平时二十五测

今天翻盘了,没有上两次的傻逼错误;

 

题解:

其实O(N)check不会T,不是严格的,而且1e6里面的约数很少,从重心跑很快;

#include<bits/stdc++.h>
using namespace std;


#define ex(i, u) for(int i = h[u]; i; i = G[i].nxt)
const int M = 1e6 + 5;
int h[M], fg ,tot, siz[M], f[M], tmp[M], n, root, vis[M];
struct edge{int v, nxt;}G[M<<1];
void add(int u, int v){G[++tot].v = v, G[tot].nxt = h[u], h[u] = tot;}

void getroot(int u, int fa){
    siz[u] = 1;
    ex(i, u){
        int v = G[i].v;
        if(v == fa) continue;
        getroot(v, u);
        siz[u] += siz[v];
        f[u] = max(f[u], siz[v]);
    }
    f[u] = max(f[u], n - siz[u]); 
    if(f[u] < f[root]) root = u;
}

int dfs(int u, int fa, int k){
    if(siz[u] == k) return 0;
    tmp[u] = 1;
    
    ex(i, u){
        if(!fg) return 0; 
        int v = G[i].v;
        if(v == fa)continue;
        if(siz[v] < k) tmp[u] += siz[v];
        else {
            tmp[u] += dfs(v, u, k);
        }
    }
    if(tmp[u] > k) fg = 0;
    if(tmp[u] == k) return 0;
    return tmp[u];
}

int read(){
    int x=0,f=1;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}


int main(){
    freopen("count.in","r",stdin);
    freopen("count.out","w",stdout);
    int cnt = 0;
    n = read();
    for(int i = 1; i < n; i++){
        int u = read(), v = read();
        add(u, v), add(v, u);
    }
    f[root] = 2e9; 
    getroot(1, 0);
    for(int i = 2; i * i <= n; i++){
        if(n % i == 0){
            fg = 1; 
            dfs(root, 0, i);
            cnt += fg;    
            if(i * i != n){
                fg = 1; dfs(root, 0, n/i);
                cnt += fg;    
    
            }
        }
    }
    printf("%d\n",cnt+2);
}
View Code

 

这道题还有一个巧妙的剪枝,不用O(N)枚举起点,对于一个点,他在传递过程的位置是1——最后,我们把1当起点,做一遍不行,就把他当第二个,当他作为最后一个后就可以停了,实测跑得飞快(这样不倍增都有60分)

 

#include<bits/stdc++.h>
using namespace std;

const int M = 1e5 + 5;

int n, a[M], m, s[M];


bool check(int k){
    int up = n, tot = 0, lst = 0;
    while(1){
        int cnt = 1;
        for(int i = 1; i <= up; i++){
            if(tot + a[i] <= k) tot += a[i];
            else tot = a[i], cnt++;
        }
        if(cnt <= m) return 1;
        tot = lst + a[up]; lst += a[up]; up--; 
        if(tot + a[1] > k) return 0;
    }    
}
int read(){
    int x=0,f=1;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}
bool check2(int k){

    int up = n, tot = 0, lst = 0;
    while(1){
        int cnt = 1;
        for(int i = 1; i <= up; ){
            int lf = i, rg = up, rec = -1;
            while(lf <= rg){
                int mid = (lf + rg) >> 1;
                if(s[mid] - s[i - 1] + tot <= k) rec = mid, lf = mid + 1;
                else rg = mid - 1;
            }
            tot = 0;
            i = rec + 1; 
            if(i == up + 1)break;
            cnt++;
        }
        if(cnt <= m) return 1;
        tot = lst + a[up]; lst += a[up]; up--; 
        if(tot + a[1] > k) return 0;
    }
}

int main(){
    freopen("dinner.in","r",stdin);
    freopen("dinner.out","w",stdout);
    n = read(); m = read();
    int ans = -1, Mx = 0, sum = 0;
    for(int i = 1; i <= n; i ++) {
        a[i] = read(), sum += a[i], Mx = max(Mx, a[i]);
        s[i] = s[i - 1] + a[i];
    }
    int lf = Mx, rg = sum;
    if(n <= 5000){
        while(lf <= rg){
            int mid = (lf + rg) >> 1;
            if(check(mid)) ans = mid, rg = mid - 1;
            else lf = mid + 1;
        }    
    }
    else {
        while(lf <= rg){
            int mid = (lf + rg) >> 1;
            if(check2(mid)) ans = mid, rg = mid - 1;
            else lf = mid + 1;
        }
    }
    printf("%d\n", ans);
    
}
View Code

 

一句话:最短路+缩边

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int M = 55, N = 3000;
int a[M][M], tot, ans, inf = 2e9, cnt, n, m, tmp[N], dis[N], h[N], stx, sty, edx, edy;
bool vis[M][M], inq[N], vis2[M][M], has[N][N];
ll fx[N];
int zl[8][2] = {{1, 2}, {1, -2}, {-1, 2}, {-1, -2}, {2, 1}, {2, -1}, {-2, 1}, {-2, -1}};
struct edge{int v,u, nxt;}G[100005];
void add(int u, int v){
    G[++tot].v = v,G[tot].u=u, G[tot].nxt = h[u], h[u] = tot;
}
inline int id(int x,int y){return x * m + y;}
inline bool check(int x, int y){
    if(x > 0 && x <= n && y > 0 && y <= m && a[x][y] != 2) return 1;
    return 0;
}
void dfs(int x, int y){
    for(int i = 0; i < 8; i++){
        int nx = x + zl[i][0], ny = y + zl[i][1];
        if(!check(nx, ny)) continue;
        if(a[nx][ny]==0 && !vis2[nx][ny]){
            vis2[nx][ny] = 1;
            tmp[++cnt] = id(nx, ny);
        }
        else if(a[nx][ny]==1 && !vis[nx][ny]){
            vis[nx][ny] = 1;
            dfs(nx, ny);
        }
    }
    
}

bool SPFA(){
    queue <int> q;
    memset(dis, 127, sizeof(dis));
    int S = id(stx, sty), T = id(edx, edy);
    dis[S] = 0, inq[S] = 1, fx[S] = 1;
    q.push(S);
    while(!q.empty()){
        int u=q.front();q.pop();inq[u]=0;
        for(int i = h[u]; i; i = G[i].nxt){
            int v = G[i].v;
            if(dis[v] > dis[u] + 1){
                dis[v] = dis[u] + 1;
                fx[v] = fx[u];
                //printf("%d %d %d\n", u, v, dis[v]);
                if(!inq[v] && v != T){
                    q.push(v), inq[v] = 1;
                }
            }
            else if(dis[v] == dis[u] + 1){
                fx[v] += fx[u];
                if(!inq[v] && v != T){
                    q.push(v), inq[v] = 1;
                }
            }
        }
    }
    return 0;
}


int read(){
    int x=0,f=1;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}

//0表示格子为空,1表示格子上有敌军,2表示格子上有友军,3表示马的位置,4表示敌方的帅。
int main(){
    freopen("chess.in","r",stdin);
    freopen("chess.out","w",stdout);
    n = read(), m = read();
    for(int i = 1; i <= n; i++) 
        for(int j = 1; j <= m; j++){    
            a[i][j] = read();
            if(a[i][j] == 3) stx = i, sty = j, a[i][j] = 0;
            if(a[i][j] == 4) edx = i, edy = j, a[i][j] = 0;    
        }
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++)
            if(!a[i][j]){
                for(int k = 0; k < 8; k++){
                    int nx = i + zl[k][0], ny = j + zl[k][1];
                    if(check(nx, ny) && !a[nx][ny])
                        add(id(i, j), id(nx, ny)); 
                }
            }    
        
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++)
        if(!vis[i][j] && a[i][j] == 1){
            memset(vis2, 0, sizeof(vis2));
            cnt = 0;
            vis[i][j] = 1;
            dfs(i, j);
            for(int k = 1; k <= cnt; k++)
                for(int z = k + 1; z <= cnt; z++){
                    if(has[tmp[k]][tmp[z]]) continue;
                    has[tmp[k]][tmp[z]] = has[tmp[z]][tmp[k]] = 1;
                    add(tmp[k], tmp[z]); add(tmp[z], tmp[k]);
                }
        }
    /*for(int i = 1; i <= tot; i+=2){
        printf("%d %d\n", G[i].u, G[i].v);
    }*/
    SPFA();
    dis[id(edx, edy)] < inf ? printf("%d\n%lld\n", dis[id(edx, edy)] - 1, fx[id(edx, edy)]) : printf("-1\n");
}
View Code

 

posted @ 2018-11-06 18:15  Ed_Sheeran  阅读(51)  评论(0编辑  收藏