UVALive 6914 Maze Mayhem 轮廓线dp Maze Mayhem
题目连接：
题意
n*m的格子，给你一个k,表示你最多可以放的障碍数,问从(1,1)不能到(n,m)
的放置障碍的方案数
题解：
轮廓线dp,1表示可以走到这里。转移有3种：走这里，不放障碍；不走这里，放障碍；放障碍
再次写轮廓线，感觉有点熟练了起来
代码
//#include <bits/stdc++.h>
#include <stdio.h>
#include <iostream>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <limits.h>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <map>
#include <stack>
#include <bitset>
#include <string>
#include <time.h>
using namespace std;
long double esp=1e-11;
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#define fi first
#define se second
#define all(a) (a).begin(),(a).end()
#define cle(a) while(!a.empty())a.pop()
#define mem(p,c) memset(p,c,sizeof(p))
#define mp(A, B) make_pair(A, B)
#define pb push_back
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
typedef long long int LL;
const long double PI = acos((long double)-1);
const LL INF=0x3f3f3f3f3f3f3f3fll;
const int MOD =1000000007ll;
const int maxn=2000100;
const int HASH=30007;
const int STATE=1<<15;
struct HASHMAP
{
int head[HASH],next[STATE],sz;
LL state[STATE];
LL dp[STATE];
void init()
{
sz=0;
memset(head,-1,sizeof(head));
}
void push(LL st,LL k,LL ans)
{
int i;
st=(st<<7)+k;
int h=st%HASH;
for(i=head[h];i!=-1;i=next[i])//这里要注意是next
if(state[i]==st)
{
dp[i]=(dp[i]+ans)%MOD;
//dp[i]+=ans;
return;
}
state[sz]=st;
dp[sz]=ans;
next[sz]=head[h];
head[h]=sz++;
}
}hm[2];
int m,f;
int mask[10];
void decode(int st)
{
st>>=7;
for(int x=0;x<m;x++)
mask[x]=st>>x&1;
}
void dfs(int st,int k,int t,LL ans)
{
if(t==m)
{
hm[f^1].push(st,k,ans);
return ;
}
if((t&&(st>>(t-1)&1))||mask[t])
dfs(st|1<<t,k,t+1,ans);
else
dfs(st,k,t+1,ans);
if(k)dfs(st,k-1,t+1,ans);
}
int main()
{
//freopen("in.txt", "r", stdin);
//freopen("inlay.in", "r", stdin);
//freopen("out.txt", "w", stdout);
//::iterator iter; %I64d
//for(int x=1;x<=n;x++)
//for(int y=1;y<=n;y++)
//scanf("%d",&a);
//printf("%d
",ans);
int T;
scanf("%d",&T);
for(int gg=1;gg<=T;gg++)
{
int n,k,get=0x7f;
scanf("%d%d%d",&n,&m,&k);
hm[1].init();
hm[1].push(1,k,1);
f=1;
for(int x=1;x<=n;x++)
{
hm[f^1].init();
for(int y=0;y<hm[f].sz;y++)
{
decode(hm[f].state[y]);
dfs(0,hm[f].state[y]&get,0,hm[f].dp[y]);
}
f^=1;
/*for(int y=0;y<hm[f].sz;y++)
{
decode(hm[f].state[y]);
for(int z=0;z<m;z++)putchar('0'+mask[z]);printf(" %d %d
",hm[f].state[y]&get,hm[f].dp[y]);
}*/
}
LL ans=0;
for(int y=0;y<hm[f].sz;y++)
{
decode(hm[f].state[y]);
if(mask[m-1]==0)
ans=(ans+hm[f].dp[y])%MOD;
}
printf("Case #%d: %lld
",gg,ans);
}
return 0;
} //