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[HNOI 2007]神奇游乐园
#include <bits/stdc++.h>
#define maxn 110
using namespace std;
typedef long long ll;
int n, m;
int a[maxn][maxn];
ll ans = -1ll << 60;
#define M 2000010
#define mod 997
struct Hashmap{
ll st[M];
int h[1000], size, nxt[M];
ll f[M];
void clear(){memset(h, 0, sizeof h); size = 0;}
void push(ll hash_, ll val){
int tmp = hash_ % mod;
for(int i = h[tmp]; i; i = nxt[i]){
if(st[i] == hash_){
f[i] = max(f[i], val);
return;
}
}
int now = ++ size;
f[now] = val;
st[now] = hash_;
nxt[now] = h[tmp];
h[tmp] = now;
}
}dp[2];
int cur, code[20], ch[20];
void Decode(ll st){
for(int i = m; i >= 0; i --)
code[i] = st & 7, st >>= 3;
}
ll Encode(){
ll ret = 0;
memset(ch, -1, sizeof ch);
ch[0] = 0; int cnt = 0;
for(int i = 0; i <= m; i ++){
if(ch[code[i]] == -1)ch[code[i]] = ++ cnt;
code[i] = ch[code[i]];
ret = ret << 3 | code[i];
}
return ret;
}
void Shift(){
for(int i = m; i >= 1; i --)
code[i] = code[i-1];
code[0] = 0;
}
inline void Change(int u, int v){
for(int i = 0; i <= m; i ++)
if(code[i] == v)
code[i] = u;
}
void DP(int i, int j){
dp[cur^1].clear();
for(int k = 1; k <= dp[cur].size; k ++){
Decode(dp[cur].st[k]);
if(j == 1){if(code[m])continue;Shift();}
int Left = code[j-1], Up = code[j];
if(Left && Up){
code[j] = code[j-1] = 0;
if(Left == Up){
ll ENCODE = Encode();
if(ENCODE == 0)ans = max(ans, dp[cur].f[k] + a[i][j]);
}
else{
Change(Left, Up);
dp[cur^1].push(Encode(), dp[cur].f[k] + a[i][j]);
}
}
else if(Left || Up){
int tmp = Left ? Left : Up;
code[j-1] = 0, code[j] = tmp;
dp[cur^1].push(Encode(), dp[cur].f[k] + a[i][j]);
code[j] = 0, code[j-1] = tmp;
dp[cur^1].push(Encode(), dp[cur].f[k] + a[i][j]);
}
else{
dp[cur^1].push(Encode(), dp[cur].f[k]);
code[j] = code[j-1] = 8;
dp[cur^1].push(Encode(), dp[cur].f[k] + a[i][j]);
}
}
cur ^= 1;
}
int main(){
#ifndef ONLINE_JUDGE
freopen("park.in","r",stdin);
freopen("park.out","w",stdout);
#endif
scanf("%d%d", &n, &m);
for(int i = 1; i <= n; i ++)
for(int j = 1; j <= m; j ++)
scanf("%d", &a[i][j]);
dp[cur].clear();
dp[cur].push(0, 0);
for(int i = 1; i <= n; i ++)
for(int j = 1; j <= m; j ++)
DP(i, j);
printf("%lld\n", ans);
return 0;
}
粘了一个模板上来=-=
基于连通性的动态规划,最小表示法很好用。
我们可以用一个压缩的数字表示一个连通情况,比如记录生成树,记录概率,记录棋盘上格子的情况,等等等
只要和连通性有关而且n很小时就可以用啦QAQ。
又忘了模板了QAQ
就是如果左边和上面是一个连通分量即Left == Up时,我们要合并连通分量,所以此时已经出现了一个圈了,这道题不能有多个圈,所以最后不放进去
code[j] = code[j-1] = 0.
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原文地址:http://www.cnblogs.com/Candyouth/p/5392609.html