From 6a7fed60f867b862d67b57c41ceecb01c886816b Mon Sep 17 00:00:00 2001
From: Kirito <1362050620@qq.com>
Date: Thu, 21 Apr 2016 18:39:51 +0800
Subject: [PATCH] From
 http://blog.csdn.net/idealism_xxm/article/details/51204130

---
 HDOJ/1501_CSDN_dp.cpp | 36 ++++++++++++++++++++++++++++++++++++
 1 file changed, 36 insertions(+)
 create mode 100644 HDOJ/1501_CSDN_dp.cpp

diff --git a/HDOJ/1501_CSDN_dp.cpp b/HDOJ/1501_CSDN_dp.cpp
new file mode 100644
index 0000000..b43ab5e
--- /dev/null
+++ b/HDOJ/1501_CSDN_dp.cpp
@@ -0,0 +1,36 @@
+#include <cstdio>
+#include <cstring>
+#include <algorithm>
+
+using namespace std;
+
+int n,m,al,bl,cl,lim;
+char a[205],b[205],c[405];
+bool dp[205][205];//dp[i][j]表示串a的前i个字符与串b的前j个字符是否能形成串c的前i+j个字符
+
+int main() {
+    int T,kase=0;
+    scanf("%d",&T);
+    while(kase<T) {
+        scanf("%s%s%s",a+1,b+1,c+1);
+        al=strlen(a+1);
+        bl=strlen(b+1);
+        cl=al+bl;
+        memset(dp,false,sizeof(dp));
+        dp[0][0]=true;
+        for(int k=1;k<=cl;++k) {//枚举串c的字符
+            lim=min(k,al);
+            for(int i=max(1,k-bl);i<=lim;++i) {//在a串上进行状态转移
+                if(dp[i-1][k-i]&&a[i]==c[k])//如果串a前i-1个字符与串b前k-i个字符能构成串c的前k个字符，并且当前a[i]==c[k]
+                    dp[i][k-i]=true;
+            }
+            lim=min(k,bl);
+            for(int j=max(1,k-al);j<=lim;++j) {//在b串上进行状态转移
+                if(dp[k-j][j-1]&&b[j]==c[k])//如果串a前k-j个字符与串b前j-1个字符能构成串c的前k个字符，并且当前b[j]==c[k]
+                    dp[k-j][j]=true;
+            }
+        }
+        printf("Data set %d: %s\n",++kase,dp[al][bl]?"yes":"no");
+    }
+    return 0;
+}