Usaco 2006 November gold

Poj 3253 fence repair STL heap operation

I would like to say that stacking in STL is the most painful operation I have seen currently.

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <cstdlib>#include <cmath>#include <utility>#include <vector>#include <queue>#include <map>#include <set>#define max(x,y) ((x)>(y)?(x):(y))#define min(x,y) ((x)>(y)?(y):(x))using namespace std;int n,a[20005];int main(){    scanf("%d",&n);    for(int i=0; i<n; i++)        scanf("%d",&a[i]);    vector< int > v(a,a+n);    make_heap(v.begin(),v.end(),greater< int >());    long long ans=0;    for(int i=1; i<=n-1; i++)    {        long long tmp=v.front();        pop_heap(v.begin(),v.end(),greater< int >());        v.pop_back();        tmp+=v.front();        pop_heap(v.begin(), v.end(),greater< int >());        v.pop_back();        v.push_back(tmp);        push_heap(v.begin(),v.end(),greater< int >());        ans+=tmp;        printf("%d\n", tmp);    }    printf("%lld\n",ans);    return 0;}

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Poj 3254 corn fields state compression recurrence

UC [] indicates the case where the adjacent two are not at the same time 1, 377, so N * 377 ^ 2

Note that bit operation Priority is lower than =!

#include <cstdio>#include <cstring>#define MOD 100000000int n,m,a[20],k;int uc[500],all;int f[20][5000];int main(){    scanf("%d%d",&n,&m);    for(int i=0; i<n; i++)        for(int j=0; j<m; j++)        {            scanf("%d",&k);            a[i]=(a[i]<<1)+k;        }    for(int i=0; i<(1<<m); i++)    {        int tmi=i,flag=0;        while(tmi)        {            if((tmi&3)==3) flag=1;            tmi>>=1;        }        if(!flag)        {            uc[all++]=i;            if((i|a[0])==a[0])                f[0][i]=1;        }    }    for(int i=1; i<n; i++)        for(int j=0; j<all; j++)        if((a[i]|uc[j])==a[i])        {            for(int k=0; k<all; k++)            if(!(uc[j]&uc[k]))            {                f[i][uc[j]]=(f[i][uc[j]]+f[i-1][uc[k]])%MOD;            }        }    int ans=0;    for(int j=0; j<all; j++)        ans=(ans+f[n-1][uc[j]])%MOD;    printf("%d\n", ans);    return 0;}

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Poj 3255 A * Algorithm for K Short Circuit

I got to know about a * for the first time.

----------------- Wikipedia ----------------------------

If G (n) represents the actual distance from the starting point to any vertex N, and H (n) Represents the estimated distance from any vertex N to the target vertex, then the formula of the * algorithm is: F (n) = g (n) + H (n ). This formula follows the following features:

• If H (n) is 0, you only need to find G (N), that is, find the shortest path from the starting point to any vertex N, then convert it into a single-source shortest path problem, that is, Dijkstra Algorithm
• If H (n) <= "n to the actual distance from the target", the optimal solution can be obtained. In addition, the smaller the H (n), the more nodes to be calculated, and the lower the algorithm efficiency.

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#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <cstdlib>#include <cmath>#include <utility>#include <vector>#include <queue>#include <map>#include <set>#define max(x,y) ((x)>(y)?(x):(y))#define min(x,y) ((x)>(y)?(y):(x))#define INF 0x3f3f3f3fusing namespace std;struct Edge{    int y,w,ne;}e[200005];int x,y,w,n,m;int be[5005],all;int h[5005];bool vis[5005];struct Point{    int x,g;    bool operator < (const Point T) const    {        return g+h[x]>T.g+h[T.x];    }};void add(int x, int y, int w){    e[all].y=y;    e[all].w=w;    e[all].ne=be[x];    be[x]=all++;}void SPFA(int s){    queue< int > q;    for(int i=0; i<=n; i++)    {        h[i]=INF;        vis[i]=0;    }    h[s]=0;    vis[s]=1;    while(!q.empty())        q.pop();    q.push(s);    while(!q.empty())    {        int u=q.front();        q.pop();        vis[u]=0;        for(int i=be[u]; i!=-1; i=e[i].ne)        {            int v=e[i].y;            if(h[v]>h[u]+e[i].w)            {                h[v]=h[u]+e[i].w;                if(!vis[v])                {                    vis[v]=1;                    q.push(v);                }            }        }    }}int Astar(int s, int t, int k){    SPFA(t);    priority_queue< Point > q;    while(!q.empty())        q.pop();    memset(vis,0,sizeof(vis));    Point cur,nxt;    cur.x=s;    cur.g=0;    q.push(cur);    while(!q.empty())    {        cur=q.top();        q.pop();        if(cur.x==t && !--k)            return cur.g;        for(int i=be[cur.x]; i!=-1; i=e[i].ne)        {            nxt.x=e[i].y;            nxt.g=cur.g+e[i].w;            q.push(nxt);        }    }    return -1;}void init(){    all=0;    memset(be,-1,sizeof(be));}int main(){    scanf("%d%d",&n,&m);    init();    for(int i=0; i<m; i++)    {        scanf("%d%d%d",&x,&y,&w);        add(x,y,w);        add(y,x,w);    }    printf("%d\n",Astar(1,n,2));    return 0;}

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Usaco 2006 November gold