Light OJ 1157 LCS Revisited (DP)

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解析：dp[i][j]是s[1..i]与t[1...j]的LCS长度，num[i][j]为不同LCS的个数。

关键点在于s[i]!=t[j]且dp[i-1][j]=dp[i][j-1]时如何转移。

如果dp[i][j]!=dp[i-1][j-1]且dp[i-1][j]=dp[i][j-1]，这时num[i][j] = num[i-1][j]+num[i][j-1];

其他情况下利用容斥的思想，num[i][j] = num[i-1][j]+num[i][j-1]-num[i-1][j-1];

[code]:

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;typedef long long LL;const int MOD = 1e6+7;char s[1005],t[1005];int n,m,dp[1001][1001];int num[1001][1001];void init(){    int i,j;    for(i = 0;i <= n;i++) dp[i][0] = 0,num[i][0]=1;    for(j = 0;j <= m;j++) dp[0][j] = 0,num[0][j]=1;}void sol(){    int i,j,k;    for(i = 1;i <= n;i++){        for(j = 1;j <= m;j++){            if(s[i]==t[j]){                dp[i][j] = dp[i-1][j-1]+1;                num[i][j] = num[i-1][j-1];            }else{                dp[i][j] = max(dp[i-1][j],dp[i][j-1]);                if(dp[i][j-1]>dp[i-1][j]){                    num[i][j] = num[i][j-1];                }else if(dp[i][j-1]<dp[i-1][j]){                    num[i][j] = num[i-1][j];                }else{                    num[i][j] = num[i-1][j]+num[i][j-1]-(dp[i][j]==dp[i-1][j-1]?num[i-1][j-1]:0);                    num[i][j] = (num[i][j]%MOD+MOD)%MOD;                }            }        }    }    printf("%d\n",num[n][m]);}int main(){    int i,j,cas;    scanf("%d",&cas);    for(int T=1;T<=cas;T++){        scanf("%s%s",s+1,t+1);        n = strlen(s+1);        m = strlen(t+1);        printf("Case %d: ",T);        init();        sol();    }    return 0;}

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