hdu 5965 扫雷 ccpc 2016 合肥站

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分析：

dp题目计dp[i][j][k]为到第i个位置，上下一共放j个雷，前面一格放k个雷个方案数。

枚举一下第i−2位的雷数与第i−1位的雷数，那么第i位的雷数可以算出来。

如果放两个或不放，数目保持不变；放一个的话前一个状态乘2。

坑爹的是这道题模的是1e8+7，WA了一下午，真是坑死了。

代码：

/*****************************************************/
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <map>
#include <set>
#include <ctime>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>

using namespace std;

#define   offcin        ios::sync_with_stdio(false)
#define   sigma_size    26
#define   lson          l,m,v<<1
#define   rson          m+1,r,v<<1|1
#define   slch          v<<1
#define   srch          v<<1|1
#define   sgetmid       int m = (l+r)>>1
#define   ll            long long
#define   ull           unsigned long long
#define   mem(x,v)      memset(x,v,sizeof(x))
#define   lowbit(x)     (x&-x)
#define   bits(a)       __builtin_popcount(a)
#define   mk            make_pair
#define   pb            push_back
#define   fi            first
#define   se            second

const int    INF    = 0x3f3f3f3f;
const ll     INFF   = 1e18;
const double pi     = acos(-1.0);
const double inf    = 1e18;
const double eps    = 1e-9;
const ll     mod    = 1e8+7;
const int    maxmat = 10;
const ull    BASE   = 133333331;

/*****************************************************/
inline void RI(int &x) {
char c;
while((c=getchar())<'0' || c>'9');
x=c-'0';
while((c=getchar())>='0' && c<='9') x=(x<<3)+(x<<1)+c-'0';
}
/*****************************************************/

const int maxn = 1e4 + 5;

ll dp[maxn][3][3];
char s[maxn];

int main(int argc, char const *argv[]) {
int T; cin>>T;
while (T --) {
s[0] = '0';
memset(dp, 0, sizeof(dp));
scanf("%s", s + 1);
int N = strlen(s) - 1;
if (N == 1) {
if (s[1] == '0') puts("1");
else if (s[1] == '1') puts("2");
else if (s[1] == '2') puts("1");
else puts("0");
}
else {
dp[0][0][0] = 1;
dp[1][0][0] = 1, dp[1][1][0] = 2, dp[1][2][0] = 1;
for (int i = 2; i <= N; i ++) {
for (int p = 0; p < 3; p ++) {
for (int q = 0; q < 3; q ++) {
int j = s[i - 1] - '0' - p - q;
if (j == 0) dp[i][j][q] = (dp[i][j][q] + dp[i - 1][q][p]) % mod;
else if (j == 1) dp[i][j][q] = (dp[i][j][q] + dp[i - 1][q][p] * 2) % mod;
else if (j == 2) dp[i][j][q] = (dp[i][j][q] + dp[i - 1][q][p]) % mod;
}
}
}
ll ans = 0;
for (int i = 0; i < 3; i ++) {
int tmp = s
- '0' - i;
if (tmp > 2 || tmp < 0) continue;
ans = (ans + dp
[i][tmp]) % mod;
}
printf("%I64d\n", ans);
}
}
return 0;
}