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.
題意：有n個節點，包括有發電站，中轉站和消費者。其中發電站有np個，消費者有nc個，剩餘的爲中轉站。有m條輸電線，分別給出m條輸電線的起點終點和輸電量，在分別給出發電站的最大發電量和消費者的最大消費量。問整個電網的最大消費量是多少？
其實這道題最值得注意的還是輸入的問題，因爲包含了各種符號。解決了這個問題，這其實就是一個裸的最大流問題。增加一個超級源點與超級匯點之後，便可以建圖，套用dinic模板了。學會了用dalao的dinic模板，感覺這種碼風真的很讓我喜歡，但是一開始運用模板的時候，端點個數竟然忘記加上增加的兩個源點和匯點，真的是好尷尬。。。。。。
#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 ;//超級源點和超級匯點，不會和其他的點衝突
for(int i = 0 ; i < m ; i ++){
int from , to , cap;
char ch = getchar();
while (ch != '(')//輸入的時候要注意！！！
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));//這裏的n+2表示的是端點數！！！
}
return 0;
}