51nod1693 水群

题目链接：51nod1693 水群

题解参考大神的博客：http://www.cnblogs.com/fighting-to-the-end/p/5874763.html

这题时限0.4秒，真的够狠的。。我用优化过的dij狂交一直TLE，最后还是死在四百多毫秒上，应该是姿势不对orz。后来看别人用spfa，然后一改就过了,这里我想收藏一下记忆化搜索的解法，真心服了。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 1000010;
const int inf = 0x3f3f3f3f;
int n;
int pri[] = {2,3,5,7,11};
int s[N][2];//1到i上一条边的状态是f的最小操作数
//j=0时，上一条边可以是i->i-1,否则是i->i*p
int dfs(int x, int f){
    if(x == 1) return 0;
    if(s[x][f]) return s[x][f];
    s[x][f] = inf;
    for(int i = 0; i < 4; i++)
        if(!(x % pri[i]))
            s[x][f] = min(s[x][f], dfs(x /pri[i], 1) + pri[i]);
    if(!f) return s[x][f];
    s[x][0] = s[x][1];
    for(int i = 1; i < 3; i++)
        s[x][f] = min(s[x][f], dfs( x + i, 0) + i);
    return s[x][f];
}
int main(){
    scanf("%d", &n);
    printf("%d\n", dfs(n, 1));
    return 0;
}

 

再贴上我这份超时了的dij，我写的好差啊，只好说我刚刚入门最短路姿势还不够orz

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
#define CLR(a,b) memset((a),(b),sizeof((a)))
using namespace std;
#define inf 0x3f3f3f3f
const int N = 1000010;
int d[N], vis[N];
int prime[] = {2, 3, 5, 7, 11};
int n;
struct qnode{
    int v,c;
    qnode(int _v=0,int _c=0):v(_v),c(_c){}
    bool operator < (const qnode &r)const{
        return r.c <c;
    }
};
void dij(){
    priority_queue<qnode>q;
    for(int i = 1; i <= n+8; ++i){
        d[i] = inf;
        vis[i] = 0;
    }
    d[1]=0;
    q.push(qnode(1, 0));
    while(!q.empty()){
        qnode t=q.top(); q.pop();
        int u=t.v;
        if(vis[u])
            continue;
        vis[u]=1;
        for(int i = 0; i< 4 && u * prime[i] <= n+8; ++i){
            int v = u * prime[i];
            int w = prime[i];
            if(!vis[v] && d[u] + w < d[v]){
                d[v] = d[u] + w;
                q.push(qnode(v, d[v]));
            }
        }
        int v = u - 1;
        int w = 1;
        if(!vis[v] && d[u] + w < d[v]){
            d[v] = d[u] + w;
            q.push(qnode(v, d[v]));
        }
    }
}
int main(){
    scanf("%d", &n);
    dij();
    printf("%d\n", d[n]);
    return 0;
}