【BZOJ 1877】【SDOI 2009】晨跑

拆点跑$MCMF最小费用最大流$

复习一下$MCMF$模板啦啦啦~~~

一些坑:更新$dist$后要接着更新$pre$,不要判断是否在队列中再更新,,,听不懂吧,听不懂就对了,因为只有我才会在这种错误上犯逗$TwT$

#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 205
#define M 20005
#define mo 410
#define read(x) x=getint()
using namespace std;
inline int getint() {int k = 0, fh = 1; char c = getchar(); for(; c < '0' || c > '9'; c = getchar()) if (c == '-') fh = -1; for(; c >= '0' && c <= '9'; c = getchar()) k = k * 10 + c - '0'; return k * fh;}
struct node {
int nxt, to, c, cost;
} E[(N << 1) + (M << 1)];
bool vis[mo];
int point[mo], pre[mo], n, m, cnt = 1, q[mo], dist[mo];
inline void ins(int x, int y, int z, int c) {
E[++cnt].nxt = point[x]; E[cnt].to = y; E[cnt].c = z; E[cnt].cost = c; point[x] = cnt;
E[++cnt].nxt = point[y]; E[cnt].to = x; E[cnt].c = 0; E[cnt].cost = -c; point[y] = cnt;
}
inline bool spfa(int s, int t) {
for(int i = 1; i <= ((n - 1) << 1); ++i)
dist[i] = 0x7fffffff;
int head = 0, tail = 1;
q[1] = s; vis[s] = 1; dist[s] = 0;
while (head != tail) {
++head; if (head == mo) head = 0;
int u = q[head];
vis[u] = 0;
for(int tmp = point[u]; tmp; tmp = E[tmp].nxt)
if (E[tmp].c > 0) {
int v = E[tmp].to;
if (dist[u] + E[tmp].cost < dist[v]) {
dist[v] = dist[u] + E[tmp].cost;
pre[v] = tmp;
if (!vis[v]) {
++tail; if (tail == mo) tail = 0;
q[tail] = v;
vis[v] = 1;
}
}
}
}
return dist[t] != 0x7fffffff;
}
inline void MCMF(int s, int t) {
int a1 = 0, a2 = 0, flow, now;
while (spfa(s, t)) {
++a1;
flow = 0x7fffffff;
for(now = pre[t]; now; now = pre[E[now ^ 1].to])
flow = min(flow, E[now].c);
for(now = pre[t]; now; now = pre[E[now ^ 1].to])
a2 += E[now].cost * flow, E[now].c -= flow, E[now ^ 1].c += flow;
}
printf("%d %d\n", a1, a2);
}
int main() {
read(n); read(m);
for(int i = 2; i < n; ++i)
ins((i << 1) - 1, i << 1, 1, 0);
int u, v, e;
for(int i = 1; i <= m; ++i) {
read(u); read(v); read(e);
if (u == 1)
if (v == n) ins(1, 2, 1, e);
else ins(1, (v << 1) - 1, 1, e);
else if (v == n)
ins(u << 1, 2, 1, e);
else
ins(u << 1, (v << 1) - 1, 1, e);
}
MCMF(1, 2);
return 0;
}

hhh

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