12166 - Equilibrium Mobile(DFS)

来源：互联网发布：魔方淘宝编辑：程序博客网时间：2024/06/10 08:33

枚举每个点,判断其余深度是否存在点的权值为(1 << (deep1 - deep2)) * x

deep1,deep2为2点的深度

写的比较麻烦

#include<cstdio>#include<vector>#include<cstring>#include<algorithm>using namespace std;typedef long long LL;const int maxn = 1000005;char str[maxn];vector<LL>G[20];LL base[20];int  pos,cnt,max_deep;void build(int deep){    max_deep = max(max_deep,deep);    if(str[pos] == '['){        pos ++;        build(deep + 1);    }    if(str[pos] >= '0' && str[pos] <= '9'){        LL v = 0;        while(str[pos] >= '0' && str[pos] <= '9'){            v = v * 10 + str[pos] - '0';            pos ++;        }        cnt ++;        G[deep].push_back(v);    }    if(str[pos] == ']'){        pos ++;    }    else if(str[pos] == ','){        pos ++;        build(deep);    }    return;}void solve(){    int ret = cnt;    for(int i = 1; i <= max_deep; i++)        sort(G[i].begin(),G[i].end());    for(int i = 1; i <= max_deep; i++){        for(int j = 0; j < G[i].size(); j++){            LL e = G[i][j];            int ans = 0;            for(int k = 1; k <= i; k++){                int d = i - k;                LL c = base[d] * e;                int num = upper_bound(G[k].begin(),G[k].end(),c) - lower_bound(G[k].begin(),G[k].end(),c);                ans += num;            }            for(int k = i + 1; k <= max_deep; k++){                int d = k - i;                if(e % base[d]) continue;                LL c = e / base[d];                int num = upper_bound(G[k].begin(),G[k].end(),c) - lower_bound(G[k].begin(),G[k].end(),c);                ans += num;            }            ret = min(ret,cnt - ans);        }    }    printf("%d\n",ret);    return;}void init(){    for(int i = 0; i <= 16; i++)        G[i].clear();    cnt = max_deep = pos = 0;}int main(){    int T;    base[0] = 1;    for(int i = 1; i <= 16; i++)        base[i] = base[i - 1] * 2;    scanf("%d",&T);    while(T--){        init();        scanf("%s",str);        if(str[0] != '['){            printf("0\n");            continue;        }        build(0);        solve();    }    return 0;}/*2[[3,7],6][[2,3],[4,5]][[40,40],[40,40]][[2,[1,1]],4]*/

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