fangguo

摘要： 思路：直接用long long 保存会WA。用下高精度加法就行了。#include#include#include#include#include#include#include#include#include#include#include#include#define pb push_back#define mp make_pair#define Maxn 50010#define Maxm 80002#define LL __int64#define Abs(x) ((x)>0?(x):(-x))#define lson(x) (x=2;j--) update(nu... 阅读全文

摘要： 思路：这是一道坑爹的动态规划，思路很容易想到，就是细节。用dp[t][i][j]，表示在第t时间，锤子停在(i,j)位置能获得的最大数量。那么只要找到一个点转移到（i,j）收益最大即可。#include#include#include#include#include#include#include#include#include#include#include#define pb push_back#define mp make_pair#define Maxn 2000010#define Maxm 80002#define LL __int64#define Abs(x) ((x)> 阅读全文

摘要： 思路：简单动态规划#include#include#include#include#include#include#include#include#include#include#include#define pb push_back#define mp make_pair#define Maxn 520#define Maxm 80002#define LL __int64#define Abs(x) ((x)>0?(x):(-x))#define lson(x) (x<<1)#define rson(x) (x<<1|1)#define inf 0x7ffff 阅读全文

摘要： 思路：简单树状数组#include#include#include#include#include#include#include#include#include#include#include#define pb push_back#define mp make_pair#define Maxn 120#define Maxm 80002#define LL __int64#define Abs(x) ((x)>0?(x):(-x))#define lson(x) (x<<1)#define rson(x) (x<<1|1)#define inf 0x7ffff 阅读全文

摘要： 思路：dp[i][j]表示区间i,j变为回文串最少的代价.#include#include#include#include#include#include#include#include#include#include#include#define pb push_back#define mp make_pair#define Maxn 2002#define Maxm 80002#define LL __int64#define Abs(x) ((x)>0?(x):(-x))#define lson(x) (x=1;j--){ if(str[i]==str[j]){ ... 阅读全文

摘要： 思路：裸的LCA#include#include#include#include#include#include#include#include#include#include#include#define pb push_back#define mp make_pair#define Maxn 40010#define Maxm 80002#define LL __int64#define Abs(x) ((x)>0?(x):(-x))#define lson(x) (x > que[Maxn];void init(){ memset(head,-1,sizeof(head)); 阅读全文

摘要： 思路： 每个点有三种状态，本身有塔，被子节点的塔覆盖，被父节点的塔覆盖。#include#include#include#include#include#include#include#include#include#include#include#define Maxn 10100#define Maxm 100010#define LL __int64#define Abs(x) ((x)>0?(x):(-x))#define lson(x) (x<<1)#define rson(x) (x<<1|1)#define inf 1000000#define Mo 阅读全文

摘要： 思路：dp[i][j]表示，以i节点为根，删去j个节点最少要断几条边。那么dp[u][j]=min(dp[u][j],dp[v][k]+dp[u][j-k]);//选取最优状态dp[u][j]=min(dp[u][j],dp[u][j-son[v]]+1);//切断与子节点相连的边对于子节点dp[v][n-son[v]]=1;dp[v][j]=min(dp[v][j],dp[v][j-n+son[v]]+1)//表示不需要由父节点过来的那条分支最有从所有状态中选举最优的dp[i][n-p];#include#include#include#include#include#include#inc 阅读全文

摘要： #include#include#include#include#include#include#include#include#include#include#include#define Maxn 100010#define Maxm 200010#define LL __int64#define Abs(x) ((x)>0?(x):(-x))#define lson(x) (x que[Maxn];void init(){ memset(head,-1,sizeof(head)); memset(vi,0,sizeof(vi)); memset(vis,0,sizeof... 阅读全文

摘要： 思路：我以前一直喜欢用根号n分段的LCA。在这题上挂了，第一次发现这样的LCA被卡。果断改用Tarjan离线算法求LCA。当前节点为u，其子节点为v。那么：当以v根的子树中含有连接子树以外点的边数为out[v]。out[v]==0,dp[u]+=m;out[v]==1,dp[u]+=1;else dp[u]+=0;最后就是dp[u]+=dp[v]。对于u点的out[u]+=out[v];最后out[u]-=cnt[u]；cnt[u]表示以u为根，在子树内的边数。#include#include#include#include#include#include#include#include#in 阅读全文