在家里果然效率是低啊….什么垃圾错误都调不出来, 竟然没发现BFS的时候数组没初始化完卡了一下午…果然还是太菜了
然后我们来看这个题 先用并查集判断是否有解
然后可以发现对于每一个时间这个图的形态都不一样
那我们可以枚举时间 对于每一个时间新建一层点
然后原本的点朝它下一个时间线的点连
的流量
对于每艘船找出当前时间 向下一个时间线的终点连
的流量
这样子不断跑最大流 直到流量
即当前时间为答案
题目链接
Codes
#include<bits/stdc++.h>
#define For(i, a, b) for(register int i = a; i <= b; ++ i)
#define go(x, i) for(register int i = head[x]; i; i = nxt[i])
#define inf (0x3f3f3f3f)
#define rel(x) (x <= 0 ? x + n + 2 : x)
using namespace std;
const int N = 50 + 10;
int n, m, k, fa[N];
int h[N], p[N][N];
int find(int x) {return fa[x] = x == fa[x] ? x : find(fa[x]);}
void merge(int x, int y) {
x = find(x), y = find(y);
if(x != y) fa[x] = y;
}
namespace Max_Flow {
const int maxn = 1e6 + 10;
int to[maxn], head[maxn], nxt[maxn], w[maxn];
int dis[maxn], S, T, e = 1, cnt;
void add(int x, int y, int z) {
to[++ e] = y;
nxt[e] = head[x];
head[x] = e;
w[e] = z;
if(z) add(y, x, 0);
}
bool bfs() {
queue<int> q;
memset(dis, 0, sizeof(dis));
dis[S] = 1, q.push(S);
while(!q.empty()) {
int k = q.front(); q.pop();
go(k, i)
if(!dis[to[i]] && w[i]) {
dis[to[i]] = dis[k] + 1;
q.push(to[i]);
}
}
return dis[T];
}
int dfs(int x, int flow) {
if(x == T || !flow)
return flow;
int used = 0;
go(x, i)
if(dis[to[i]] == dis[x] + 1 && w[i]) {
int di = dfs(to[i], min(flow, w[i]));
w[i] -= di, w[i ^ 1] += di;
flow -= di, used += di;
if(!flow) break;
}
return used;
}
int dinic() {
int res = 0, flow;
while(bfs())
while(flow = dfs(S, inf))
res += flow;
return res;
}
}
int main() {
#ifndef ONLINE_JUDGE
freopen("2754.in", "r", stdin);
freopen("2754.out", "w", stdout);
#endif
int res = 0;
scanf("%d%d%d", &n, &m, &k);
Max_Flow::S = rel(0), Max_Flow::T = rel(-1);
For(i, 1, n + 2) fa[i] = i;
For(i, 1, m) {
scanf("%d%d", &h[i], &p[i][0]);
For(j, 1, p[i][0]) {
scanf("%d", &p[i][j]), p[i][j] = rel(p[i][j]);
if(j >= 2) merge(p[i][j], p[i][j - 1]);
}
}
if(find(n + 2) != find(n + 1)) return puts("0"), 0;
for(int ans = 1; ; ++ ans) {
For(i, 0, n)
Max_Flow::add((ans - 1) * (n + 2) + rel(i), ans * (n + 2) + rel(i), inf);
For(i, 1, m) {
int from = (ans - 1) % p[i][0] + 1, to = ans % p[i][0] + 1, x, y;
if(p[i][from] == rel(-1)) x = p[i][from];
else x = (ans - 1) * (n + 2) + rel(p[i][from]);
if(p[i][to] == rel(-1)) y = p[i][to];
else y = ans * (n + 2) + rel(p[i][to]);
Max_Flow::add(x, y, h[i]);
}
res += Max_Flow::dinic();
if(res >= k) return printf("%d\n", ans), 0;
}
return 0;
}