UVaLive 6832 Bit String Reordering (模拟)

题意：给定一个01序列，然后让你你最少的操作数把这它变成目标。

析：由于01必须是交替出现的，那么我们就算两次，然后取最值。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define debug puts("+++++")
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 2e5 + 5;
const LL mod = 1e9 + 7;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
inline int gcd(int a, int b){  return b == 0 ? a : gcd(b, a%b); }
inline int lcm(int a, int b){  return a * b / gcd(a, b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
int a[20], b[20];
int c[20], d[20];

int solve(){
    int ans = 0;
    memcpy(a, d, sizeof d);
    for(int i = 0; i < n; ++i){
        if(a[i] == c[i])  continue;
        for(int j = i+1; j < n; ++j)
            if(a[j] == c[i]){ ans += j-i; a[j] = !c[i];  break; }
    }
    return ans;
}

int main(){
    while(scanf("%d %d", &n, &m) == 2){
        int sum = 0;
        for(int i = 0; i < n; ++i)  scanf("%d", a+i), sum += a[i];
        for(int i = 0; i < m; ++i)  scanf("%d", b+i);
        memcpy(d, a, sizeof a);
        int cnt = 0, s = 0;
        bool ok = true;
        for(int i = 0; i < m; ++i, ok = !ok)
            for(int j = 0; j < b[i]; ++j)
                c[cnt++] = ok, s += ok;
        int ans = INF;
        if(s == sum)  ans = Min(ans, solve());
        s = 0;
        for(int i = 0; i < n; ++i)  c[i] = !c[i],s += c[i];
        if(s == sum) ans = Min(ans, solve());
        printf("%d\n", ans);
    }
    return 0;
}