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题意：给出一个长字符串，再给一个短字符串，进行匹配，如果第i个恰好匹配，则 +8,；如果不匹配，可以给长或短字符串添加-，先后匹配，这样-3，
连续的长字符串添加-，需要减去一个4；也可不给添加-，则-5。
析：dp[i][j][0] 表示第一个字符串第 i 个位置，和第二个字符串的第 j 个位置相匹配，dp[i][j][1] 表示第一个字符串第 i 个位置，和第二个字符串的第 j 个位置加_相匹配.
那么怎么转移呢？如果s1[i] == s2[j] 那么 dp[i+1][j+1][0] = Max(dp[i+1][j+1][0], Max(dp[i][j][0], dp[i][j][1])+8);意思就是匹配的加8分，
如果不相等，dp[i+1][j+1][0] = Max(dp[i+1][j+1][0], Max(dp[i][j][0], dp[i][j][1])-5);不匹配，减 5 呗。
dp[i][j+1][1] = Max(dp[i][j+1][1], Max(dp[i][j][0]-7, dp[i][j][1]-3));
dp[i+1][j][1] = Max(dp[i+1][j][1], Max(dp[i][j][0]-7, dp[i][j][1]-3));
同样的意思。注意，这个题说学生的串不过50，但是你开小于100就会WA。。。。真坑啊
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e6 + 5;
const int mod = 1e8;
const char *mark = "+-*";
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, -1, 1, 1, -1};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
char s1[110], s2[110];
int dp[110][110][2];
int main(){
int T; cin >> T;
while(T--){
scanf("%s", s1); scanf("%s", s2);
int len1 = strlen(s1);
int len2 = strlen(s2);
for(int i = 0; i <= len1; ++i)
for(int j = 0; j <= len2; ++j)
for(int k = 0; k < 2; ++k)
dp[i][j][k] = -INF;
for(int i = 0; i <= len1; ++i) dp[i][0][0] = 0;
for(int i = 0; i < len1; ++i){
for(int j = 0; j < len2; ++j){
if(s1[i] == s2[j]) dp[i+1][j+1][0] = Max(dp[i+1][j+1][0], Max(dp[i][j][0], dp[i][j][1])+8);
else dp[i+1][j+1][0] = Max(dp[i+1][j+1][0], Max(dp[i][j][0], dp[i][j][1])-5);
dp[i][j+1][1] = Max(dp[i][j+1][1], Max(dp[i][j][0]-7, dp[i][j][1]-3));
dp[i+1][j][1] = Max(dp[i+1][j][1], Max(dp[i][j][0]-7, dp[i][j][1]-3));
}
}
int ans = -INF;
for(int i = 0; i <= len1; ++i)
ans = Max(ans, Max(dp[i][len2][0], dp[i][len2][1]));
printf("%d\n", ans);
}
return 0;
}
UVaLive 6697 Homework Evaluation (DP)
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原文地址：http://www.cnblogs.com/dwtfukgv/p/5801244.html
