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) <= p max(u)
of power, may consume an amount 0 <= c(u) <= min(s(u),c max(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) <= l max(u,v) of power delivered by u to v. Let Con=Σ uc(u) be the power consumed in the net. The problem is to
compute the maximum value of Con.
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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 c max(u)=y. The label x/y of power transport line (u,v) shows that l(u,v)=x and l max(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
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 c max(u)=y. The label x/y of power transport line (u,v) shows that l(u,v)=x and l max(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.
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 l max(u,v). Follow np doublets (u)z, where u is the identifier of a power station and 0 <= z <= 10000 is the value of p max(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 c max(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 6Hint
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.
题意：有n个节点，包括有发电站，中转站和消费者。其中发电站有np个，消费者有nc个，剩余的为中转站。有m条输电线，分别给出m条输电线的起点终点和输电量，在分别给出发电站的最大发电量和消费者的最大消费量。问整个电网的最大消费量是多少？
其实这道题最值得注意的还是输入的问题，因为包含了各种符号。解决了这个问题，这其实就是一个裸的最大流问题。增加一个超级源点与超级汇点之后，便可以建图，套用dinic模板了。学会了用dalao的dinic模板，感觉这种码风真的很让我喜欢，但是一开始运用模板的时候，端点个数竟然忘记加上增加的两个源点和汇点，真的是好尴尬。。。。。。
#include<stdio.h>
#include<iostream>
#include<set>
#include<queue>
#include<algorithm>
#include<math.h>
#include<string.h>
#include<string>
#include<vector>
using namespace std;
const int inf = 0x3f3f3f3f;
const int MX = 105;
const int MXE = 4 * MX * MX;
struct MaxFlow {//裸的最大流模板，直接套用即可
struct Edge {
int v, w, nxt;
} edge[MXE];
int tot, num, s, t;
int head[MX];
void init() {
memset(head, -1, sizeof(head));
tot = 0;
}
void add(int u, int v, int w) {
edge[tot].v = v;
edge[tot].w = w;
edge[tot].nxt = head[u];
head[u] = tot++;
edge[tot].v = u;
edge[tot].w = 0;
edge[tot].nxt = head[v];
head[v] = tot++;
}
int d[MX], vis[MX], gap[MX];
void bfs() {
memset(d, 0, sizeof(d));
memset(gap, 0, sizeof(gap));
memset(vis, 0, sizeof(vis));
queue<int>q;
q.push(t);
vis[t] = 1;
while (!q.empty()) {
int u = q.front();
q.pop();
for (int i = head[u]; ~i; i = edge[i].nxt) {
int v = edge[i].v;
if (!vis[v]) {
d[v] = d[u] + 1;
gap[d[v]]++;
q.push(v);
vis[v] = 1;
}
}
}
}
int last[MX];
int dfs(int u, int f) {
if (u == t) return f;
int sap = 0;
for (int i = last[u]; ~i; i = edge[i].nxt) {
int v = edge[i].v;
if (edge[i].w > 0 && d[u] == d[v] + 1) {
last[u] = i;
int tmp = dfs(v, min(f - sap, edge[i].w));
edge[i].w -= tmp;
edge[i ^ 1].w += tmp;
sap += tmp;
if (sap == f) return sap;
}
}
if (d[s] >= num) return sap;
if (!(--gap[d[u]])) d[s] = num;
++gap[++d[u]];
last[u] = head[u];
return sap;
}
int solve(int st, int ed, int n) {
int flow = 0;
num = n;
s = st;
t = ed;
bfs();
memcpy(last, head, sizeof(head));
while (d[s] < num) flow += dfs(s, inf);
return flow;
}
} F;
int main()
{
int n , np , nc , m ;
while(~scanf("%d %d %d %d" , &n , &np , &nc , &m)){
F.init();
int s = 101 , t = 102 ;//超级源点和超级汇点，不会和其他的点冲突
for(int i = 0 ; i < m ; i ++){
int from , to , cap;
char ch = getchar();
while (ch != '(')//输入的时候要注意！！！
ch = getchar();
scanf("%d,%d)%d" , &from , &to , &cap);
F.add(from , to , cap);
}
for(int i = 0 ; i < np ; i ++){
int x , y ;
char ch = getchar();
while (ch != '(')
ch = getchar();
scanf("%d)%d",&x , &y);
F.add(s , x , y);
}
for(int i = 0 ; i < nc ; i ++){
int x , y ;
char ch = getchar();
while (ch != '(')
ch = getchar();
scanf("%d)%d",&x , &y);
F.add(x , t , y);
}
printf("%d\n",F.solve(s , t , n+2));//这里的n+2表示的是端点数！！！
}
return 0;
}