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            poj1459

            Power Network

            Time Limit: 2000MS Memory Limit: 32768K
            Total Submissions: 16422 Accepted: 8712

            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

            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.


            哎,糾結(jié)死了,我對(duì)網(wǎng)絡(luò)流這方面理解的還不行

            如果自己寫(xiě)代碼的話還是有點(diǎn)難度,所以找個(gè)好的模版還是很重要的

            額,模版也比較糾結(jié),好多中算法

            找了個(gè)比較簡(jiǎn)單的算法 Edmonds_karp 

            時(shí)間復(fù)雜度為O(V*E^2)

            Edmonds-Karp算法就是利用寬度優(yōu)先不斷地找一條從s到t的可改進(jìn)路,然后改進(jìn)流量,一直到找不到可改進(jìn)路為止。

            由于用寬度優(yōu)先,每次找到的可改進(jìn)路是最短的可改進(jìn)路,通過(guò)分析可以知道其復(fù)雜度為O(VE^2)。

            代碼好丑
              1#include<stdio.h>
              2#include<string.h>
              3#include<math.h>
              4#define MAX 105
              5int map[MAX][MAX],flow[MAX][MAX],c[MAX][MAX];
              6int n,nc,np,nt,s,t;
              7int sum;
              8int min(int a,int b)
              9{
             10    if (a<b) return a;else return b;
             11}

             12void Edmonds_Karp()
             13{
             14    int l1[MAX],l2[MAX],q[MAX];
             15    int u,v,head,tail;
             16    do 
             17    {
             18        memset(l1,0,sizeof(l1));
             19        memset(l2,0,sizeof(l2));//初始化所有標(biāo)號(hào)為0
             20        l1[s]=0;l2[s]=0x7fffffff;
             21        head=0;tail=1;
             22        q[tail]=s;
             23        while (head<tail&&l2[t]==0)//q未空且匯點(diǎn)未標(biāo)號(hào)
             24        {
             25            head++;
             26            u=q[head];
             27            for (v=1;v<=n ;v++ )
             28            {
             29                if (flow[u][v]<c[u][v]&&l2[v]==0)//未標(biāo)號(hào)且有可行流
             30                {
             31                    tail++;
             32                    q[tail]=v;
             33                    l2[v]=min(c[u][v]-flow[u][v],l2[u]);
             34                    //l2[v]記錄s到v增廣路中最小的可改進(jìn)流
             35                    l1[v]=u;//記錄前驅(qū)
             36                }

             37            }

             38        }

             39        if (l2[t]>0)//匯點(diǎn)未標(biāo)號(hào)
             40        {
             41            v=t;
             42            u=l1[v];
             43            while (v!=s)
             44            {
             45                flow[u][v]+=l2[t];
             46                flow[v][u]=-flow[u][v];
             47                v=u;
             48                u=l1[v];
             49            }

             50        }

             51    }

             52    while (l2[t]!=0);//直到匯點(diǎn)未標(biāo)號(hào)
             53}

             54void init()
             55{
             56    int i,j,a,b,w,x;
             57    char ch1;
             58    s=1;t=n+2;
             59    memset(map,0,sizeof(map));
             60    for (i=1;i<=nt ;i++ )
             61    {
             62        scanf("%c",&ch1);
             63        while (ch1!='(')
             64        {
             65            scanf("%c",&ch1);
             66        }

             67        scanf("%d",&a);a=a+2;
             68        scanf("%c",&ch1);scanf("%d",&b);b=b+2;
             69        scanf("%c",&ch1);scanf("%d",&w);
             70        map[a][b]=w;
             71    }

             72    for (i=1;i<=np ;i++ )
             73    {
             74        scanf("%c",&ch1);
             75        while (ch1!='(')
             76        {
             77            scanf("%c",&ch1);
             78        }

             79        scanf("%d",&a);a=a+2;
             80        scanf("%c",&ch1);scanf("%d",&w);
             81        map[s][a]=w;//map[a][s]=-w;
             82    }

             83    for (i=1;i<=nc ;i++ )
             84    {
             85        scanf("%c",&ch1);
             86        while (ch1!='(')
             87        {
             88            scanf("%c",&ch1);
             89        }

             90        scanf("%d",&a);a=a+2;
             91        scanf("%c",&ch1);scanf("%d",&w);
             92        map[a][t]=w;//map[t][a]=-w;
             93    }

             94    n=n+2;
             95    /*/for (i=1;i<=n ;i++ )
             96    {
             97        for (j=1;j<=n;j++ )
             98        {
             99            printf("%d ",map[i][j]);
            100        }
            101        printf("\n");
            102    }*/

            103    for (i=1;i<=n ;i++ )
            104    {
            105        for (j=1;j<=n;j++ )
            106        {
            107            c[i][j]=map[i][j];
            108        }

            109    }

            110}

            111int main()
            112{
            113    int i;
            114    while (scanf("%d%d%d%d",&n,&np,&nc,&nt)!=EOF)
            115    {
            116        memset(flow,0,sizeof(flow));
            117        init();
            118        Edmonds_Karp(1,n);
            119        sum=0;
            120        for (i=1;i<=n;i++ )
            121        {
            122            sum+=flow[1][i];
            123        }

            124        printf("%d\n",sum);
            125    }

            126    return 0;
            127}

            128



            posted on 2012-02-23 17:34 jh818012 閱讀(131) 評(píng)論(0)  編輯 收藏 引用


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            • 1.?re: poj1426
            • 我嚓,,輝哥,,居然搜到你的題解了
            • --season
            • 2.?re: poj3083
            • @王私江
              (8+i)&3 相當(dāng)于是 取余3的意思 因?yàn)?3 的 二進(jìn)制是 000011 和(8+i)
            • --游客
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            • @王私江
              0ms
            • --jh818012
            • 4.?re: poj3414
            • 200+行,跑了多少ms呢?我的130+行哦,你菜啦,哈哈。
            • --王私江
            • 5.?re: poj1426
            • 評(píng)論內(nèi)容較長(zhǎng),點(diǎn)擊標(biāo)題查看
            • --王私江
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