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题目链接:http://poj.org/problem?id=1050
发现这个题没有写过题解,现在补上吧,思路挺经典的。
思路就是枚举所有的连续的连续的行,比如1 2 3 4 12 23 34 45 123 234 345...然后把这些行对应列相加缩成一行,之后就是求最大子序列和了。
1 /* 2 ━━━━━┒ギリギリ♂ eye! 3 ┓┏┓┏┓┃キリキリ♂ mind! 4 ┛┗┛┗┛┃\○/ 5 ┓┏┓┏┓┃ / 6 ┛┗┛┗┛┃ノ) 7 ┓┏┓┏┓┃ 8 ┛┗┛┗┛┃ 9 ┓┏┓┏┓┃ 10 ┛┗┛┗┛┃ 11 ┓┏┓┏┓┃ 12 ┛┗┛┗┛┃ 13 ┓┏┓┏┓┃ 14 ┃┃┃┃┃┃ 15 ┻┻┻┻┻┻ 16 */ 17 #include <algorithm> 18 #include <iostream> 19 #include <iomanip> 20 #include <cstring> 21 #include <climits> 22 #include <complex> 23 #include <fstream> 24 #include <cassert> 25 #include <cstdio> 26 #include <bitset> 27 #include <vector> 28 #include <deque> 29 #include <queue> 30 #include <stack> 31 #include <ctime> 32 #include <set> 33 #include <map> 34 #include <cmath> 35 using namespace std; 36 #define fr first 37 #define sc second 38 #define cl clear 39 #define BUG puts("here!!!") 40 #define W(a) while(a--) 41 #define pb(a) push_back(a) 42 #define Rint(a) scanf("%d", &a) 43 #define Rll(a) scanf("%lld", &a) 44 #define Rs(a) scanf("%s", a) 45 #define Cin(a) cin >> a 46 #define FRead() freopen("in", "r", stdin) 47 #define FWrite() freopen("out", "w", stdout) 48 #define Rep(i, len) for(int i = 0; i < (len); i++) 49 #define For(i, a, len) for(int i = (a); i < (len); i++) 50 #define Cls(a) memset((a), 0, sizeof(a)) 51 #define Clr(a, x) memset((a), (x), sizeof(a)) 52 #define Full(a) memset((a), 0x7f7f7f, sizeof(a)) 53 #define lrt rt << 1 54 #define rrt rt << 1 | 1 55 #define pi 3.14159265359 56 #define RT return 57 #define lowbit(x) x & (-x) 58 #define onenum(x) __builtin_popcount(x) 59 typedef long long LL; 60 typedef long double LD; 61 typedef unsigned long long ULL; 62 typedef pair<int, int> pii; 63 typedef pair<string, int> psi; 64 typedef pair<LL, LL> pll; 65 typedef map<string, int> msi; 66 typedef vector<int> vi; 67 typedef vector<LL> vl; 68 typedef vector<vl> vvl; 69 typedef vector<bool> vb; 70 71 const int maxn = 110; 72 int G[maxn][maxn]; 73 int n; 74 int dp[maxn]; 75 76 int lss() { 77 int tmp = -0x7f7f7f, ret = -0x7f7f7f; 78 For(i, 1, n+1) { 79 if(tmp > 0) tmp += dp[i]; 80 else tmp = dp[i]; 81 ret = max(tmp, ret); 82 } 83 return ret; 84 } 85 86 int main() { 87 // FRead(); 88 while(~Rint(n)) { 89 For(i, 1, n+1) For(j, 1, n+1) Rint(G[i][j]); 90 int ret = 0; 91 For(i, 1, n+1) { 92 Cls(dp); 93 For(j, i, n+1) { 94 For(k, 1, n+1) dp[k] += G[j][k]; 95 ret = max(ret, lss()); 96 } 97 } 98 printf("%d\n", ret); 99 } 100 RT 0; 101 }
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原文地址:http://www.cnblogs.com/vincentX/p/5595849.html