文章地址:http://henuly.top/?p=782
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
题目链接
根据题意建图跑最大流即可。
AC代码:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define pb push_back
#define mp make_pair
#define lowbit(x) (x&(-x))
#define XDebug(x) cout<<#x<<"="<<x<<endl;
#define ArrayDebug(x,i) cout<<#x<<"["<<i<<"]="<<x[i]<<endl;
#define print(x) out(x);putchar('\n')
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;
typedef pair<ll,ll> PLL;
const int INF = 0x3f3f3f3f;
const int maxn = 1e2 + 5;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double pi = asin(1.0) * 2;
const double e = 2.718281828459;
template <class T>
inline bool read(T &ret) {
char c;
int sgn;
if (c = getchar(), c == EOF) {
return false;
}
while (c != '-' && (c < '0' || c > '9')) {
c = getchar();
}
sgn = (c == '-') ? -1 : 1;
ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9') {
ret = ret * 10 + (c - '0');
}
ret *= sgn;
return true;
}
template <class T>
inline void out(T x) {
if (x < 0) {
putchar('-');
x = -x;
}
if (x > 9) {
out(x / 10);
}
putchar(x % 10 + '0');
}
struct Link {
int V, Weight, Next;
Link(int _V = 0, int _Weight = 0, int _Next = 0): V(_V), Weight(_Weight), Next(_Next) {}
};
Link edges[1000005];
int Head[maxn << 1];
int Tot;
int Depth[maxn << 1];
int Current[maxn << 1];
int N, NP, NC, M;
void Init() {
Tot = 0;
memset(Head, -1, sizeof(Head));
}
void AddEdge(int U, int V, int Weight, int ReverseWeight = 0) {
edges[Tot].V = V;
edges[Tot].Weight = Weight;
edges[Tot].Next = Head[U];
Head[U] = Tot++;
edges[Tot].V = U;
edges[Tot].Weight = ReverseWeight;
edges[Tot].Next = Head[V];
Head[V] = Tot++;
}
bool Bfs(int Start, int End) {
memset(Depth, -1, sizeof(Depth));
std::queue<int> Que;
Depth[Start] = 0;
Que.push(Start);
while (!Que.empty()) {
int Vertex = Que.front();
Que.pop();
for (int i = Head[Vertex]; i != -1; i = edges[i].Next) {
if (Depth[edges[i].V] == -1 && edges[i].Weight > 0) {
Depth[edges[i].V] = Depth[Vertex] + 1;
Que.push(edges[i].V);
}
}
}
return Depth[End] != -1;
}
int Dfs(int Vertex, int End, int NowFlow) {
if (Vertex == End || NowFlow == 0) {
return NowFlow;
}
int UsableFlow = 0, FindFlow;
for (int &i = Current[Vertex]; i != -1; i = edges[i].Next) {
if (edges[i].Weight > 0 && Depth[edges[i].V] == Depth[Vertex] + 1) {
FindFlow = Dfs(edges[i].V, End, std::min(NowFlow - UsableFlow, edges[i].Weight));
if (FindFlow > 0) {
edges[i].Weight -= FindFlow;
edges[i ^ 1].Weight += FindFlow;
UsableFlow += FindFlow;
if (UsableFlow == NowFlow) {
return NowFlow;
}
}
}
}
if (!UsableFlow) {
Depth[Vertex] = -2;
}
return UsableFlow;
}
int Dinic(int Start, int End) {
int MaxFlow = 0;
while (Bfs(Start, End)) {
for (int i = 0; i <= N + 1; ++i) {
Current[i] = Head[i];
}
MaxFlow += Dfs(Start, End, INF);
}
return MaxFlow;
}
int main(int argc, char *argv[]) {
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
while (cin >> N >> NP >> NC >> M) {
Init();
for (int i = 1, U, V, C; i <= M; ++i) {
char Waste;
cin >> Waste >> U >> Waste >> V >> Waste >> C;
U++; V++;
AddEdge(U, V, C);
}
for (int i = 1, X, C; i <= NP; ++i) {
char Waste;
cin >> Waste >> X >> Waste >> C;
X++;
AddEdge(0, X, C);
}
for (int i = 1, X, C; i <= NC; ++i) {
char Waste;
cin >> Waste >> X >> Waste >> C;
X++;
AddEdge(X, N + 1, C);
}
cout << Dinic(0, N + 1) << endl;
}
#ifndef ONLINE_JUDGE
fclose(stdin);
fclose(stdout);
system("gedit out.txt");
#endif
return 0;
}