算法入门经典训练指南88页练习
::这道题只要把原矩阵扩大4倍,那么其跟最大子矩阵的题目就很类似,把二维转化成一维,求最大的序列和,不过这个序列的长度不能超过n。
长度不能超过n? 那这道题又跟hdu 3415 HDU 3415 Max Sum of Max-K-sub-sequence (单调队列)
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <queue> 6 using namespace std; 7 #define REP(i,n) for(int i=0; i<(n); i++) 8 #define FOR(i,s,t) for(int i=(s); i<=t; i++) 9 typedef long long ll; 10 const int INF = 1<<30; 11 const int N = 155; 12 int S[N][N],n,_; 13 int sum[N]; 14 15 void show(){ 16 FOR(i,1,2*n){ 17 FOR(j,1,2*n) 18 printf("%4d",S[i][j]); 19 cout<<endl; 20 } 21 } 22 23 void solve(){ 24 scanf("%d", &n); 25 26 //矩阵扩大 27 FOR(i,1,n) FOR(j,1,n){ 28 scanf("%d", &S[i][j]); 29 S[i+n][j] = S[i][j]; 30 S[i][j] += S[i-1][j]; 31 S[i][j+n] = S[i][j]; 32 33 } 34 FOR(i,n+1,2*n) FOR(j,1,n){ 35 S[i][j] += S[i-1][j]; 36 S[i][j+n] = S[i][j]; 37 } 38 39 int ans = -INF; 40 FOR(i,0,n) FOR(j,i+1,i+n){//枚举上下边界,从右往左扫描 41 int q[N<<1],head=0,tail=0; 42 for(int k = 1; k<2*n; k++){ 43 sum[k] =S[j][k] - S[i][k] + sum[k-1]; 44 while(tail>head && sum[k] <= sum[q[tail]]) tail--;//用单调队列优化 45 q[++tail] = k; 46 while(tail>head && k - q[head+1]>n) head++; 47 ans = max(ans , sum[k] - sum[q[head+1]]); 48 } 49 } 50 printf("%d\n",ans); 51 } 52 53 int main(){ 54 // freopen("in.txt","r",stdin); 55 scanf("%d", &_); 56 while(_--) solve(); 57 return 0; 58 }
UVA 10827 Maximum sum on a torus,布布扣,bubuko.com
UVA 10827 Maximum sum on a torus
原文:http://www.cnblogs.com/zyx1314/p/3760249.html