zoukankan      html  css  js  c++  java
  • POJ 1459 Power Network

    Power Network

    Time Limit: 2000ms
    Memory Limit: 32768KB
    This problem will be judged on PKU. Original ID: 1459
    64-bit integer IO format: %lld      Java class name: Main
     
     
    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.
     

    Source

     
    解题:最大流。。主要是题目意思难懂
     
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <cmath>
     5 #include <algorithm>
     6 #include <climits>
     7 #include <vector>
     8 #include <queue>
     9 #include <cstdlib>
    10 #include <string>
    11 #include <set>
    12 #include <stack>
    13 #define LL long long
    14 #define pii pair<int,int>
    15 #define INF 0x3f3f3f3f
    16 using namespace std;
    17 const int maxn = 105;
    18 int e[maxn][maxn],n,m,p,c,maxflow;
    19 int d[maxn],q[100000],S,T,hd,tail;
    20 bool bfs(){
    21     memset(d,-1,sizeof(d));
    22     d[S] = 0;
    23     q[tail++] = S;
    24     while(hd < tail){
    25         int u = q[hd++];
    26         for(int i = 1; i <= n; i++){
    27             if(d[i] == -1 && e[u][i] > 0){
    28                 d[i] = d[u]+1;
    29                 q[tail++] = i;
    30             }
    31         }
    32     }
    33     return d[T] > 0;
    34 }
    35 int dfs(int u,int low){
    36     int a = 0;
    37     if(u == T) return low;
    38     for(int i = 1; i <= n; i++){
    39         if(e[u][i] > 0 && d[i] == d[u]+1 && (a = dfs(i,min(low,e[u][i])))){
    40             e[u][i] -= a;
    41             e[i][u] += a;
    42             return a;
    43         }
    44     }
    45     return 0;
    46 }
    47 int main() {
    48     char ch;
    49     int u,v,w;
    50     while(~scanf("%d %d %d %d",&n,&p,&c,&m)){
    51         memset(e,0,sizeof(e));
    52         for(int i = 1; i <= m; i++){
    53             cin>>ch>>u>>ch>>v>>ch>>w;
    54             e[u+1][v+1] += w;
    55         }
    56         S = n+1;
    57         T = n+2;
    58         n += 2;
    59         for(int i = 1; i <= p; i++){
    60             cin>>ch>>v>>ch>>w;
    61             e[S][v+1] += w;
    62         }
    63         for(int i = 1; i <= c; i++){
    64             cin>>ch>>u>>ch>>w;
    65             e[u+1][T] += w;
    66         }
    67         int ans = maxflow = 0;
    68         while(bfs()) while(maxflow = dfs(S,INF)) ans += maxflow;
    69         printf("%d
    ",ans);
    70     }
    71     return 0;
    72 }
    View Code

    多路增广优化

     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <cmath>
     5 #include <algorithm>
     6 #include <climits>
     7 #include <vector>
     8 #include <queue>
     9 #include <cstdlib>
    10 #include <string>
    11 #include <set>
    12 #include <stack>
    13 #define LL long long
    14 #define pii pair<int,int>
    15 #define INF 0x3f3f3f3f
    16 using namespace std;
    17 const int maxn = 105;
    18 int e[maxn][maxn],n,m,p,c,maxflow;
    19 int d[maxn],q[100000],S,T,hd,tail;
    20 bool bfs() {
    21     memset(d,-1,sizeof(d));
    22     d[S] = 0;
    23     q[tail++] = S;
    24     while(hd < tail) {
    25         int u = q[hd++];
    26         for(int i = 1; i <= n; i++) {
    27             if(d[i] == -1 && e[u][i] > 0) {
    28                 d[i] = d[u]+1;
    29                 q[tail++] = i;
    30             }
    31         }
    32     }
    33     return d[T] > 0;
    34 }
    35 int dfs(int u,int low) {
    36     int a = 0,tmp = 0;
    37     if(u == T) return low;
    38     for(int i = 1; i <= n; i++) {
    39         if(e[u][i] > 0 && d[i] == d[u]+1 && (a = dfs(i,min(low,e[u][i])))) {
    40             e[u][i] -= a;
    41             e[i][u] += a;
    42             tmp += a;
    43             low -= a;
    44             if(!low) break;
    45         }
    46     }
    47     return tmp;
    48 }
    49 int main() {
    50     char ch;
    51     int u,v,w;
    52     while(~scanf("%d %d %d %d",&n,&p,&c,&m)) {
    53         memset(e,0,sizeof(e));
    54         for(int i = 1; i <= m; i++) {
    55             cin>>ch>>u>>ch>>v>>ch>>w;
    56             e[u+1][v+1] += w;
    57         }
    58         S = n+1;
    59         T = n+2;
    60         n += 2;
    61         for(int i = 1; i <= p; i++) {
    62             cin>>ch>>v>>ch>>w;
    63             e[S][v+1] += w;
    64         }
    65         for(int i = 1; i <= c; i++) {
    66             cin>>ch>>u>>ch>>w;
    67             e[u+1][T] += w;
    68         }
    69         int ans =  0;
    70         while(bfs()) ans += dfs(S,INF);
    71         printf("%d
    ",ans);
    72     }
    73     return 0;
    74 }
    View Code
  • 相关阅读:
    bzoj 1086 [SCOI2005]王室联邦——思路
    bzoj 3809 Gty的二逼妹子序列——莫队+分块
    bzoj 3781 小B的询问——分块
    bzoj 3309 DZY Loves Math——反演+线性筛
    洛谷 1079 Vigenère 密码——模拟水题
    洛谷 1082 同余方程——exgcd(水题)
    bzoj2662 [BeiJing wc2012]冻结 ——分层图
    bzoj2242 [SDOI2011]计算器——BSGS
    bzoj1008 [HNOI2008]越狱——快速幂
    bzoj1607 [Usaco2008 Dec]Patting Heads 轻拍牛头——暴力
  • 原文地址:https://www.cnblogs.com/crackpotisback/p/3969645.html
Copyright © 2011-2022 走看看