(SWERC 2017)

A Cakey McCakeFace

O(N^2) 求出所有的差值即可，map 存下

#include<bits/stdc++.h>using namespace std;
typedef long long LL;
const int maxn = 2000+10;
LL a[maxn];
int main(void){map<LL,int> ma;int N,M;LL u;scanf("%d%d",&N,&M);for(int i =  1;i <= N; ++i){scanf("%lld",&a[i]);// ma[u]++;}for(int i =1;i <= M; ++i){scanf("%lld",&u);for(int i = 1;i <= N; ++i){if(u >= a[i]){ma[u-a[i]]++;}}// ma[u]++;}LL num = 0;LL Min = 0;for(auto c: ma){if(c.second > num){num = c.second;Min = c.first;}}cout<<Min<<endl;return 0;
}

C Macaron

题意：

用1∗2,1∗11*2,1*11∗2,1∗1 的矩阵覆盖n∗mn*mn∗m 的矩形的方法数, n<=8,M<=1e18n <= 8,M <= 1e18n<=8,M<=1e18

分析:

本题是 1033 骨牌覆盖 V2的扩展, 本题的状态比较麻烦，可以用dfs确定状态iii 之后可以转移的状态jjj，其中1代表覆盖，0代表未覆盖


// 矩阵快速幂
// 注意修改maxn 的值，要不然容易Tconst int maxn = 260;
int n;
struct Matrix{int n,m;Matrix(int nn = 1,int mm = 1):n(nn),m(mm){ memset(a,0,sizeof(a));};int  a[maxn][maxn];
};
void print(const Matrix &a)
{for(int i = 1;i <= a.n; ++i,cout<<endl)for(int j= 1;j <= a.m; ++j)cout<<a.a[i][j]<<" ";
}
Matrix operator*(Matrix a,Matrix b)
{Matrix c(a.n,b.m);for(int i = 1;i <= a.n; ++i){for(int j = 1;j <= b.m; ++j){for(int k = 1;k <= a.m; ++k){c.a[i][j] = (1ll*c.a[i][j]+1ll*a.a[i][k] * b.a[k][j])%mod;}}}
//  print(c);return c;
}
// 状态压缩LL MM[maxn][maxn];
LL N,M;
// a 代表是a的递推，now代表当前行的状态，nxt代表下一行的状态
void dfs(int  a,int  now,int  nxt){// cout<<a<<endl;int tmpnow = now,tmpnxt = nxt;int one[10],two[10];memset(one,0,sizeof(one));memset(two,0,sizeof(two));int cnt = 0;while(tmpnow > 0){one[cnt++] = tmpnow&1;tmpnow >>= 1;}bool flag = true;for(int i = 0;i < N; ++i){if(!one[i]){flag = false;break;}}if((now & NN) == NN){MM[a][nxt]++;return ;}cnt = 0;while(tmpnxt  > 0){two[cnt++] = tmpnxt&1;tmpnxt >>= 1;}for(int i = 0;i < N; ++i){if(!one[i]){dfs(a,now|(1<<i),nxt);dfs(a,now|(1<<i),nxt|(1<<i));if(i + 1 < N&& !one[i+1]){dfs(a,now|(1<<i)|(1<<(i+1)),nxt);}break;}}}
int NN;
Matrix ans(NN,NN);
Matrix B(NN,NN);
void solve(){B.n = B.m = ans.n = ans.m = NN;for(int i = 1;i <= NN; ++i){for(int j = 1;j <= NN; ++j){B.a[i][j] = MM[i-1][j-1];}}for(int i = 1;i <= NN; ++i) ans.a[i][i] = 1;while(M > 0){if(M & 1)ans = ans*B;B = B*B;M >>= 1;}cout<<ans.a[1][1]<<endl;
}
int main(void)
{scanf("%lld%lld",&N,&M);// cout<<N<<" "<<M<<endl;NN = 1<<N;// cout<<N<<" "<<NN<<endl;for(int i = 0;i < NN; ++i){dfs(i,i,0);}solve();return 0;
}

D Candy Chain

E Ingredients

题意

n种菜，每种菜由其它菜加一种食材组合而成，消耗 w,收获 v 点声望，不由其他菜组合而成的为基础菜，生产需要消耗 0 ，收获 0 点声望
每一种菜可能由多条方式合成，使用消耗最小的那一个
问：消耗最小，得到声望最多是多少

分析

跑最短路，然后背包，最短路建图的时候是单向边，可以自己假设一个超级源点


#include<bits/stdc++.h>using namespace std;
#define Pb   push_back
typedef long long LL;
const LL maxn = 1e6+100;
const LL maxm = 30+3;
// char ar[maxm];
// typedef pair<LL,LL> P;
char  dish[maxn];
LL C[maxn],V[maxn];
char ar[maxm],br[maxm];
bool vis[maxn];
LL F[maxn];
typedef pair<int,int> P;
struct Edge{int u,v,d;int val;Edge(int uu,int vv,int dd,int va):u(uu),v(vv),d(dd),val(va){}
};
struct Dijstra{#define maxn 123456#define INF  123456789int N,M,S,T;typedef pair<int,int> P;vector<Edge> edges;vector<int> G[maxn];bool done[maxn];int d[maxn];// int p[maxn];void init(){for(int i = 1;i <= N; ++i) G[i].clear();edges.clear();// scanf("%d %d %d %d",&N,&M,&S,&T);//  cout<<N<<M<<S<<T<<endl;// int u,v,w;// for(int i = 1;i <= M; ++i){//     scanf("%d %d %d",&u,&v,&w);//     AddEdge(u,v,w);//     AddEdge(v,u,w);// }}void AddEdge(int u,int v,int d,int val){edges.push_back(Edge(u,v,d,val));int m = edges.size();G[u].push_back(m-1);    }void solve(){priority_queue<P,vector<P>,greater<P>> Q;for(int i = 1;i <= N; ++i) d[i] = INF;// cout<<N<<endl;d[S] = 0;memset(done,0,sizeof(done));Q.push(P(0,S));while(!Q.empty()){P x = Q.top(); Q.pop();int u = x.second;// cout<<u<<endl;if(done[u]) continue;done[u] = true;for(int i = 0;i < (int)G[u].size(); ++i){Edge &e = edges[G[u][i]];if(!done[e.v]&&d[e.v] >= d[u]+e.d){if(d[e.v] > d[u]+e.d)d[e.v] = d[u] + e.d,V[e.v] = V[u]+e.val,Q.push(P(d[e.v],e.v));else    if(V[e.v] < V[u]+e.val)V[e.v] = V[u]+e.val,Q.push(P(d[e.v],e.v));}}}// printf("%d\n",d[T]);}
};
Dijstra Dij;
bool deg[maxn];
int  main(void){LL M,N;scanf("%lld%lld",&M,&N);map<string,LL> ma;Dij.init();LL cnt = 0;LL a,b;string s1,s2;for(LL i = 1;i <= N; ++i){scanf("%s",dish);scanf("%s%s",ar,br);scanf("%lld%lld",&a,&b);s1 = dish;s2 = ar;LL u = ma[s1];LL v = ma[s2];if(u == 0) ma[s1] =++cnt,u = cnt;if(v == 0)ma[s2] =++cnt,v = cnt;// cout<<"v "<<v<<" "<<u<<endl;Dij.AddEdge(v,u,a,b);// Dij.AddEdge(u,v,a,b);deg[u]++;        }for(LL i = 1;i <= cnt; ++i){// cout<<deg[i]<<" "<<endl;if(deg[i] == 0)Dij.AddEdge(cnt+1,i,0,0);}// cout<<cnt<<endl;Dij.S = cnt+1;Dij.N = cnt+1;Dij.solve();// cout<<cnt<<endl;// 背包for(int i = 1;i <= cnt; ++i)C[i] = Dij.d[i];for(LL i = 1;i <= cnt; ++i){// if(V[i] == 0) continue;for(LL  j = M;j >= C[i]; --j){F[j] = max(F[j],F[j-C[i]]+V[i]);}}// for(int i = 1;i <= cnt; ++i)//     cout<<C[i]<<" "<<V[i]<<endl;LL idx = 0;for(LL i = 0;i <= M; ++i){if(F[M] > F[idx]){idx = i;}}cout<<F[idx]<<endl;cout<<idx<<endl;return 0;
}

F Shattered Cake

面积不变

J Frosting on the Cake

#include<bits/stdc++.h>using namespace std;const int maxn = 1e5+10;
typedef long long LL;
LL A[maxn];
LL sum[4];
LL ans[4];
int main(void){int n;scanf("%d",&n);// LL tmp = 0;for(int i = 1;i <= n; ++i){scanf("%lld",&A[i]);// tmp += A[i];sum[(i-1)%3+1] += A[i];}// cout<<sum[1]<<" "<<sum[2]<<" "<<sum[3]<<endl;LL u,v;u = v = 0;LL ans[10] = {0};for(int i = 1;i <= n; ++i){scanf( "%lld",&u);for(int j = 1;j <= 3; ++j){ans[(i+j)%3] +=  sum[j]*u;}v = u;}cout<<ans[0]<<" "<<ans[1]<<" " <<ans[2]<<endl;return 0;
}

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(SWERC 2017)

文章目录

(SWERC 2017)

A Cakey McCakeFace

C Macaron

题意：

分析:

D Candy Chain

E Ingredients

题意

分析

F Shattered Cake

J Frosting on the Cake

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