poj - 1273 - Drainage Ditches（最大流）

题意：M个点，N条有向路，每条沟（路）有最大排水量，问从点1到点M的最大排水量是多少(0 <= N <= 200, 2 <= M <= 200, 0 <= 每条沟的容量 <= 10000000)。

题目链接：http://poj.org/problem?id=1273

——>>LJ白书增广路算法的模板题。。。

1、不是求最小值，可以汇流的；2、测试数据有重边（开始没想到，WA了一次）。

(920K, 0MS)

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>

using namespace std;

const int maxn = 200 + 10;
const int INF = 0x3f3f3f3f;
int cap[maxn][maxn], a[maxn], flow[maxn][maxn], p[maxn], M;

void init(){
memset(cap, 0, sizeof(cap));
}

int EK(int s, int t){
queue<int> qu;
while(!qu.empty()) qu.pop();
memset(flow, 0, sizeof(flow));
int f = 0;
for(;;){
memset(a, 0, sizeof(a));
a[s] = INF;
qu.push(s);
while(!qu.empty()){
int u = qu.front(); qu.pop();
for(int v = 1; v <= M; v++) if(!a[v] && cap[u][v] > flow[u][v]){
p[v] = u; qu.push(v);
a[v] = min(a[u], cap[u][v] - flow[u][v]);
}
}
if(a[t] == 0) break;
for(int u = t; u != s; u = p[u]){
flow[p[u]][u] += a[t];
flow[u][p[u]] -= a[t];
}
f += a[t];
}
return f;
}

int main()
{
int N, S, E, C, i;
while(scanf("%d%d", &N, &M) == 2){
init();
for(i = 0; i < N; i++){
scanf("%d%d%d", &S, &E, &C);
if(!cap[S][E]) cap[S][E] = C;
else cap[S][E] += C;
}
printf("%d\n", EK(1, M));
}
return 0;
}

用Dinic算法邻接表实现，可直接解决重边问题。

(724K, 0MS)

#include <cstdio>
#include <vector>
#include <cstring>
#include <queue>

using namespace std;

const int maxn = 200 + 10;
const int INF = 0x3f3f3f3f;

struct Edge{
int u;
int v;
int cap;
int flow;
};

struct Dinic{
int n, m, s, t;
vector<Edge> edges;
vector<int> G[maxn];
bool vis[maxn];
int d[maxn];
int cur[maxn];

void addEdge(int uu, int vv, int cap){
edges.push_back((Edge){uu, vv, cap, 0});
edges.push_back((Edge){vv, uu, 0, 0});
m = edges.size();
G[uu].push_back(m-2);
G[vv].push_back(m-1);
}

bool bfs(){
memset(vis, 0, sizeof(vis));
queue<int> qu;
qu.push(s);
d[s] = 0;
vis[s] = 1;
while(!qu.empty()){
int x = qu.front(); qu.pop();
int si = G[x].size();
for(int i = 0; i < si; i++){
Edge& e = edges[G[x][i]];
if(!vis[e.v] && e.cap > e.flow){
vis[e.v] = 1;
d[e.v] = d[x] + 1;
qu.push(e.v);
}
}
}
return vis[t];
}

int dfs(int x, int a){
if(x == t || a == 0) return a;
int flow = 0, f;
int si = G[x].size();
for(int& i = cur[x]; i < si; i++){
Edge& e = edges[G[x][i]];
if(d[x] + 1 == d[e.v] && (f = dfs(e.v, min(a, e.cap-e.flow))) > 0){
e.flow += f;
edges[G[x][i]^1].flow -= f;
flow += f;
a -= f;
if(a == 0) 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;
}
};

int main()
{
int N, M, S, E, C, i;
while(scanf("%d%d", &N, &M) == 2){
Dinic dic;
for(i = 0; i < N; i++){
scanf("%d%d%d", &S, &E, &C);
dic.addEdge(S, E, C);
}
printf("%d\n", dic.Maxflow(1, M));
}
return 0;
}