【LOJ】#2384. 「HNOI2013」切糕
题解
神仙网络流啊……
naive的我一直想把每个纵轴拆点,每个纵轴建R个点(大概是要跑费用流吧……)……然后第二个限制就gg了,什么也想不出来,菜啊TAT
后来我发现大神们的建图都是,一个原点,一个汇点,一段长条,每一段就是一个点,流量是值,那么最小割就是最小值了,很神奇
然后每相邻的两个纵轴z向z - D连一条边,z再向z + D连一条边
感觉很神奇,后来我想了一下,这样并不能保证割了z之后一定割了相邻的[z - D,z+D]的点,而是可以保证,如果割错了这个割一定会被换掉!错误的答案是有限的,重复的错误不会出现……所以,最后就是正确的答案了呗。。。
最小割保证了答案最小
代码
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#define enter putchar('\n')
#define space putchar(' ')
#define MAXN 70005
//#define ivorysi
using namespace std;
typedef long long int64;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0';
c = getchar();
}
res = res * f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) out(x / 10);
putchar('0' + x % 10);
}
struct node {
int to,next,cap;
}E[4000005];
int head[MAXN],sumE = 1,S,T,last[MAXN],Ncnt;
int dis[MAXN],gap[MAXN],P,Q,R,D;
int val[45][45][45],MK[45][45][45];
int dx[] = {0,-1};
int dy[] = {-1,0};
void add(int u,int v,int c) {
E[++sumE].to = v;
E[sumE].next = head[u];
E[sumE].cap = c;
head[u] = sumE;
}
void addtwo(int u,int v,int c) {
add(u,v,c);add(v,u,0);
}
int sap(int u,int aug) {
if(u == T) return aug;
int flow = 0;
for(int i = last[u] ; i ; last[u] = i = E[i].next) {
int v = E[i].to;
if(E[i].cap) {
if(dis[v] + 1 == dis[u]) {
int t = sap(v,min(aug - flow,E[i].cap));
flow += t;
E[i].cap -= t;
E[i ^ 1].cap += t;
if(flow == aug) return flow;
if(dis[S] >= T) return flow;
}
}
}
--gap[dis[u]];if(!gap[dis[u]]) dis[S] = T;
++gap[++dis[u]];last[u] = head[u];
return flow;
}
void Solve() {
read(P);read(Q);read(R);
read(D);
for(int i = 1 ; i <= R ; ++i) {
for(int j = 1 ; j <= P ; ++j) {
for(int k = 1 ; k <= Q ; ++k) {
read(val[j][k][i]);
}
}
}
S = ++Ncnt;
T = P * Q * (R + 1) + 10;
for(int i = 1 ; i <= P ; ++i) {
for(int j = 1 ; j <= Q ; ++j) {
++Ncnt;addtwo(S,Ncnt,0x7fffffff);
for(int k = 1 ; k <= R; ++k) {
++Ncnt;
MK[i][j][k] = Ncnt;
addtwo(Ncnt - 1,Ncnt,val[i][j][k]);
}
addtwo(Ncnt,T,0x7fffffff);
}
}
for(int i = 1 ; i <= P ; ++i) {
for(int j = 1 ; j <= Q ; ++j) {
for(int h = 0 ; h <= 1 ; ++h) {
int tx = i + dx[h],ty = j + dy[h];
if(tx <= 0 || ty <= 0) continue;
for(int k = 1 ; k <= R ; ++k) {
if(k - D >= 1) addtwo(MK[i][j][k],MK[tx][ty][k - D],0x7fffffff);
if(k + D <= R) addtwo(MK[tx][ty][k + D],MK[i][j][k],0x7fffffff);
}
}
}
}
for(int i = 1 ; i <= T ; ++i) last[i] = head[i];
int ans = 0;
while(dis[S] < T) ans += sap(S,0x7fffffff);
out(ans);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}