Merge pull request #23 from KiritoTRw/master

ACM Templates Added

.ACM-Templates/LCIS.cpp

@@ -0,0 +1,29 @@
+/// LCIS 最长公共上升子序列  
+namespace LCIS  
+{  
+  
+const int MAXLEN_A = 500;  
+const int MAXLEN_B = 500;  
+int dp[MAXLEN_A+5][MAXLEN_B+5];  
+int deal(const char* a,const char* b)  
+{  
+    int lena=strlen(a);  
+    int lenb=strlen(b);  
+    for(int i=1;i<=lenb;i++)  
+    {  
+        int k=0;  
+        for(int j=1;j<=lena;j++)  
+        {  
+            dp[i][j]=dp[i-1][j];/// when b[i-1] != a[j-1]  
+            if(b[i-1]>a[j-1]) k=max(k,dp[i-1][j]);  
+            else if(b[i-1]==a[j-1]) dp[i][j]=k+1;  
+        }  
+    }  
+    int ans=0;  
+    for(int i=1;i<=lena;i++) ans=max(ans,dp[lenb][i]);  
+    return ans;  
+}  
+  
+}  
+//End of namespace LCIS  
+

.ACM-Templates/LCS.cpp

@@ -0,0 +1,30 @@
+/// LCS 最长子序列  
+namespace LCS  
+{  
+const int MAXLEN_A = 500;  
+const int MAXLEN_B = 500;  
+int dp[MAXLEN_A+5][MAXLEN_B+5];  
+int deal(const char* a,const char* b)  
+{  
+    int lena=strlen(a);  
+    int lenb=strlen(b);  
+    for(int i=0;i<lenb;i++)  
+    {  
+        for(int j=0;j<lena;j++)  
+        {  
+            if(i==0) dp[i][j]=0;  
+            else if(b[i]==a[j])  
+            {  
+                if(j!=0) dp[i][j]=dp[i-1][j-1]+1;  
+                else dp[i][j]=1;  
+            }  
+            else  
+            {  
+                if(j!=0) dp[i][j]=max(dp[i-1][j],dp[i][j-1]);  
+                else dp[i][j]=dp[i-1][j];  
+            }  
+        }  
+    }  
+    return dp[lenb-1][lena-1];  
+}  
+}//End of namespace LCS  

.ACM-Templates/LIS.cpp

@@ -0,0 +1,30 @@
+#include <stdio.h>  
+#define  MAXN 1010  
+int seq[MAXN];  
+int seqlen[MAXN];  
+int main()  
+{  
+    int N;  
+    while(scanf("%d",&N)==1)  
+    {  
+        int i,j,k,max,maxlen=1;  
+        for(i=1; i<=N; i++)  
+            seqlen[i]=1;               //seqlen数组存以第i个数为终点的最长上升序列  
+        for(i=1; i<=N; i++)  
+            scanf("%d",&seq[i]);       //seq数组保存序列数组  
+        for (i=2; i<=N; i++)  
+        {  
+            max=0;  
+            for (j=1; j<=i-1; j++)  
+            {  
+                if(seq[j]<seq[i]&&seqlen[j]>max)  //在前i-1个序列中，寻找以终点小于seq[i]的最长的子序列，即最优子状态  
+                    max=seqlen[j];  
+            }  
+            seqlen[i]=max+1;  
+            if(seqlen[i]>maxlen)           //seqlen中保存的是第i个数为终点的最长上升序列，找出这个数组中最大的值即为最优序列长度  
+                maxlen=seqlen[i];  
+        }  
+        printf("%d\n",maxlen);  
+    }  
+    return 0;  
+}  

.ACM-Templates/RMQ-ST.cpp

@@ -0,0 +1,34 @@
+#include <iostream>
+#include <algorithm>
+using namespace std;
+
+namespace RMQ_ST
+{
+const int MAXN=10000;
+int f[MAXN][MAXN];
+int a[MAXN];
+int n;
+void init()
+{
+    for(int i = 1;i<=n;i++)
+    {
+        f[i][0]=a[i];
+    }
+    for(int j=1;(1<<j)<=n;j++)
+    {
+        for(int i=1;i+(i<<j)-1<=n;i++)
+        {
+            f[i][j]=max(f[i][j-1],f[i+(1<<j-1)][j-1]);
+        }
+    }
+}
+
+int rmq(int L,int R)
+{
+    int k=0;
+    while((1<<(k+1)<=R-L+1)) k++;
+    return max(f[L][k],f[R-(1<<k)+1][k]);
+}
+
+
+}/// End of namespace RMQ_ST

.ACM-Templates/floyd.cpp

@@ -0,0 +1,18 @@
+for(int k=1;k<=n;k++)    
+{    
+    for(int f=1;f<=n;f++)    
+    {    
+        for(int t=1;t<=n;t++)    
+        {    
+            if(f==t||f==k||t==k) continue;    
+            if(m[f][k]!=INF&&m[k][t]!=INF)    
+            {    
+                int total=m[f][k]+m[k][t];    
+                if(total<m[f][t]||m[f][t]==INF)    
+                {    
+                    m[f][t]=total;    
+                }    
+            }    
+        }    
+    }    
+}

.ACM-Templates/isPrime.cpp

@@ -0,0 +1,20 @@
+//Written by Coffee. 判断素数  
+bool isPrime(int num)  
+{  
+    if (num == 2 || num == 3)  
+    {  
+        return true;  
+    }  
+    if (num % 6 != 1 && num % 6 != 5)  
+    {  
+        return false;  
+    }  
+    for (int i = 5; i*i <= num; i += 6)  
+    {  
+        if (num % i == 0 || num % (i+2) == 0)  
+        {  
+            return false;  
+        }  
+    }  
+    return true;  
+}  

.Tools/Readme.md

@@ -0,0 +1 @@
+#Tools

HDOJ/1698_chendl111.cpp

@@ -0,0 +1,59 @@
+#include<cstdio> 
+#include<algorithm>
+#include<iostream>
+using namespace std;
+#define lchild rt << 1, l, m
+#define rchild rt << 1 | 1, m + 1, r
+const int N=400010;
+int tree[N],lazy[N],t,ans=1,L,R,delta,n,m;
+
+//区间求和 
+void push_up(int rt){
+	tree[rt]=tree[rt << 1]+tree[rt << 1 | 1];
+}
+
+void push_down(int rt, int len) {
+	if(lazy[rt])
+	{
+		lazy[rt << 1]=lazy[rt << 1 | 1] = lazy[rt];
+    	tree[rt << 1] = lazy[rt << 1] * (len - (len >> 1));
+    	tree[rt << 1 | 1] = lazy[rt << 1 | 1] * (len >> 1);
+    	lazy[rt] = 0;
+	}	
+}
+
+void build(int rt = 1, int l = 1, int r = n) {
+	lazy[rt]=0;
+    if (l == r) { tree[rt]=1; return; }//注意对树的初始赋值 
+    int m = (l + r) >> 1;
+    build(lchild); build(rchild);
+    push_up(rt);
+}
+
+void update(int L, int R, int delta, int rt = 1, int l = 1, int r = n) {
+    if (L <= l && r <= R) {
+        tree[rt] = delta * (r - l + 1);//数据更新 
+        lazy[rt] = delta;
+        return;
+    }
+    push_down(rt, r - l + 1);
+    int m = (l + r) >> 1;
+    if (L <= m) update(L, R, delta, lchild);
+    if (R > m)  update(L, R, delta, rchild);
+    push_up(rt);
+}
+
+int main()
+{
+	for(scanf("%d",&t);t--;)
+	{
+		scanf("%d%d",&n,&m);
+		build();
+		for(int i=0;i<m;i++) 
+		{
+			scanf("%d%d%d",&L,&R,&delta);
+			update(L,R,delta);
+		}
+		printf("Case %d: The total value of the hook is %d.\n",ans++,tree[1]);
+	} 
+}

HDOJ/1698_libin56842.cpp

@@ -0,0 +1,80 @@
+#include <stdio.h>
+#include <string.h>
+#include <algorithm>
+using namespace std;
+const int maxn = 100000+10;
+
+int n,sum;
+
+struct node
+{
+    int l,r,n;
+} a[maxn<<2];
+
+void init(int l,int r,int i)
+{
+    a[i].l = l;
+    a[i].r = r;
+    a[i].n = 1;
+    if(l!=r)
+    {
+        int mid = (l+r)>>1;
+        init(l,mid,2*i);
+        init(mid+1,r,2*i+1);
+    }
+}
+
+void insert(int i,int x,int y,int m)
+{
+    if(a[i].n == m)//相同则不用修改了
+    return ;
+    if(a[i].l == x && a[i].r == y)//找到了区间，直接更新
+    {
+        a[i].n = m;
+        return ;
+    }
+    if(a[i].n != -1)//如果该区间只有一种颜色
+    {
+        a[2*i].n = a[2*i+1].n = a[i].n;//由于后面必定对子树操作，所以更新子树的值等于父亲的值
+        a[i].n = -1;//由于该区域颜色与修改不同，而且不是给定区域，所以该区域必定为杂色
+    }
+    //父区间为杂色时对所有子节点进行操作
+    int mid = (a[i].l+a[i].r)>>1;
+    if(x>mid)
+    insert(2*i+1,x,y,m);
+    else if(y<=mid)
+    insert(2*i,x,y,m);
+    else
+    {
+        insert(2*i,x,mid,m);
+        insert(2*i+1,mid+1,y,m);
+    }
+}
+
+int find(int i)//区间求和
+{
+    if(a[i].n != -1)//纯色直接算这个区间
+    return (a[i].r - a[i].l+1)*a[i].n;
+    else//不存则左右子树去找
+    return find(i*2)+find(i*2+1);
+}
+
+int main()
+{
+    int t,i,k,x,y,m;
+    int cas = 1;
+    scanf("%d",&t);
+    while(t--)
+    {
+        scanf("%d%d",&n,&k);
+        init(1,n,1);
+        while(k--)
+        {
+            scanf("%d%d%d",&x,&y,&m);
+            insert(1,x,y,m);
+        }
+        printf("Case %d: The total value of the hook is %d.\n",cas++,find(1));
+    }
+
+    return 0;
+}