• <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
            国产精品99久久99久久久| 久久亚洲AV永久无码精品| 一本久久久久久久| 国产精品免费久久久久影院| 国产高潮国产高潮久久久| 久久久久久夜精品精品免费啦| 久久99国内精品自在现线| 国产伊人久久| 久久久91人妻无码精品蜜桃HD| 九九热久久免费视频| 久久久久无码精品国产app| 亚洲中文久久精品无码| 色偷偷888欧美精品久久久| 无码任你躁久久久久久老妇App| 久久人人爽人人爽人人AV东京热| 国产精品久久波多野结衣| 久久婷婷五月综合97色直播| 国产成人香蕉久久久久| av色综合久久天堂av色综合在| 久久国产免费| 亚洲精品成人久久久| 久久99久国产麻精品66| 94久久国产乱子伦精品免费| 久久久精品人妻一区二区三区蜜桃 | 99久久99久久精品国产片果冻 | 久久WWW免费人成—看片| 嫩草影院久久国产精品| 国产美女亚洲精品久久久综合| 亚洲人成电影网站久久| 99久久无码一区人妻a黑| 欧美牲交A欧牲交aⅴ久久| 国色天香久久久久久久小说| 国内精品久久久久影院亚洲| 91精品国产综合久久久久久| 亚洲AV日韩AV永久无码久久| 99久久精品国内| 国产精品久久久久蜜芽| 国产精品福利一区二区久久| 婷婷久久精品国产| 久久久久久久精品妇女99| 国产高潮久久免费观看|