• <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

            Dearboy, a goods victualer, now comes to a big problem, and he needs your help. In his sale area there are N shopkeepers (marked from 1 to N) which stocks goods from him.Dearboy has M supply places (marked from 1 to M), each provides K different kinds of goods (marked from 1 to K). Once shopkeepers order goods, Dearboy should arrange which supply place provide how much amount of goods to shopkeepers to cut down the total cost of transport.

            It's known that the cost to transport one unit goods for different kinds from different supply places to different shopkeepers may be different. Given each supply places' storage of K kinds of goods, N shopkeepers' order of K kinds of goods and the cost to transport goods for different kinds from different supply places to different shopkeepers, you should tell how to arrange the goods supply to minimize the total cost of transport.

            Input

            The input consists of multiple test cases. The first line of each test case contains three integers N, M, K (0 < N, M, K < 50), which are described above. The next N lines give the shopkeepers' orders, with each line containing K integers (there integers are belong to [0, 3]), which represents the amount of goods each shopkeeper needs. The next M lines give the supply places' storage, with each line containing K integers (there integers are also belong to [0, 3]), which represents the amount of goods stored in that supply place.

            Then come K integer matrices (each with the size N * M), the integer (this integer is belong to (0, 100)) at the i-th row, j-th column in the k-th matrix represents the cost to transport one unit of k-th goods from the j-th supply place to the i-th shopkeeper.

            The input is terminated with three "0"s. This test case should not be processed.

            Output

            For each test case, if Dearboy can satisfy all the needs of all the shopkeepers, print in one line an integer, which is the minimum cost; otherwise just output "-1".

            Sample Input

            1 3 3
            1 1 1
            0 1 1
            1 2 2
            1 0 1
            1 2 3
            1 1 1
            2 1 1
            1 1 1
            3
            2
            20
            0 0 0
            

            Sample Output

            4
            -1
            


            二、分析
                  一個的最小費用最大流問題,詳細算法:最小費用最大流。
            三、代碼

              1#include<iostream>
              2#include<queue>
              3using namespace std;
              4int n, m, kind;
              5int s, t;
              6int order[51][51];
              7int store[51][51];
              8int cost[51][51][51];
              9int c[102][102];
             10int f[102][102];
             11int b[102][102];
             12int p[102];
             13int d[102];
             14bool visit[102]; //表示spfa中點是否在隊列中
             15void spfa() //求Gf的最短路
             16{
             17    queue<int> q;
             18    memset(visit, 0sizeof(visit));
             19    q.push(s);
             20    visit[s] = true;
             21    while(!q.empty())
             22    {
             23        int u = q.front();
             24        visit[u] = false;
             25        q.pop();
             26        for(int v=0; v<=n+m+1; v++)
             27            if(c[u][v] > f[u][v] && d[v] > d[u] + b[u][v])
             28            {
             29                d[v] = d[u] + b[u][v];
             30                p[v] = u;
             31                if(!visit[v])
             32                {
             33                    q.push(v);
             34                    visit[v] = true;
             35                }

             36            }

             37    }

             38}

             39void mcmf()
             40{
             41    while(1)
             42    {
             43        memset(p, -1sizeof(p));
             44        for(int i=1; i<=n+m+1; i++)
             45            d[i] = 100000;
             46        d[s] = 0;
             47        spfa();
             48        if(p[t] == -1//表示已無增廣路
             49            break;
             50        int minf = INT_MAX;
             51        int it = t;
             52        while(p[it] != -1)
             53        {
             54            minf = min(minf, c[p[it]][it] - f[p[it]][it]);
             55            it = p[it];
             56        }

             57        it = t;
             58        while(p[it] != -1)
             59        {
             60            f[p[it]][it] += minf;
             61            f[it][p[it]] = -f[p[it]][it];
             62            it = p[it];
             63        }

             64    }

             65}

             66int main()
             67{
             68    while(1)
             69    {
             70        scanf("%d%d%d"&n, &m, &kind);
             71        if(n==0 && m==0 && kind==0)
             72            break;
             73        for(int i=1; i<=n; i++)
             74            for(int j=1; j<=kind; j++)
             75                scanf("%d"&order[i][j]);
             76        for(int i=1; i<=m; i++)
             77            for(int j=1; j<=kind; j++)
             78                scanf("%d"&store[i][j]);
             79        for(int i=1; i<=kind; i++)
             80            for(int j=1; j<=n; j++)
             81                for(int k=1; k<=m; k++)
             82                    scanf("%d"&cost[i][k][j]);
             83        s = 0; t = m+n+1;
             84        int res = 0;
             85        bool flag = true;
             86        for(int i=1; i<=kind; i++)
             87        {
             88            memset(c, 0sizeof(c));
             89            for(int j=1; j<=m; j++)
             90                c[s][j] = store[j][i];
             91            for(int j=1; j<=m; j++)
             92                for(int k=1; k<=n; k++)
             93                    c[j][k+m] = store[j][i];
             94            for(int j=1; j<=n; j++)
             95                c[j+m][t] = order[j][i];
             96            memset(b, 0sizeof(b));
             97            for(int j=1; j<=m; j++)
             98                for(int k=1; k<=n; k++)
             99                {
            100                    b[j][k+m] = cost[i][j][k];
            101                    b[k+m][j] = -b[j][k+m]; //負費用,表示回流會減小費用
            102                }

            103            memset(f, 0sizeof(f));
            104            mcmf();
            105            for(int j=1; j<=n; j++)
            106                if(c[j+m][t] != f[j+m][t])
            107                {
            108                    flag = false;
            109                    break;
            110                }

            111            if(!flag) break;
            112            for(int j=1; j<=m; j++)
            113                for(int k=1; k<=n; k++)
            114                    res += f[j][m+k] * b[j][m+k];
            115        }

            116        if(flag)
            117            printf("%d\n", res);
            118        else
            119            printf("-1\n");
            120    }

            121}
            posted on 2009-06-30 22:09 Icyflame 閱讀(3345) 評論(1)  編輯 收藏 引用 所屬分類: 解題報告
            国产精品一区二区久久| 人妻精品久久无码专区精东影业| 2021少妇久久久久久久久久| 国产高潮国产高潮久久久| 波多野结衣中文字幕久久| 久久精品无码一区二区app| 久久亚洲AV无码精品色午夜| 2021久久精品国产99国产精品| 久久青青国产| A狠狠久久蜜臀婷色中文网| 久久亚洲电影| 99久久精品国内| 99久久香蕉国产线看观香 | 狠狠久久综合伊人不卡| 亚洲va久久久噜噜噜久久狠狠| 午夜不卡888久久| 久久亚洲精品国产精品| 尹人香蕉久久99天天拍| 国产成人精品久久亚洲高清不卡| 久久精品国产乱子伦| 亚洲中文字幕伊人久久无码| 99久久精品国产一区二区三区 | 区久久AAA片69亚洲 | 欧美一区二区三区久久综合| 国产午夜福利精品久久| 久久国产乱子伦免费精品| 亚洲伊人久久成综合人影院| 久久亚洲欧洲国产综合| 国产精品美女久久久久av爽| 久久综合久久综合九色| 精品国产91久久久久久久| 亚洲国产精品无码成人片久久| 亚洲Av无码国产情品久久| 久久久久久一区国产精品| 色综合久久中文综合网| 久久精品视频免费| 大美女久久久久久j久久| 国内精品久久久久久麻豆| 亚洲一区二区三区日本久久九| 国产精品日韩深夜福利久久| 国产—久久香蕉国产线看观看 |