130725hdu杭电多校第二场解题报告

1002 Warm up

这题不难，关键是怎么去缩点和扩栈。

#include <cstdio>#include <cstring>#include <vector>#include<queue>#include<cmath>#include <algorithm>using namespace std;#define MAXN 200006#define MAXM 2000006#pragma comment(linker, "/STACK:1024000000,1024000000")struct node {    int v, w, pre;} edge[MAXM];int pos[MAXN], nEdge; //图的存储：链式前向星（池子法）struct Bridge {    int u, v;} bridge[MAXM];  //用来记录桥int tot; //桥的个数int fa[MAXN], cc; //fa：各个点所属的缩点（连通块），cc连通块的个数int dfn[MAXN], low[MAXN], Time; //时间戳int stack[MAXN], top;   //用于维护连通块的int n, m;   //点的个数和边的条数void connect(int u, int v, int w) {    nEdge++;    edge[nEdge].pre = pos[u];    edge[nEdge].v = v;    edge[nEdge].w = w;    pos[u] = nEdge;}vector<int> graph[MAXN];void tarjan(int cur, int from, int from_edge) {    low[cur] = dfn[cur] = Time++;    stack[++top] = cur;    for (int p=pos[cur]; p; p=edge[p].pre) {        int v = edge[p].v;        if (v == from&&abs(p-from_edge)<=1) continue;  //注意一下这里        if (!dfn[v]) {            tarjan(v, cur, p);            if (low[v] < low[cur]) low[cur] = low[v];            if (low[v] > dfn[cur]) {                bridge[tot].u = cur;                bridge[tot++].v = v;                cc++;                graph[cc].clear();                do {                    fa[stack[top]] = cc;                } while (stack[top--] != v);            }        } else if (low[cur] > dfn[v]) low[cur] = dfn[v];    }}int bfs(int x,int flag){    queue<int> q;    int y;    q.push(x);    memset(dfn,0,sizeof(dfn));    memset(low,0,sizeof(low));    low[x]=0;    dfn[x]=1;    while(q.size()>0)    {        y=q.front();   //     printf("%d\n",y);        q.pop();        for(int i=0;i<graph[y].size();i++)        {            int z=graph[y][i];            if(!dfn[z])            {                dfn[z]=1;                low[z]=low[y]+1;                q.push(z);            }        }    }    if(flag)    return y;    return low[y];}int main() {    while(scanf("%d%d", &n, &m),n+m)    {    memset(pos, 0, sizeof(pos));    nEdge = 0;    int u, v, w;    for (int i=0; i<m; i++) {        scanf("%d%d", &u, &v);        connect(u, v, 1);        connect(v, u, 1);    }    memset(dfn, 0, sizeof(dfn));    memset(fa, -1, sizeof(fa));    cc = tot = 0;    for (int i=1; i<=n; i++)        if (!dfn[i]) {            top = Time = 1;            tarjan(i, -1, -3);            ++cc;            graph[cc].clear();            for (int j=1; j<=n; j++)                if (dfn[j] && fa[j]==-1) fa[j] = cc;        }        for(int i=0;i<tot;i++)        {            graph[fa[bridge[i].u]].push_back(fa[bridge[i].v]);            graph[fa[bridge[i].v]].push_back(fa[bridge[i].u]);        }        int y=bfs(1,1);        int ans2=bfs(y,0);      //  printf("%d %d\n",cc,ans2);        printf("%d\n",cc-1-(ans2));    }    return 0;}

1008 Palindrome Sub-Array

求n*m的矩阵中的回文方阵，由于写了四个for，所以很担心会超时，所以加了l判断长度来压缩时间，最后竟然压到了100ms，一种很开心的感觉。

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <string>#include <algorithm>#include <vector>#include <queue>#include <map>using namespace std;int s[305][305];int n,m;int l;int ans;int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        for(int i=0; i<n; i++)        {            for(int j=0; j<m; j++)            {                scanf("%d",&s[i][j]);            }        }        ans = 1;        for(int i=0; i<n; i++)        {            for(int j=0; j<m; j++)            {                if(min(n-i,m-j) <= ans)                {                    break;                }                l = min(n-i,m-j);                while(l > ans)//判断是否是回文方阵                {                    if(s[i][j]==s[i+l-1][j]&&s[i+l-1][j]==s[i][j+l-1]&&s[i][j+l-1]==s[i+l-1][j+l-1])                    {                        int q = 0;                        for(int k = 1; k * 2 < l; k++)                        {                            if(s[i+k][j]==s[i+l-1-k][j]&&s[i+l-1-k][j]==s[i+k][j+l-1]&&s[i+k][j+l-1]==s[i+l-1-k][j+l-1])                            {                                if(s[i][j+k]==s[i+l-1][j+k]&&s[i+l-1][j+k]==s[i][j+l-1-k]&&s[i][j+l-1-k]==s[i+l-1][j+l-1-k])                                {                                    if(s[i+k][j+k]==s[i+l-1-k][j+k]&&s[i+l-1-k][j+k]==s[i+k][j+l-1-k]&&s[i+k][j+l-1-k]==s[i+l-1-k][j+l-1-k])                                    {                                        continue;                                    }                                    else                                    {                                        q = 1;                                        break;                                    }                                }                                else                                {                                    q = 1;                                    break;                                }                            }                            else                            {                                q = 1;                                break;                            }                            for(int c = k - 1; c > 0; c--)                            {                                if(s[i+k][j+c]==s[i+l-1-k][j+c]&&s[i+l-1-k][j+c]==s[i+k][j+l-1-c]&&s[i+k][j+l-1-c]==s[i+l-1-k][j+l-1-c])                                {                                    if(s[i+c][j+k]==s[i+l-1-c][j+k]&&s[i+l-1-c][j+k]==s[i+c][j+l-1-k]&&s[i+c][j+l-1-k]==s[i+l-1-c][j+l-1-k])                                    {                                        continue;                                    }                                    else                                    {                                        q = 1;                                        break;                                    }                                }                                else                                {                                    q = 1;                                    break;                                }                            }                            if(q == 1)                            {                                break;                            }                        }                        if(q == 0)                        {                            ans = l;                            //cout<<i<<' '<<j<<' '<<l<<'c'<<endl;                        }                    }                    l--;                }            }        }        printf("%d\n",ans);    }    return 0;}

1009 Warm up 2

本场最水的一道题，不多说了。

#include <iostream>#include <cstdio>#include <algorithm>#include <string>#include <cmath>#include <cstring>#include <queue>#include <set>#include <vector>#include <stack>#include <map>#include <iomanip>#define PI acos(-1.0)#define Max 2505#define inf 1<<28#define LL(x) ( x << 1 )#define RR(x) ( x << 1 | 1 )#define REP(i,s,t) for( int i = ( s ) ; i <= ( t ) ; ++ i )#define ll long long#define mem(a,b) memset(a,b,sizeof(a))#define mp(a,b) make_pair(a,b)#define PII pair<int,int>using namespace std;inline void RD(int &ret) {    char c;    do {        c = getchar();    } while(c < '0' || c > '9') ;    ret = c - '0';    while((c=getchar()) >= '0' && c <= '9')        ret = ret * 10 + ( c - '0' );}int Map[1111][1111] ;int link[1111] ;bool vis[1111] ;int n , m ,s, v;int dfs(int now){    for (int i = 1 ;i <= m ;i ++)    {        if(Map[now][i])        if(!vis[i])        {            vis[i] = 1 ;            if(link[i] == -1 || dfs(link[i]))            {                link[i] = now ;                return 1 ;            }        }    }    return 0 ;}int x1[1111], y1[1111] ;int x2[11111],y2[1111] ;bool check(int i , int j){    if(x1[i] <= x2[j] && x2[j] <= x1[i] + 1){        if(y2[j] <= y1[i] && y2[j] + 1 >= y1[i]){            return 1 ;        }    }    return 0 ;}int main() {    while(scanf("%d%d",&n,&m) , (n +  m)){        mem(Map,0) ;        mem(link , -1) ;        for (int i = 1 ; i <= n ;i ++ ){            scanf("%d%d",&x1[i] ,&y1[i]) ;        }        for (int i = 1 ; i <= m ;i ++ ){            scanf("%d%d",&x2[i] , &y2[i]) ;        }        for (int i = 1 ; i <= n ; i ++ ){            for (int j = 1 ; j <= m ;j ++ ){                Map[i][j] = check(i , j ) ;            }        }        int ans = 0 ;        for (int i = 1 ; i <= n ;i ++){            mem(vis, 0 ) ;            ans += dfs(i) ;        }        cout << n + m - ans << endl;    }    return 0 ;}