版权声明：本博客的博文版权为2018韦泽鑫所有，复制文章转载时记得附上本博客的地址哦 https://blog.csdn.net/qq_40167327/article/details/85171095
T1
这道题bfs（按[连通器原理](https://baike.baidu.com/item/%E8%BF%9E%E9%80%9A%E5%99%A8/1534580)），注意当h=n+1时要搜上面的边
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int a[1111][1111],q1[1000002],q2[1000002];
int walk1[5] = {0,1,0,-1,0};
int walk2[5] = {0,0,1,0,-1};
int n,m,h;
void bfs(int x,int y)
{
int head = 1;
int tail = 2;
q1[1] = x;
q2[1] = y;
a[x][y] = 2;
while (head < tail)
{
for (int i=1;i<=4;i++)
{
int tx = q1[head] + walk1[i];
int ty = q2[head] + walk2[i];
if(tx >= 1 && ty >= 1 && tx<=n && ty<=m && !a[tx][ty] && tx>=n-h+1)
{
q1[tail] = tx;
q2[tail++] = ty;
a[tx][ty] = 2;
}
}
head++;
}
}
int main()
{
freopen("cruise.in","r",stdin);
freopen("cruise.out","w",stdout);
cin>>n>>m>>h;
if(h > n) h = n;
memset(a,0,sizeof(a));
for (int i=1;i<=n;i++)
for (int j=1;j<=m;j++)
{
char s;
cin>>s;
if(s == '1') a[i][j] = 1;
}
int x = 0;
int y = 0;
for (int i=n;i>=n-h+1;i--)
for (int j=1;j<=m;j++)
if(!a[i][j])
{
if(i > 1 && j > 1 && i < n && j < m) continue;
x = i;
y = j;
bfs(x,y);
}
for (int i=1;i<=n;i++)
{
for (int j=1;j<=m;j++)
cout<<a[i][j];
cout<<endl;
}
fclose(stdin);
fclose(stdout);
return 0;
}T2
这道题O（n*t），只需要遇到折点就记录就行，注意前后两个数相等的情况
#include<cstdio>
using namespace std;
int a[250002];
int main()
{
freopen("rose.in","r",stdin);
freopen("rose.out","w",stdout);
int t,n;
scanf("%d%d",&t,&n);
for (int i=1;i<=t;i++)
{
bool p = true;
for (int j=1;j<=n;j++)
scanf("%d",&a[j]);
int l=1;
int w=1,x=0,y=0,z=0;
for (int j=2;j<=n;j++)
if(l % 2 == 0 && a[j]>a[j-1])
{
l++;
if(l == 3) y = j-1;
}
else if(l % 2 == 1 && a[j]<a[j-1])
{
l++;
if(l == 2) x = j-1;
if(l == 4) z = j-1;
}
else if(a[j] == a[j-1])
{
p = false;
break;
}
if(p)
{
if(l == 4 && w != x && x != y && y !=z) printf("TAK\n");
else printf("NIE\n");
}
else printf("NIE\n");
}
fclose(stdin);
fclose(stdout);
return 0;
}T3
这道题Floyd，暴力算出不删点和删2~n-1的点的各个最短路，判断即可
#include<cmath>
#include<cstdio>
long long int n,m,p[101];
long double map[101][101],chu[101][101],cha[101][101];
int main()
{
freopen("fool.in","r",stdin);
freopen("fool.out","w",stdout);
scanf("%lld%lld",&n,&m);
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
chu[i][j]=-217483647.00;cha[i][j]=chu[i][j];map[i][j]=chu[i][j];
}
}
for(int i=1;i<=n;i++)
{
scanf("%LF",&map[i][i]);map[i][i]=1-map[i][i]*0.01;chu[i][i]=map[i][i];cha[i][i]=map[i][i];
}
for(int i=1;i<=m;i++)
{
long long int x,y;
scanf("%lld%lld",&x,&y);
map[x][y]=map[y][y];chu[x][y]=map[x][y];cha[x][y]=map[x][y];
map[y][x]=map[x][x];chu[y][x]=map[y][x];cha[y][x]=map[y][x];
}
p[1]=1;p[n]=1;
for(int k=1;k<=n;k++)
{
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
if(i!=j&&i!=k&&j!=k&&chu[i][k]*chu[k][j]>chu[i][j]&&chu[i][k]>=0&&chu[k][j]>=0)
{
chu[i][j]=chu[i][k]*chu[k][j];
}
}
}
}
long long int xc=chu[1][n]*1000000;
for(int a=2;a<n;a++)
{
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
map[i][j]=cha[i][j];
}
}
for(int i=1;i<=n;i++)map[a][i]=-10;
for(int i=1;i<=n;i++)map[i][a]=-10;
for(int k=1;k<=n;k++)
{
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
if(i!=j&&i!=k&&j!=k&&map[i][k]*map[k][j]>map[i][j]&&map[i][k]>=0&&map[k][j]>=0)
{
map[i][j]=map[i][k]*map[k][j];
}
}
}
}
long long int cx=map[1][n]*1000000;
if(xc>cx)
{
p[a]=1;
}
}
for(int i=1;i<=n;i++)printf("%lld\n",p[i]);
return 0;
}T4
这道题动态转移方程，f[i][j]=max(f[i][j-1],f[i-1][j])+a[i][j]即可
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#define mod 1000000007
using namespace std;
long long int f[2001][2001],n,m,a[2001][2001],zyf[2001][2001];
int main()
{
freopen("chess.in","r",stdin);
freopen("chess.out","w",stdout);
scanf("%lld%lld",&n,&m);
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
scanf("%lld",&a[i][j]);f[i][j]=-2147483647;
}
}
zyf[1][1]=1;
f[1][1]=a[1][1];
for(int i=2;i<=m;i++)
{
f[1][i]=a[1][i]+f[1][i-1];zyf[1][i]=1;
}
for(int i=2;i<=n;i++)
{
f[i][1]=a[i][1]+f[i-1][1];zyf[i][1]=1;
}
for(int i=2;i<=n;i++)
{
for(int j=2;j<=m;j++)
{
f[i][j]=max(f[i][j-1],f[i-1][j])+a[i][j];
if(f[i][j-1]>f[i-1][j])
{
zyf[i][j]=zyf[i][j-1]%mod;
}
if(f[i][j-1]<f[i-1][j])
{
zyf[i][j]=zyf[i-1][j]%mod;
}
if(f[i][j-1]==f[i-1][j])
{
zyf[i][j]=(zyf[i-1][j]+zyf[i][j-1])%mod;
}
}
}
printf("%lld\n%lld",f[n][m],zyf[n][m]);
return 0;
}