DP——Two ( HDU 5791 ) ( 2016 Multi-University Training Contest 5 1011 )
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题目链接:
http://acm.hdu.edu.cn/showproblem.php?pid=5791分析:
给出两串数字,求求它们有多少相同的子序列。题解:
确定DP方程:
DP[i][j]表示A串前i个字符和B串前j个字符之间由多少相同的子序列,
dp[i][j] = (dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1])%MOD;//加上前面的串的相同的子序列数量if(dp[i][j] < 0) //<注释> dp[i][j] += MOD;//加个MOD变正if(a[i] == b[j]) dp[i][j] = (dp[i][j]+dp[i-1][j-1]+1)%MOD; //如果当前字符相等,那么前面的dp[i-1][j-1]就能和这个新的相等字符匹配一遍,所以需要加上dp[i-1][j-1]+1
- AC代码:
#include <cstdio>#include <iostream>#include <cstdlib>#include <cstring>#include <algorithm>#include <cmath>#include <cctype>#include <stack>#include <map>#include <set>#include <queue>using namespace std;typedef pair<int,int> Pii;typedef long long LL;typedef unsigned long long ULL;typedef long double LD;int a[1234];int b[1234];int N,M;const int MOD = 1000000007;int dp[1234][1234];int solve(){ memset(dp, 0, sizeof(dp)); for(int i=1;i<=N;i++) { for(int j=1;j<=M;j++) { dp[i][j] = (dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1])%MOD; if(dp[i][j] < 0) dp[i][j] += MOD; if(a[i] == b[j]) dp[i][j] = (dp[i][j]+dp[i-1][j-1]+1)%MOD; } } return dp[N][M];}int main(){ while(~scanf("%d%d", &N, &M) ) { for(int i=1;i<=N;i++) scanf("%d", &a[i]); for(int j=1;j<=M;j++) scanf("%d", &b[j]); LL ans = solve(); cout << ans << endl; } return 0;}
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