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 ;
}