费用流,拆边。阴间输入。
https://www.luogu.com.cn/problem/P4012
题意
有一个网格和一些能向上或者向右行走的机器人,机器人第一次通过网格边时可以获得奖励,但是之后到的机器人不会得到奖励(即每个奖励只能被获得一次),有一些机器人起点和能容纳一定数量机器人的终点,问机器人走完之后最多能获得多少奖励?
Tutorial
不难想到最大费用流,需要稍稍处理一下奖励问题:
将每个奖励拆成两条边:
一条容量为1,费用为奖励数;
一条容量为inf,费用为0.
spfa找最长路,这样就保证了只有第一个通过的能得到奖励,并且一定优先走有奖励的那条边。
图建起来是比较奇怪的,有两行要反过来。为了确保建出了想要的图,一些傻瓜式debug还是需要的,比如:printf("(%d, %d)->(%d, %d)\n", j-1, i, j, i);
点击查看代码
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <vector>
#define endl '\n'
#define IOS \
ios::sync_with_stdio(0); \
cin.tie(0); \
cout.tie(0);
#define P pair<int, int>
#define endl '\n'
using namespace std;
#define int long long
typedef long long ll;
const int maxn = 16 * 16 * 2 + 10;
const ll inf = 1e18;
int n1, n2, cnt_edge = 1, S, T;
int head[maxn];
ll dis[maxn];
bool vis[maxn];
struct edge {
int to, nxt;
ll flow, cost;
} e[(maxn * maxn) << 2];
inline void add(int u, int v, ll w, ll c) {
e[++cnt_edge].nxt = head[u];
head[u] = cnt_edge;
e[cnt_edge].to = v;
e[cnt_edge].flow = w;
e[cnt_edge].cost = c;
}
inline void addflow(int u, int v, ll w, ll c) {
add(u, v, w, c);
add(v, u, 0, -c);
}
inline bool spfa(int on) {
memset(vis, 0, sizeof(vis));
if (on == 1)
for (int i = 0; i <= T; i++) dis[i] = inf;
else
for (int i = 0; i <= T; i++) dis[i] = -inf;
queue<int> q;
q.push(S);
dis[S] = 0;
vis[S] = 1;
while (!q.empty()) {
int x = q.front();
q.pop();
vis[x] = 0;
for (int i = head[x]; i; i = e[i].nxt) {
int y = e[i].to;
// cout << "->" << y << endl;
if ((on == 1 && e[i].flow && dis[y] > dis[x] + e[i].cost) ||
(on == -1 && e[i].flow && dis[y] < dis[x] + e[i].cost)) {
dis[y] = dis[x] + e[i].cost;
if (!vis[y]) q.push(y), vis[y] = 1;
}
}
}
if (on == 1)
return dis[T] != inf;
else
return dis[T] != -inf;
}
ll dfs(int x, ll lim) {
vis[x] = 1;
if (x == T || lim <= 0) return lim;
ll res = lim;
for (int i = head[x]; i; i = e[i].nxt) {
int y = e[i].to;
if (dis[y] != dis[x] + e[i].cost || e[i].flow <= 0 || vis[y]) continue;
ll tmp = dfs(y, min(res, e[i].flow));
res -= tmp;
e[i].flow -= tmp;
e[i ^ 1].flow += tmp;
if (res <= 0) break;
}
return lim - res;
}
inline ll Dinic(int on) {
ll res = 0, cost = 0;
while (spfa(on)) {
ll flow = dfs(S, inf);
res += flow, cost += flow * dis[T];
}
return cost;
}
int id(int x, int y, int in) {
return 2 * ((n1 * 2 + x - 2) * (x - 1) / 2 + y - 1) + in;
}
int p, q;
// int cnt = 0;
int id(int x, int y) {
// if(cnt) cout << x << " " << y << " :" << (q+1) * (y) + x << endl;
// cnt ^= 1;
return (q + 1) * (y) + x;
}
signed main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> n1 >> n2 >> p >> q;
S = (p + 1) * (q + 1), T = S + 1;
int m;
for (int i = 0; i <= p; i++) {
for (int j = 1; j <= q; j++) {
cin >> m;
// printf("(%d, %d)->(%d, %d)\n", j-1, i, j, i);
addflow(id(j - 1, i), id(j, i), 1, m);
addflow(id(j - 1, i), id(j, i), inf, 0);
}
}
for (int i = 0; i <= q; i++) {
for (int j = 1; j <= p; j++) {
cin >> m;
// printf("(%d, %d)->(%d, %d)\n", i, j-1, i, j);
addflow(id(i, j - 1), id(i, j), 1, m);
addflow(id(i, j - 1), id(i, j), inf, 0);
}
}
for (int i = 0, k, x, y; i < n1; i++) {
cin >> k >> x >> y;
swap(x, y);
// printf("S->(%d, %d)\n", x, y);
addflow(S, id(x, y), k, 0);
}
for (int i = 0, k, x, y; i < n2; i++) {
cin >> k >> x >> y;
swap(x, y);
// printf("(%d, %d)->T\n", x, y);
addflow(id(x, y), T, k, 0);
}
// cout << S << " " << T << endl;
cout << Dinic(-1);
return 0;
}