文章地址:http://henuly.top/?p=780
Description:
You, the head of Department of Security, recently received a top-secret information that a group of terrorists is planning to transport some WMD 1 from one city (the source) to another one (the destination). You know their date, source and destination, and they are using the highway network.
The highway network consists of bidirectional highways, connecting two distinct city. A vehicle can only enter/exit the highway network at cities only.
You may locate some SA (special agents) in some selected cities, so that when the terrorists enter a city under observation (that is, SA is in this city), they would be caught immediately.
It is possible to locate SA in all cities, but since controlling a city with SA may cost your department a certain amount of money, which might vary from city to city, and your budget might not be able to bear the full cost of controlling all cities, you must identify a set of cities, that:
- all traffic of the terrorists must pass at least one city of the set.
- sum of cost of controlling all cities in the set is minimal.
You may assume that it is always possible to get from source of the terrorists to their destination.
1 Weapon of Mass Destruction
Input:
There are several test cases.
The first line of a single test case contains two integer N and M ( 2 <= N <= 200; 1 <= M <= 20000), the number of cities and the number of highways. Cities are numbered from 1 to N.
The second line contains two integer S,D ( 1 <= S,D <= N), the number of the source and the number of the destination.
The following N lines contains costs. Of these lines the ith one contains exactly one integer, the cost of locating SA in the ith city to put it under observation. You may assume that the cost is positive and not exceeding 107.
The followingM lines tells you about highway network. Each of these lines contains two integers A and B, indicating a bidirectional highway between A and B.
Please process until EOF (End Of File).
Output:
For each test case you should output exactly one line, containing one integer, the sum of cost of your selected set.
See samples for detailed information.
Sample Input:
5 6
5 3
5
2
3
4
12
1 5
5 4
2 3
2 4
4 3
2 1
Sample Output:
3
题目链接
这道题目要求的是最小割点，由于最大流最小割中割的是边而不是点，所以考虑拆点，这样割掉拆点之间的边就相当于割掉这个点。
AC代码:
#include <bits/stdc++.h>
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 = 4e4 + 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[maxn << 2];
int Head[maxn << 2];
int Tot;
int N, E;
int Depth[maxn << 2];
int Current[maxn << 2];
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 = 1; i <= N + N; ++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 (~scanf("%d%d", &N, &E)) {
Init();
int S, T;
read(S); read(T);
for (int i = 1, X; i <= N; ++i) {
read(X);
AddEdge(i, i + N, X);
}
for (int i = 1, U, V; i <= E; ++i) {
read(U); read(V);
AddEdge(U + N, V, INF);
AddEdge(V + N, U, INF);
}
print(Dinic(S, T + N));
}
#ifndef ONLINE_JUDGE
fclose(stdin);
fclose(stdout);
system("gedit out.txt");
#endif
return 0;
}1