区间DP问题（矩阵连乘，石子合并，括号匹配）
矩阵连乘源码：
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <cstring>
#include <climits>
#include <cmath>
#include <cctype>
typedef long long ll;
using namespace std;
int m[110][110];
int a[110];
int n;
int course(int i,int j)
{
if(m[i][j] != -1)//已经算出来这个值，直接返回
{
return m[i][j];
}
if(i == j)
{
return 0;
}
if(i == j-1)
{
m[i][j] = a[i] * a[i+1] * a[i+2];
return m[i][j];
}
int u = course(i,i) + course(i+1,j) + a[i] * a[i+1] * a[j+1];
int k,t;
for(k=i+1; k<j; k++)
{
t = course(i,k) + course(k+1,j) + a[i] * a[k+1] * a[j+1];
if(t < u)
u = t;
}
m[i][j] = u;
return m[i][j];
}
int main()
{
int i;
while(scanf("%d",&n) != EOF)
{
memset(m,-1,sizeof(m));//设置备忘录。刚开始初始值为-1
for(i=1; i<=n+1; i++)
{
scanf("%d",&a[i]);
}
int sum = course(1,n);
printf("%d\n",sum);
}
return 0;
}
题目地址：http://acm.nyist.net/JudgeOnline/problem.php?pid=536
思路：在模板上的简单改动
AC代码：
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <cstring>
#include <climits>
#include <cmath>
#include <cctype>
typedef long long ll;
using namespace std;
int m[110][110];
int a[110];
int n;
int course(int i,int j)
{
if(m[i][j] != -1)//已经算出来这个值，直接返回
{
return m[i][j];
}
if(i == j)
{
return 0;
}
if(i == j-1)
{
m[i][j] = a[i] * a[i+1] * a[i+2];
return m[i][j];
}
int u = course(i,i) + course(i+1,j) + a[i] * a[i+1] * a[j+1];
int k,t;
for(k=i+1; k<j; k++)
{
t = course(i,k) + course(k+1,j) + a[i] * a[k+1] * a[j+1];
if(t < u)
u = t;
}
m[i][j] = u;
return m[i][j];
}
int main()
{
int i;
while(scanf("%d",&n) != EOF)
{
int b;
memset(m,-1,sizeof(m));//设置备忘录。刚开始初始值为-1
for(i=1; i<=n-1; i++)
{
scanf("%d%d",&a[i],&b);
}
scanf("%d%d",&a[n],&a[n+1]);
int sum = course(1,n);
printf("%d\n",sum);
}
return 0;
}
石子合并
题目地址：http://acm.nyist.net/JudgeOnline/problem.php?pid=737
思路：
从最中间的一条线向右上角斜着算，和矩阵连乘类似，可以用四边形不等式优化
四边形不等式详解地址：http://blog.163.com/dqx_wl/blog/static/2396821452015111133052112/
AC代码：
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <cstring>
#include <climits>
#include <cmath>
#include <cctype>
typedef long long ll;
using namespace std;
int a[210];
int dp[210][210];
int s[210][210];
int main()
{
int n,i,j,k,r;
while(scanf("%d",&n) != EOF)
{
for(i=1; i<=n; i++)
{
scanf("%d",&a[i]);
}
for(i=1; i<=n; i++)
{
s[i][i] = a[i];
for(j=i+1; j<=n; j++)
{
s[i][j] = s[i][j-1] + a[j];//得出合并区间i到j的结果
}
}
for(r=2; r<=n; r++)
{
for(i=1; i<=n-r+1; i++)
{
j = i + r - 1;
dp[i][j] = INT_MAX;
for(k=i; k<j; k++)
{
if(dp[i][j] > dp[i][k] + dp[k+1][j])
{
dp[i][j] = dp[i][k] + dp[k+1][j];
}
}
dp[i][j] += s[i][j];
}
}
printf("%d\n",dp[1][n]);
}
return 0;
}
题目地址：点击打开链接
思路：注意数组是从a+1开始存的
AC代码：
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <cstring>
#include <climits>
#include <cmath>
#include <cctype>
typedef long long ll;
using namespace std;
char a[110];
int dp[110][110];
int main()
{
int i,j,k,l;
while(scanf("%s",a+1))
{
memset(dp,0,sizeof(dp));
if(!strcmp(a+1,"end"))
break;
int len = strlen(a+1);
for(l=2; l<=len; l++)
{
for(i=1; i<=len-l+1; i++)
{
j = i + l - 1;
if((a[i] == '(' && a[j] == ')') || ( a[i] == '[' && a[j] == ']'))
{
dp[i][j] = dp[i+1][j-1] + 2;
}
for(k=i; k<j; k++)
{
dp[i][j] = max(dp[i][j],dp[i][k] + dp[k+1][j]);
}
}
}
printf("%d\n",dp[1][len]);
}
return 0;
}