标签：style blog http color io os for sp div
TYVJ1125 JR‘s chop
http://www.tyvj.cn/Problem_Show.aspx?id=1125
先将每根筷子按长度排序
设dp[i][j][0]为前i个筷子取j对且不取第i个的最小解
dp[i][j][1]为前i个筷子取j对且取第i个的最小解
dp[i][j][0] = min(dp[]i-1[j][0],dp[i-1][j][1])
dp[i][j][1] = min(dp[k][j-1][0],dp[k][j-1][1])+b[i][k+1] | (j-1)*2<=k<=i-2
b[i][k] 为第i个和第k个筷子的平方差的和
1 #include <cstdio> 2 #include <cstring> 3 #include <cstdlib> 4 #include <iostream> 5 #include <algorithm> 6 using namespace std; 7 const int maxn = 105; 8 const int maxv = 1005; 9 const int INF = 0x3f3f3f3f; 10 const int mod = 1e9+7; 11 typedef long long LL; 12 int a[maxn],dp[maxn][maxn][2]; 13 int b[maxn][maxn]; 14 int main() 15 { 16 // freopen("in.txt","r",stdin); 17 int n,m; 18 while(cin>>n>>m) 19 { 20 m+=3; 21 if(n<m*2){ 22 printf("-1\n"); 23 break; 24 } 25 memset(dp,0x3f,sizeof(dp)); 26 memset(b,0,sizeof(b)); 27 for(int i = 1;i<=n;++i)scanf("%d",a+i); 28 sort(a+1,a+1+n); 29 for(int i = 1;i<=n;++i)for(int j = 1;j<=i;++j) 30 b[i][j] = (a[i]-a[j])*(a[i]-a[j]); 31 dp[2][0][1] = 0; 32 dp[2][1][1] = b[2][1]; 33 for(int i = 3;i<=n;++i) 34 for(int j = 1;j<=i/2;++j) 35 { 36 dp[i][j][0] = min(dp[i-1][j][0],dp[i-1][j][1]); 37 for(int k = (j-1)*2;k<=i-2;k++) 38 { 39 int t = min(dp[k][j-1][1],dp[k][j-1][0])+b[i][k+1]; 40 dp[i][j][1] = min(t,dp[i][j][1]); 41 } 42 } 43 printf("%d\n",min(dp[n][m][0],dp[n][m][1])); 44 } 45 return 0; 46 }
TYVJ 1213 嵌套矩形
http://www.tyvj.cn/Problem_Show.aspx?id=1213
水题
1 #include <cstdio> 2 #include <cstring> 3 #include <cstdlib> 4 #include <iostream> 5 #include <algorithm> 6 using namespace std; 7 const int maxn = 2005; 8 const int maxv = 1005; 9 const int INF = 0x3f3f3f3f; 10 const int mod = 1e9+7; 11 typedef long long LL; 12 struct node 13 { 14 int x; 15 int y; 16 }a[maxn]; 17 int cmp(const void *a,const void *b) 18 { 19 struct node *aa = (struct node *)a; 20 struct node *bb = (struct node *)b; 21 if(aa->x==bb->x)return aa->y-bb->y; 22 return aa->x-bb->x; 23 } 24 int dp[maxn]; 25 int main() 26 { 27 // freopen("in.txt","r",stdin); 28 int n;scanf("%d",&n); 29 for(int i = 1;i<=n;++i) 30 { 31 int x,y; 32 scanf("%d%d",&x,&y); 33 if(x>y)swap(x,y); 34 a[i].x = x; 35 a[i].y = y; 36 } 37 qsort(a+1,n,sizeof(a[1]),cmp); 38 39 for(int i = 1;i<=n;++i) 40 { 41 dp[i] = 1; 42 for(int j = 1;j<i;++j)if(a[i].x>a[j].x && a[i].y>a[j].y) 43 dp[i] = max(dp[i],dp[j]+1); 44 } 45 int ans = 0; 46 for(int i = 1;i<=n;++i) 47 ans = max(ans,dp[i]); 48 cout<<ans<<endl; 49 return 0; 50 }
TYVJ1095 美元
http://www.tyvj.cn/Problem_Show.aspx?id=1095
水题
设dp[i][0]表示第i天以美元为单位最多的钱
dp[i][1]表示第i天以马克为单位最多的钱
1 #include <cstdio> 2 #include <cstring> 3 #include <cstdlib> 4 #include <iostream> 5 #include <algorithm> 6 using namespace std; 7 const int maxn = 2005; 8 const int maxv = 1005; 9 const int INF = 0x3f3f3f3f; 10 const int mod = 1e9+7; 11 typedef long long LL; 12 double dp[maxn][2]; 13 int main() 14 { 15 // freopen("in.txt","r",stdin); 16 int n;cin>>n; 17 for(int i = 1;i<=n;++i) 18 { 19 int t;scanf("%d",&t); 20 if(i==1) 21 { 22 dp[i][0] = 100; 23 dp[i][1] = t*1.0; 24 continue; 25 } 26 dp[i][0] = max(dp[i-1][0],dp[i-1][1]/t*100); 27 dp[i][1] = max(dp[i-1][1],dp[i-1][0]/100*t); 28 } 29 printf("%.2lf\n",dp[n][0]); 30 return 0; 31 }
标签：style blog http color io os for sp div
原文地址：http://www.cnblogs.com/GJKACAC/p/4044298.html
