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【BZOJ1497】【NOI2006】最大获利

2018-01-15 09:45 267 查看
【题目链接】

点击打开链接

【思路要点】

直接解最大权闭合子图问题即可。
时间复杂度\(O(Dinic(N+M,N+3*M))\)。

【代码】

#include<bits/stdc++.h>
using namespace std;
const int MAXP = 60005;
const int INF = 1e9;
template <typename T> void read(T &x) {
x = 0; int f = 1;
char c = getchar();
for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
x *= f;
}
template <typename T> void write(T x) {
if (x < 0) x = -x, putchar('-');
if (x > 9) write(x / 10);
putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) {
write(x);
puts("");
}
struct edge {int dest, flow; unsigned home; };
int n, m, s, t, tot;
int dist[MAXP];
unsigned curr[MAXP];
vector <edge> a[MAXP];
int dinic(int pos, int limit) {
if (pos == t) return limit;
int used = 0, tmp;
for (unsigned &i = curr[pos]; i < a[pos].size(); i++) {
if (a[pos][i].flow && dist[pos] + 1 == dist[a[pos][i].dest] && (tmp = dinic(a[pos][i].dest, min(limit - used, a[pos][i].flow)))) {
used += tmp;
a[pos][i].flow -= tmp;
a[a[pos][i].dest][a[pos][i].home].flow += tmp;
}
if (used == limit) return used;
}
return used;
}
bool bfs() {
static int q[MAXP];
static int l = 0, r = -1;
for (int i = 0; i <= r; i++)
dist[q[i]] = 0;
l = 0, r = -1;
q[++r] = s; dist[s] = 1;
while (l <= r) {
int tmp = q[l++];
for (unsigned i = 0; i < a[tmp].size(); i++)
if (a[tmp][i].flow && dist[a[tmp][i].dest] == 0) {
dist[a[tmp][i].dest] = dist[tmp] + 1;
q[++r] = a[tmp][i].dest;
}
}
return dist[t] != 0;
}
void addedge(int s, int t, int flow) {
a[s].push_back((edge) {t, flow, a[t].size()});
a[t].push_back((edge) {s, 0, a[s].size() - 1});
}
int main() {
read(n), read(m);
s = 0, t = n + m + 1;
int ans = 0;
for (int i = 1; i <= n; i++) {
int x; read(x);
addedge(i, t, x);
}
for (int i = 1; i <= m; i++) {
int x, y, z;
read(x), read(y), read(z);
ans += z;
addedge(s, n + i, z);
addedge(n + i, x, INF);
addedge(n + i, y, INF);
}
while (bfs()) {
memset(curr, 0, sizeof(curr));
ans -= dinic(s, INF);
}
printf("%d\n", ans);
return 0;
}
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