DP套DP HDOJ 4899 Hero meet devil（国王的子民的DNA）

#include <bits/stdc++.h>

const int S = (1 << 15) + 5;
const int MOD = 1e9 + 7;

char color[] = "ATGC";
char s[20];
int pre[20], lcs[20];
int dp[2][S], add[S][4];
int ans[20];
int n, m;

void add_mod(int &a, int b) {
a += b;
if (a >= MOD) {
a -= MOD;
}
}

int bit_count(int x) {
return x ? bit_count (x >> 1) + (x & 1) : x;
}

void init() {
n = strlen (s + 1);
for (int state=0; state<(1<<n); ++state) {  //状压枚举S与T公共点的组合
pre[0] = 0;
for (int i=1; i<=n; ++i) {
pre[i] = pre[i-1] + ((state>>(i-1)) & 1);  //S匹配到前i个时LCS的长度
}
for (int k=0; k<4; ++k) {  //T的某一个位置是color[k]，新的LCS的长度和公共点位置
for (int i=1; i<=n; ++i) {
if (s[i] == color[k]) {
lcs[i] = pre[i-1] + 1;
} else {
lcs[i] = std::max (lcs[i-1], pre[i]);
}
}

int &tmp = add[state][k] = 0;  //对于state状态，加一个颜色k后新的state
for (int i=1; i<=n; ++i) {
tmp |= ((lcs[i]!=lcs[i-1]) << (i-1));
}
}
}
}

void solve() {
int now = 0;
memset (dp[now], 0, sizeof (dp[now]));
dp[now][0] = 1;
for (int i=1; i<=m; ++i) {
now ^= 1;
memset (dp[now], 0, sizeof (dp[now]));
for (int state=0; state<(1<<n); ++state) {
for (int k=0; k<4; ++k) {
add_mod (dp[now][add[state][k]], dp[now^1][state]);
}
}
}
memset (ans, 0, sizeof (ans));
for (int state=0; state<(1<<n); ++state) {
add_mod (ans[bit_count (state)], dp[now][state]);
}
for (int i=0; i<=n; ++i) {
printf ("%d\n", ans[i]);
}
}

int main() {
int T;
scanf ("%d", &T);
while (T--) {
scanf ("%s", s + 1);
scanf ("%d", &m);
init ();

solve ();
}
return 0;
}