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HDU 2476 String painter (区间DP)
题意:给定两个串,问你从第一个串刷成第二个串最少要几次。
析:我们可以先求一个空串变成第二个串,然后再求第一个串的,dp[i][j] 表示 i-j 这个区间已经和第二个串相同了,最少要几次,区间dp么,
然后再求和第一个的。
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> #include <sstream> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const LL LNF = 1e16; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 100 + 10; const int mod = 1e9 + 7; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; 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 bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } int dp[maxn][maxn], ans[maxn]; char a[maxn], b[maxn]; int main(){ while(scanf("%s", a) == 1){ scanf("%s", b); memset(dp, 0, sizeof dp); n = strlen(a); for(int i = 0; i < n; ++i) dp[i][i] = 1; for(int l = 1; l < n; ++l) for(int i = 0; i + l < n; ++i){ int j = i + l; dp[i][j] = dp[i+1][j] + 1; for(int k = i+1; k <= j; ++k) dp[i][j] = min(dp[i][j], dp[i+1][k] + dp[k+1][j] + (b[i] != b[k])); } for(int i = 0; i < n; ++i){ ans[i] = dp[0][i]; if(a[i] == b[i]) ans[i] = i == 0 ? 0 : ans[i-1]; for(int j = 0; j < i; ++j) ans[i] = min(ans[i], ans[j] + dp[j+1][i]); } printf("%d\n", ans[n-1]); } return 0; }
HDU 2476 String painter (区间DP)
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