UVALive 6914

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题目描述：

一个8*8的格子.给你一个最多可以放的障碍物的数目.一个人只能往下或者往右走.问把00点和(n-1)(m-1)点能够隔开的总放障碍的方法数.

题解：

首先我们注意到n很小,可以状压. 如果我们不状压怎么样?f[i][j][k]隔断到i,j点,用了k个障碍. 那么i,j放不放障碍呢?放的话,好转移.不放的话,我们要考虑它的上方和左方放不放点.有一个放和两个都放好弄.两个都不放就不会了. 于是我们再增加一维sta. 用来状压i,j点还没放的时候那一层的状态情况.含义和铺砖一样. 一样就可以只看当前点要不要放来转移.

重点：

8果断用一层的状压来简化思路(不状压发现不会转移….)

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <cmath>#include <ctype.h>#include <limits.h>#include <cstdlib>#include <algorithm>#include <vector>#include <queue>#include <map>#include <stack>#include <set>#include <bitset>#define CLR(a) memset(a, 0, sizeof(a))#define REP(i, a, b) for(ll i = a;i < b;i++)#define REP_D(i, a, b) for(ll i = a;i <= b;i++)typedef long long ll;using namespace std;const ll maxS = (1<<8);const ll maxn = 10;const ll MOD = 1e9+7.1;ll f[maxn][maxn][maxn*maxn][maxS];ll n, m, limit;void solve(){    ll key = (1<<m);    CLR(f);    f[0][0][0][(key-1)^1] = 1;    for(ll i = 0; i<n; i++)    {        for(ll j = 0; j<m; j++)        {            for(ll k = 0; k<=limit; k++)            {                for(ll s = 0; s<key; s++)                {                    if(j==0)                    {                        ll newS = s;                        f[i][j+1][k][newS]=(f[i][j+1][k][newS] + f[i][j][k][s])%MOD;                        newS = (s|(1));                        f[i][j+1][k+1][newS] = (f[i][j+1][k+1][newS] + f[i][j][k][s])%MOD;                    }                    else if(j==m-1)                    {                        ll newS = s;                        if((1<<(j-1)&s)==0 || ((1<<j)&s)==0)                        {                            ll t = (1<<j);                            t = ~t;                            newS = (s&(t));                        }                        else                        {                            newS = (s|(1<<j));                        }                        f[i+1][0][k][newS] = (f[i+1][0][k][newS] + f[i][j][k][s])%MOD;                        newS = (s|(1<<j));                        f[i+1][0][k+1][newS] = (f[i+1][0][k+1][newS] + f[i][j][k][s])%MOD;                    }                    else                    {                        ll newS = s;                        if((1<<(j-1)&s)==0 || ((1<<j)&s)==0)                        {                            ll t = (1<<j);                            t = ~t;                            newS = (s&(t));                        }                        else                        {                            newS = (s|(1<<j));                        }                        f[i][j+1][k][newS] = (f[i][j+1][k][newS] + f[i][j][k][s])%MOD;                        newS = (s|(1<<j));                        f[i][j+1][k+1][newS] = (f[i][j+1][k+1][newS] + f[i][j][k][s])%MOD;                    }                }            }        }    }    //printf("-------------%lld \n", f[0][2][0][4]);    ll ans = 0;    for(ll s = 0; s < key; s++)    {        if((s&(1<<(m-1)) ))        {            for(ll i = 0; i<=limit; i++)            {                ans = (ans + f[n][0][i][s])%MOD;            }        }    }    printf("%lld\n", ans);}ll C[20][20];void getC(){    C[0][0] = 1;    for(ll i = 1; i<=8; i++)    {        C[i][0] = 1;        for(ll j = 1; j<i; j++)        {            C[i][j] = (C[i-1][j] + C[i-1][j-1])%MOD;        }        C[i][i] = 1;    }}int main(){    //freopen("9Iin.txt", "r", stdin);    //freopen("9Iout.txt", "w", stdout);    ll t;    getC();    scanf("%lld", &t);    REP_D(_,1,t)    {        printf("Case #%lld: ", _);        scanf("%lld%lld%lld", &n, &m, &limit);        if(m==1)        {            ll ans = 0;            for(ll i = 1; i<=limit; i++)            {                ans = (ans + C[n][i])%MOD;            }            printf("%lld\n", ans);        }        else        {            solve();        }    }    return 0;}