre></blockquote>
    <noscript id="pjuwb"></noscript>
          <sup id="pjuwb"><pre id="pjuwb"></pre></sup>
            <dd id="pjuwb"></dd>
            <abbr id="pjuwb"></abbr>
            posts - 18,  comments - 5,  trackbacks - 0
            一、題目描述

            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) <= pmax(u) of power, may consume an amount 0 <= c(u) <= min(s(u),cmax(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) <= lmax(u,v) of power delivered by u to v. Let Con=Σuc(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 pmax(u)=y. The label x/y of consumer u shows that c(u)=x and cmax(u)=y. The label x/y of power transport line (u,v) shows that l(u,v)=x and lmax(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 lmax(u,v). Follow np doublets (u)z, where u is the identifier of a power station and 0 <= z <= 10000 is the value of pmax(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 cmax(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


            二、分析
                  增加點n為s,點n+1為t,求最大流,使用Push-Relabel算法,具體算法:最大流問題
            三、代碼

             1#include<iostream>
             2using namespace std;
             3#define MAXN 202
             4int s, t;
             5int n, np, nc, m;
             6char str[50];
             7int c[MAXN][MAXN];
             8int f[MAXN][MAXN];
             9int e[MAXN];
            10int h[MAXN];
            11void push(int u, int v)
            12{
            13    int d = min(e[u], c[u][v] - f[u][v]);
            14    f[u][v] += d;
            15    f[v][u] = -f[u][v];
            16    e[u] -= d;
            17    e[v] += d;
            18}

            19bool relabel(int u)
            20{
            21    int mh = INT_MAX;
            22    for(int i=0; i<n+2; i++)
            23    {
            24        if(c[u][i] > f[u][i])
            25            mh = min(mh, h[i]);
            26    }

            27    if(mh == INT_MAX)
            28        return false//殘留網絡中無從u出發的路
            29    h[u] = mh + 1;
            30    for(int i=0; i<n+2; i++)
            31    {
            32        if(e[u] == 0//已無余流,不需再次push
            33            break;
            34        if(h[i] == mh && c[u][i] > f[u][i]) //push的條件
            35            push(u, i);
            36    }

            37    return true;
            38}

            39void init_preflow()
            40{
            41    memset(h, 0sizeof(h));
            42    memset(e, 0sizeof(e));
            43    h[s] = n+2;
            44    for(int i=0; i<n+2; i++)
            45    {
            46        if(c[s][i] == 0)
            47            continue;
            48        f[s][i] = c[s][i];
            49        f[i][s] = -f[s][i];
            50        e[i] = c[s][i];
            51        e[s] -= c[s][i];
            52    }

            53}

            54void push_relabel()
            55{
            56    init_preflow();
            57    bool flag = true//表示是否還有relabel操作
            58    while(flag)
            59    {
            60        flag = false;
            61        for(int i=0; i<n; i++)
            62            if(e[i] > 0)
            63                flag = flag || relabel(i);
            64    }

            65}

            66int main()
            67{
            68    while(scanf("%d%d%d%d"&n, &np, &nc, &m) != EOF)
            69    {
            70        s = n; t = n+1;
            71        memset(c, 0sizeof(c));
            72        memset(f, 0sizeof(f));
            73        while(m--)
            74        {
            75            scanf("%s"&str);
            76            int u=0, v=0, z=0;
            77            sscanf(str, "(%d,%d)%d"&u, &v, &z);
            78            c[u][v] = z;
            79        }

            80        for(int i=0; i<np+nc; i++)
            81        {
            82            scanf("%s"&str);
            83            int u=0, z=0;
            84            sscanf(str, "(%d)%d"&u, &z);
            85            if(i < np)
            86                c[s][u] = z;
            87            else if(i >= np && i < np + nc)
            88                c[u][t] = z;
            89        }

            90        push_relabel();
            91        printf("%d\n", e[t]);
            92    }

            93}
            posted on 2009-06-24 19:38 Icyflame 閱讀(2108) 評論(1)  編輯 收藏 引用 所屬分類: 解題報告
            亚洲国产欧美国产综合久久| 精品久久久久中文字幕一区| 国产aⅴ激情无码久久| 久久久无码精品亚洲日韩京东传媒 | 国产精品99久久久久久猫咪| 欧美久久亚洲精品| 成人国内精品久久久久影院VR| 久久久高清免费视频| 久久免费看黄a级毛片| 久久国产精品99久久久久久老狼 | 国产AV影片久久久久久| 无码八A片人妻少妇久久| 99久久精品免费看国产免费| 亚洲精品WWW久久久久久| 国产高潮久久免费观看| 天天躁日日躁狠狠久久| 青青草原综合久久大伊人导航| 久久久久久久综合日本亚洲| 亚洲人成网亚洲欧洲无码久久 | 久久er国产精品免费观看8| 久久亚洲国产成人精品性色| 婷婷久久精品国产| 99久久无码一区人妻| 亚洲一本综合久久| 久久夜色精品国产亚洲| 欧美精品乱码99久久蜜桃| 国产午夜精品理论片久久| 久久久久国产精品| 国产精品99久久精品| 久久久精品2019免费观看| 久久婷婷国产剧情内射白浆| 亚洲国产综合久久天堂| 亚洲国产小视频精品久久久三级 | 久久久久国产精品三级网| 狠狠精品久久久无码中文字幕| 99精品国产在热久久无毒不卡| 国产麻豆精品久久一二三| 精品一区二区久久久久久久网站| 亚洲色欲久久久综合网东京热 | 中文国产成人精品久久不卡| 久久久久久国产精品无码下载|