题意:N个点,M条边的有向图,边有权(容量),求从点1到点N的最大流(2 <= N <= 15, 0 <= M <= 1000,1 <= 每条边的容量 <= 1000)。
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3549
——>>……练练……
#include <cstdio>
#include <vector>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
const int maxn = 15 + 10;
const int INF = 0x3f3f3f3f;
int N, M;
struct Edge{
int u, v, cap, flow;
Edge(int u = 0, int v = 0, int cap = 0, int flow = 0):
u(u), v(v), cap(cap), flow(flow){}
};
struct Dinic{
int m, s, t;
vector<Edge> edges;
vector<int> G[maxn];
int d[maxn];
bool vis[maxn];
int cur[maxn];
void init(){
edges.clear();
for(int i = 1; i <= N; i++) G[i].clear();
}
void addEdge(int u, int v, int cap){
edges.push_back(Edge(u, v, cap, 0));
edges.push_back(Edge(v, u, 0, 0));
m = edges.size();
G[u].push_back(m-2);
G[v].push_back(m-1);
}
bool bfs(){
memset(vis, 0, sizeof(vis));
queue<int> qu;
qu.push(s);
vis[s] = 1;
d[s] = 0;
while(!qu.empty()){
int u = qu.front(); qu.pop();
int sz = G[u].size();
for(int i = 0; i < sz; i++){
Edge& e = edges[G[u][i]];
if(!vis[e.v] && e.cap > e.flow){
d[e.v] = d[u] + 1;
vis[e.v] = 1;
qu.push(e.v);
}
}
}
return vis[t];
}
int dfs(int u, int a){
if(u == t || a == 0) return a;
int sz = G[u].size(), f, flow = 0;
for(int& i = cur[u]; i < sz; i++){
Edge& e = edges[G[u][i]];
if(d[e.v] == d[u] + 1 && (f = dfs(e.v, min(a, e.cap-e.flow))) > 0){
cur[u] = i;
e.flow += f;
edges[G[u][i]^1].flow -= f;
flow += f;
a -= f;
if(!a) break;
}
}
return flow;
}
int Maxflow(int s, int t){
this->s = s;
this->t = t;
int flow = 0;
while(bfs()){
memset(cur, 0, sizeof(cur));
flow += dfs(s, INF);
}
return flow;
}
}din;
void read(){
int X, Y, C;
scanf("%d%d", &N, &M);
din.init();
for(int i = 0; i < M; i++){
scanf("%d%d%d", &X, &Y, &C);
din.addEdge(X, Y, C);
}
}
int main()
{
int T, kase = 1;
scanf("%d", &T);
while(T--){
read();
printf("Case %d: %d\n", kase++, din.Maxflow(1, N));
}
return 0;
}