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P3356 火星探险问题

P3356 火星探险问题

题面

题目链接

解法

问题很容易转化成最大费用最大流,拆点分别考虑即可。

如何输出路径呢?只需要跑一次$dfs$或$bfs$,从$S$到$T$走反边流量非零的边即可,记得每次走完将反边的流量$-1$。

AC代码

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/**
* @file: P3356.cpp
* @author: yaoxi-std
* @url: https://www.luogu.com.cn/problem/P3356
*/
// #pragma GCC optimize ("O2")
// #pragma GCC optimize ("Ofast", "inline", "-ffast-math")
// #pragma GCC target ("avx,sse2,sse3,sse4,mmx")
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define resetIO(x) \
freopen(#x ".in", "r", stdin), freopen(#x ".out", "w", stdout)
#define debug(fmt, ...) \
fprintf(stderr, "[%s:%d] " fmt "\n", __FILE__, __LINE__, ##__VA_ARGS__)
template <class _Tp>
inline _Tp& read(_Tp& x) {
bool sign = false;
char ch = getchar();
long double tmp = 1;
for (; !isdigit(ch); ch = getchar())
sign |= (ch == '-');
for (x = 0; isdigit(ch); ch = getchar())
x = x * 10 + (ch ^ 48);
if (ch == '.')
for (ch = getchar(); isdigit(ch); ch = getchar())
tmp /= 10.0, x += tmp * (ch ^ 48);
return sign ? (x = -x) : x;
}
template <class _Tp>
inline void write(_Tp x) {
if (x < 0)
putchar('-'), x = -x;
if (x > 9)
write(x / 10);
putchar((x % 10) ^ 48);
}
template <const int MAXV, const int MAXE>
struct MCMF {
const int INF = 0x3f3f3f3f3f3f3f3f;
struct Edge {
int v, flow, cost;
} edge[MAXE * 2];
int tot = 1, head[MAXV], nxt[MAXE];
int flow, cost, cur[MAXV], dis[MAXV];
bool vis[MAXV];
void addedge(int u, int v, int flow, int cost) {
edge[++tot] = {v, flow, cost};
nxt[tot] = head[u], head[u] = tot;
edge[++tot] = {u, 0, -cost};
nxt[tot] = head[v], head[v] = tot;
}
bool spfa(int s, int t) {
fill(vis, vis + MAXV, 0);
fill(dis, dis + MAXV, -INF);
queue<int> que;
que.push(s);
dis[s] = 0;
vis[s] = 1;
while (!que.empty()) {
int u = que.front();
que.pop();
vis[u] = 0;
for (int i = head[u]; i; i = nxt[i]) {
int v = edge[i].v;
if (edge[i].flow && dis[v] < dis[u] + edge[i].cost) {
dis[v] = dis[u] + edge[i].cost;
if (!vis[v]) {
que.push(v);
vis[v] = 1;
}
}
}
}
return dis[t] != -INF;
}
int augment(int u, int t, int mx) {
if (u == t || mx == 0)
return mx;
vis[u] = 1;
int ret = 0;
for (int &i = cur[u]; i; i = nxt[i]) {
int v = edge[i].v;
if (vis[v] || dis[v] != dis[u] + edge[i].cost)
continue;
int tmp = augment(v, t, min(mx, edge[i].flow));
cost += tmp * edge[i].cost;
mx -= tmp, ret += tmp;
edge[i].flow -= tmp, edge[i ^ 1].flow += tmp;
if (mx == 0)
break;
}
vis[u] = 0;
return ret;
}
pair<int, int> mcmf(int s, int t) {
while (spfa(s, t)) {
copy(head, head + MAXV, cur);
flow += augment(s, t, INF);
}
return make_pair(flow, cost);
}
};
const int MAXN = 40;
const int MAXV = 4e4 + 10;
const int MAXE = 8e4 + 10;
const int INF = 0x3f3f3f3f3f3f3f3f;
int n, p, q, s, t, num, mp[MAXN][MAXN];
int px[MAXV], py[MAXV], pt[MAXN][MAXN][2];
MCMF<MAXV, MAXE> network;
void print(int cnt, int u) {
for (int i = network.head[u]; i; i = network.nxt[i]) {
int v = network.edge[i].v;
if ((i & 1) == 0 && network.edge[i ^ 1].flow) {
if (px[u] && py[u] && px[v] && py[v] && ((px[u] ^ px[v]) || (py[u] ^ py[v])))
printf("%lld %lld\n", cnt, py[v] - py[u]);
--network.edge[i ^ 1].flow;
print(cnt, v);
break;
}
}
}
signed main() {
read(n), read(p), read(q);
for (int i = 1; i <= q; ++i)
for (int j = 1; j <= p; ++j)
read(mp[i][j]);
s = ++num, t = ++num;
for (int i = 1; i <= q; ++i) {
for (int j = 1; j <= p; ++j) {
pt[i][j][0] = ++num;
px[num] = i, py[num] = j;
pt[i][j][1] = ++num;
px[num] = i, py[num] = j;
}
}
network.addedge(s, pt[1][1][0], n, 0);
network.addedge(pt[q][p][1], t, n, 0);
for (int i = 1; i <= q; ++i) {
for (int j = 1; j <= p; ++j) {
if (mp[i][j] != 1)
network.addedge(pt[i][j][0], pt[i][j][1], INF, 0);
if (mp[i][j] == 2)
network.addedge(pt[i][j][0], pt[i][j][1], 1, 1);
if (i < q)
network.addedge(pt[i][j][1], pt[i + 1][j][0], INF, 0);
if (j < p)
network.addedge(pt[i][j][1], pt[i][j + 1][0], INF, 0);
}
}
network.mcmf(s, t);
for (int i = 1; i <= n; ++i)
print(i, s);
return 0;
}