2018北京冬令营测试
2018北京冬令营测试
Problem A 题解
我们发现“所有数加起来等于线索上的权值”这个东西可以用流量平衡来解决，所以可以跑上下界费用流。
建图就是，把上下线索放左边，左右线索放右边，源点向所有上下线索连一个上下界均为对应权值的边，同理左右线索向汇点连一个上下界均为对应权值的边；对于空格就是把左边的一个节点和右边的一个节点连起来，上下界均为这个空格的初始权值。
然后考虑调整，以上建的每条边都可以调整流量（正向表示权值加，反向表示权值减），注意反向要限制流量为对应初始权值减 \(1\)，这样来保证最终都是正整数。如果代价为 \(-1\)，可以直接不建调整的边。
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <algorithm>
#include <map>
using namespace std;
#define rep(i, s, t) for(int i = (s), mi = (t); i <= mi; i++)
#define dwn(i, s, t) for(int i = (s), mi = (t); i >= mi; i--)
int read() {
int x = 0, f = 1; char c = getchar();
while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }
while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }
return x * f;
}
#define maxn 1810
#define maxm 14420
#define LL long long
#define oo 2147483647
#define ool (1ll << 60)
struct Edge {
int from, to, flow, cost;
Edge() {}
Edge(int _1, int _2, int _3, int _4): from(_1), to(_2), flow(_3), cost(_4) {}
};
struct ZKW {
int n, m, s, t, head[maxn], nxt[maxm];
LL cost, ans;
Edge es[maxm];
LL d[maxn];
int Q[maxn], hd, tl;
bool inq[maxn];
bool vis[maxn];
void init() {
m = 0; memset(head, -1, sizeof(head));
return ;
}
void setn(int _) {
n = _;
return ;
}
void AddEdge(int a, int b, int c, int d) {
es[m] = Edge(a, b, c, d); nxt[m] = head[a]; head[a] = m++;
es[m] = Edge(b, a, 0, -d); nxt[m] = head[b]; head[b] = m++;
return ;
}
int Nxt(int x) { return (x + 1) % maxn; }
bool BFS() {
rep(i, 1, n) d[i] = ool;
d[t] = 0;
hd = tl = 1; Q[tl = Nxt(tl)] = t; inq[t] = 1;
while(hd != tl) {
int u = Q[hd = Nxt(hd)]; inq[u] = 0;
for(int i = head[u]; i != -1; i = nxt[i]) {
Edge& e = es[i^1];
if(e.flow && d[e.from] > d[u] + e.cost) {
d[e.from] = d[u] + e.cost;
if(!inq[e.from]) inq[e.from] = 1, Q[tl = Nxt(tl)] = e.from;
}
}
}
if(d[s] == ool) return 0;
cost = d[s];
return 1;
}
int DFS(int u, int a) {
if(u == t || !a) return ans += cost * a, a;
if(vis[u]) return 0;
vis[u] = 1;
int flow = 0, f;
for(int i = head[u]; i != -1; i = nxt[i]) {
Edge& e = es[i];
if(d[e.to] == d[u] - e.cost && (f = DFS(e.to, min(a, e.flow)))) {
flow += f; a -= f;
e.flow -= f; es[i^1].flow += f;
if(!a) return flow;
}
}
return flow;
}
int MaxFlow(int _s, int _t) {
s = _s; t = _t;
int flow = 0, f;
while(BFS())
do {
rep(i, 1, n) vis[i] = 0;
f = DFS(s, oo);
flow += f;
} while(f);
return flow;
}
} sol;
#define maxr 35
struct Grid {
int tp, v1, v2, c1, c2;
} gs[maxr][maxr];
int CntP, ind[maxn];
struct Point {
int id;
Point(): id(0) {}
int p() { return id ? id : id = ++CntP; }
} ud[maxr][maxr], lr[maxr][maxr], S, T, SS, TT;
int main() {
freopen("kakuro.in", "r", stdin);
freopen("kakuro.out", "w", stdout);
int n = read(), m = read();
rep(i, 1, n) rep(j, 1, m) gs[i][j].tp = read();
rep(i, 1, n) rep(j, 1, m) {
if(gs[i][j].tp == 1 || gs[i][j].tp == 3) gs[i][j].v1 = read();
if(gs[i][j].tp == 2 || gs[i][j].tp == 3) gs[i][j].v2 = read();
if(gs[i][j].tp == 4) gs[i][j].v1 = read();
}
rep(i, 1, n) rep(j, 1, m) {
if(gs[i][j].tp == 1 || gs[i][j].tp == 3) gs[i][j].c1 = read();
if(gs[i][j].tp == 2 || gs[i][j].tp == 3) gs[i][j].c2 = read();
if(gs[i][j].tp == 4) gs[i][j].c1 = read();
}
sol.init();
rep(i, 1, n) rep(j, 1, m) {
if(gs[i][j].tp == 1 || gs[i][j].tp == 3) {
ind[S.p()] -= gs[i][j].v1;
ind[ud[i][j].p()] += gs[i][j].v1;
if(gs[i][j].c1 >= 0) {
sol.AddEdge(S.p(), ud[i][j].p(), oo, gs[i][j].c1);
sol.AddEdge(ud[i][j].p(), S.p(), gs[i][j].v1 - 1, gs[i][j].c1);
}
}
if(gs[i][j].tp == 2 || gs[i][j].tp == 3) {
ind[lr[i][j].p()] -= gs[i][j].v2;
ind[T.p()] += gs[i][j].v2;
if(gs[i][j].c2 >= 0) {
sol.AddEdge(lr[i][j].p(), T.p(), oo, gs[i][j].c2);
sol.AddEdge(T.p(), lr[i][j].p(), gs[i][j].v2 - 1, gs[i][j].c2);
}
}
if(gs[i][j].tp == 4) {
int x1 = i, y1 = j, x2 = i, y2 = j;
while(gs[x1][y1].tp == 4) x1--; // ud
while(gs[x2][y2].tp == 4) y2--; // lr
ind[ud[x1][y1].p()] -= gs[i][j].v1;
ind[lr[x2][y2].p()] += gs[i][j].v1;
if(gs[i][j].c1 >= 0) {
sol.AddEdge(ud[x1][y1].p(), lr[x2][y2].p(), oo, gs[i][j].c1);
sol.AddEdge(lr[x2][y2].p(), ud[x1][y1].p(), gs[i][j].v1 - 1, gs[i][j].c1);
}
}
}
sol.AddEdge(T.p(), S.p(), oo, 0);
int sum = 0;
rep(i, 1, CntP) {
if(ind[i] > 0) sol.AddEdge(SS.p(), i, ind[i], 0), sum += ind[i];
if(ind[i] < 0) sol.AddEdge(i, TT.p(), -ind[i], 0);
}
sol.setn(CntP);
int flow = sol.MaxFlow(SS.p(), TT.p());
if(flow < sum) return puts("-1"), 0;
printf("%lld\n", sol.ans);
fclose(stdin);
fclose(stdout);
return 0;
}
Problem B 题解
我们将线索看成点，空格就是连接恰好两个点的无向边。把这个图建出来，随便选一棵生成树，非树边就随机一下权值，树边权值就可以确定了（可通过非树边的权值和点权从深往浅递推）。
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <algorithm>
#include <map>
using namespace std;
#define rep(i, s, t) for(int i = (s), mi = (t); i <= mi; i++)
#define dwn(i, s, t) for(int i = (s), mi = (t); i >= mi; i--)
#define LL long long
LL read() {
LL x = 0, f = 1; char c = getchar();
while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }
while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }
return x * f;
}
#define maxr 210
#define maxn 80010
#define maxm 160010
LL get_rand() {
LL x = 0;
rep(i, 0, 11) x = x << 5 | (rand() & (1 << 5) - 1);
return x;
}
int r, c;
struct Grid {
int tp; LL v1, v2;
} gs[maxr][maxr];
int CntP, eid[maxr][maxr];
LL val[maxn];
struct Point {
int id;
Point(): id(0) {}
int p() { return id ? id : id = ++CntP; }
} ud[maxr][maxr], lr[maxr][maxr];
int M, m, head[maxn], nxt[maxm], to[maxm], id[maxm];
void AddEdge(int a, int b, int i) {
to[++m] = b; id[m] = i; nxt[m] = head[a]; head[a] = m;
swap(a, b);
to[++m] = b; id[m] = i; nxt[m] = head[a]; head[a] = m;
return ;
}
short vis[maxn], OK;
LL Ans[maxn];
void dfs(int u, int fa) {
vis[u] = 2;
for(int e = head[u]; e; e = nxt[e]) if(to[e] != fa) {
if(vis[to[e]] == 2) {
Ans[id[e]] = get_rand();
val[u] ^= Ans[id[e]]; val[to[e]] ^= Ans[id[e]];
}
else if(!vis[to[e]]) {
dfs(to[e], u); Ans[id[e]] = val[to[e]];
val[u] ^= Ans[id[e]]; val[to[e]] ^= Ans[id[e]];
if(val[to[e]]) OK = 0;
}
}
if(!fa && val[u]) OK = 0;
vis[u] = 1;
return ;
}
map <LL, bool> has;
void work() {
rep(i, 1, r) rep(j, 1, c) ud[i][j].id = lr[i][j].id = 0;
r = read(); c = read();
rep(i, 1, r) rep(j, 1, c) gs[i][j].tp = read();
rep(i, 1, r) rep(j, 1, c) {
if(gs[i][j].tp == 1 || gs[i][j].tp == 3) gs[i][j].v1 = read();
if(gs[i][j].tp == 2 || gs[i][j].tp == 3) gs[i][j].v2 = read();
}
CntP = M = m = 0; memset(head, 0, sizeof(head));
rep(i, 1, r) rep(j, 1, c) if(gs[i][j].tp == 4) {
int x1 = i, y1 = j, x2 = i, y2 = j;
while(gs[x1][y1].tp == 4) x1--; // ud
while(gs[x2][y2].tp == 4) y2--; // lr
val[ud[x1][y1].p()] = gs[x1][y1].v1;
val[lr[x2][y2].p()] = gs[x2][y2].v2;
AddEdge(ud[x1][y1].p(), lr[x2][y2].p(), eid[i][j] = ++M);
}
memset(vis, 0, sizeof(vis)); OK = 1;
rep(i, 1, CntP) if(!vis[i]) {
dfs(i, 0);
if(!OK) return (void)puts("-1");
}
has.clear();
rep(i, 1, M) {
if(!Ans[i] || has.count(Ans[i])) return (void)puts("-1");
has[Ans[i]] = 1;
}
rep(i, 1, r) {
int lst = 0;
rep(j, 1, c) if(gs[i][j].tp == 4) lst = max(lst, j);
rep(j, 1, c) if(gs[i][j].tp == 4) printf("%lld%c", Ans[eid[i][j]], j < lst ? ' ' : '\n');
}
return ;
}
int main() {
freopen("cross.in", "r", stdin);
freopen("cross.out", "w", stdout);
int T = read();
while(T--) work();
fclose(stdin);
fclose(stdout);
return 0;
}
Problem C 题解
这个每条信息的候选数都是一段连续的区间最多去掉一个数字【需要分情况讨论，当 \(len \le 2\)（\(len\) 表示线索限制的区间长度）时候选数字是单点】。于是我们可以用一个扫描线从上往下扫描这个网格，遇到一个纵向的线索的开头就加入数据结构，结尾就从数据结构删除，然后对于一个横向的线索相当于区间查询信息。我们开一个线段树套线段树，外层位置内层值域，维护左右端点的总和（如果候选数字是贴着 \(k\) 的区间就存左端点，是贴着 \(1\) 的区间就存右断电）、删除的单点，以及如果候选数字是单点就维护一下这个单点信息。
具体实现我也不会，就是不知道如何求出删掉哪个值，感觉这种分类讨论题非常恶心
