Light OJ 1157 LCS Revisited (DP)

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
[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;
}