本文介绍了求解最大连续子序列和的经典问题及其多种解决方法，包括直接枚举、前缀和预处理、分治算法、线段树优化以及动态规划等，并扩展到二维矩阵的情况。
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Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 161486 Accepted Submission(s): 37830
Problem Description
Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.
Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).
Output
For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.
Sample Input
2 5 6 -1 5 4 -7 7 0 6 -1 1 -6 7 -5
Sample Output
Case 1: 14 1 4 Case 2: 7 1 6
方法一：直接枚举。O（n^3），超时
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define N 100010
int a[N];
int main()
{
//freopen("in.txt", "r", stdin);
int T, n, w = 0;
int i, j, k;
scanf("%d", &T);
while(T--)
{
scanf("%d", &n);
for(i=0; i<n; i++)
scanf("%d", &a[i]);
int max_s = -(1<<30);
int sum, left, right;
for(i=0; i<n; i++){
for(j=i; j<n; j++){
sum = 0;
for(k=i; k<=j; k++)
sum += a[k];
if(sum > max_s){
max_s = sum;
left = i;
right = j;
}
}
}
printf("Case %d:\n%d %d %d\n", ++w, max_s, left + 1, right + 1);
if(T) printf("\n");
}
return 0;
}方法二：对前i项和进行预处理，O（n^2），超时。
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define N 100010
int a[N];
int sum[N];
int main()
{
//freopen("in.txt", "r", stdin);
int T, n, w = 0;
int i, j, k;
scanf("%d", &T);
while(T--)
{
memset(sum, 0, sizeof(sum));
scanf("%d", &n);
for(i=1; i<=n; i++){
scanf("%d", &a[i]);
sum[i] = sum[i-1] + a[i];
}
int max_s = -(1<<30);
int s, left, right;
for(i=1; i<=n; i++){
for(j=i; j<=n; j++){
s = sum[j] - sum[i-1];
if(s > max_s){
max_s = s;
left = i;
right = j;
}
}
}
printf("Case %d:\n%d %d %d\n", ++w, max_s, left, right);
if(T) printf("\n");
}
return 0;
}方法三：分治，O（nlg(n)），可AC。
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define N 100010
int a[N];
typedef struct NODE
{
int val, l, r;
}Node;
Node Max_Ans(int *A, int l, int r)
{
Node t;
if(1 == r - l){
t.val = A[l];
t.l = l;
t.r = l;
return t;
}
Node t1, t2, t3;
int mid = l + (r - l) / 2;
t1 = Max_Ans(A, l, mid);
t2 = Max_Ans(A, mid, r);
if(t1.val >= t2.val) t = t1;
else t = t2;
int i, v, L, R;
v = 0; L = A[mid -1]; t3.l = mid - 1;
for(i=mid-1; i>=l; i--){
v += A[i];
if(v >= L){
L = v;
t3.l = i;
}
}
v = 0; R = A[mid]; t3.r = mid;
for(i=mid; i<r; i++){
v += A[i];
if(v > R){
R = v;
t3.r = i;
}
}
t3.val = L + R;
if(t3.val >= t.val)
t = t3;
return t;
}
int main()
{
//freopen("in.txt", "r", stdin);
int T, w = 0;
int n;
Node max_s;
int i, k;
scanf("%d", &T);
while(T--)
{
scanf("%d", &n);
for(i=0; i<n; i++){
scanf("%d", &a[i]);
}
max_s = Max_Ans(a, 0, n);
printf("Case %d:\n%d %d %d\n", ++w, max_s.val, max_s.l + 1, max_s.r + 1);
if(T) printf("\n");
}
return 0;
}方法四：线段树，可AC。
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define ms(x,y) memset(x,y,sizeof(x))
#define MAX(x,y) ((x)>(y)?(x):(y))
#define LL long long
const int MAXN=100000+10;
const int INF=1<<30;
using namespace std;
int sum[MAXN<<2];
int msum[MAXN<<2];
int lsum[MAXN<<2];
int rsum[MAXN<<2];
int ll[MAXN<<2];
int lr[MAXN<<2];
int ml[MAXN<<2];
int mr[MAXN<<2];
int rl[MAXN<<2];
int rr[MAXN<<2];
void up(int root)
{
int lroot = root<<1;
int rroot = root<<1|1;
sum[root] = sum[lroot] + sum[rroot];
lsum[root] = lsum[lroot];
rsum[root] = rsum[rroot];
//维护前缀
ll[root] = ll[lroot];
lr[root] = lr[lroot];
int sl = sum[lroot] + lsum[rroot];
if(sl > lsum[root]){
lsum[root] = sl;
lr[root] = lr[rroot];
}
//维护后缀
rl[root] = rl[rroot];
rr[root] = rr[rroot];
int sr = sum[rroot] + rsum[lroot];
if(sr >= rsum[root]){
rsum[root] = sr;
rl[root] = rl[lroot];
}
//维护区间最值
ml[root] = ml[lroot];
mr[root] = mr[lroot];
msum[root] = msum[lroot];
if(msum[root] < rsum[lroot] + lsum[rroot]){
msum[root] = rsum[lroot] + lsum[rroot];
ml[root] = rl[lroot];
mr[root] = lr[rroot];
}
if(msum[root] < msum[rroot]){
msum[root] = msum[rroot];
ml[root] = ml[rroot];
mr[root] = mr[rroot];
}
}
void Build(int root, int left, int right)
{
if(left == right){
scanf("%d", &sum[root]);
lsum[root] = sum[root]; ll[root] = left; lr[root] = right;
msum[root] = sum[root]; ml[root] = left; mr[root] = right;
rsum[root] = sum[root]; rl[root] = left; rr[root] = right;
return;
}
int mid = (left + right)>>1;
Build(root<<1, left, mid);
Build(root<<1|1, mid+1, right);
up(root);
}
struct Node
{
int msum, lsum, rsum, sum;
int ll, lr, ml, mr, rl, rr;
};
Node query(int root, int left, int right, int l, int r)
{
if(l == left && right == r){
Node N;
N.sum = sum[root];
N.msum = msum[root];
N.lsum = lsum[root];
N.rsum = rsum[root];
N.ll = ll[root];
N.lr = lr[root];
N.ml = ml[root];
N.mr = mr[root];
N.rl = rl[root];
N.rr = rr[root];
return N;
}
int mid = (left + right)>>1;
Node res,res1,res2;
int lroot = root<<1;
int rroot = root<<1|1;
int markl = 0, markr = 0;
if(r <= mid) return res = query(lroot, left, mid, l, r);
if(l > mid) return res2 = query(rroot, mid+1, right, l, r);
else{
res = query(lroot, left, mid, l, mid);
res2 = query(rroot, mid+1, right, mid+1, r);
res1.sum = res.sum + res2.sum;
res1.lsum = res.lsum;
res1.rsum = res2.rsum;
res1.ll = res.ll;
res1.lr = res.lr;
LL sl = res.sum + res2.lsum;
if(sl > res1.lsum){
res1.lsum = sl;
res1.lr = res2.lr;
}
res1.rl = res2.rl;
res1.rr = res2.rr;
LL sr = res2.sum + res.rsum;
if(sr >= res1.rsum){
res1.rsum = sr;
res1.rl = res.rl;
}
res1.msum = res.msum;
res1.ml = res.ml;
res1.mr = res.mr;
if(res1.msum < res.rsum + res2.lsum){
res1.msum = res.rsum + res2.lsum;
res1.ml = res.rl;
res1.mr = res2.lr;
}
if(res1.msum < res2.msum){
res1.msum = res2.msum;
res1.ml = res2.ml;
res1.mr = res2.mr;
}
return res1;
}
}
int main()
{
//freopen("in.txt", "r", stdin);
int T, n, w = 0;
int i, j, k;
Node max_s;
scanf("%d", &T);
while(T--)
{
scanf("%d", &n);
Build(1, 1, n);
max_s = query(1, 1, n, 1, n);
printf("Case %d:\n%d %d %d\n", ++w, max_s.msum, max_s.ml, max_s.mr);
if(T) printf("\n");
}
return 0;
}方法五：扫一遍，累加sum， 当sum < 0 时，置sum 为0。如：
-1 5 7 -9 2 -7 -2 1 3 可分成
-1 | 5 12 3 5 | -7 | -2 | 1 4
竖线为分界。比较每个区间，取最大值。
其实还可看做是DP：sum[i] = max{sum[i-1] + a[i], a[i]}; sum[i]表示以a[i]结尾的最大连续和，最后再在sum[]里面找最大值就行了。
思路来自：http://blog.youkuaiyun.com/hcbbt/article/details/10454947
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
//freopen("in.txt", "r", stdin);
int T, w = 0;
int n, num;
int beg, end, max_s, tmp;
int i, k;
scanf("%d", &T);
while(T--)
{
scanf("%d", &n);
max_s = -(1<<30);
tmp = 0;
beg = end = k = 1;
for(i=1; i<=n; i++){
scanf("%d", &num);
tmp += num;
if(tmp > max_s){
max_s = tmp;
beg = k;
end = i;
}
if(tmp < 0){
tmp = 0;
k = i + 1;
}
}
printf("Case %d:\n%d %d %d\n", ++w, max_s, beg, end);
if(T) printf("\n");
}
return 0;
}二维最大连续子序列和：
To the Max
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 42224 | Accepted: 22464 |
Description
Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*1 or greater located within the whole array. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the sub-rectangle with the largest sum is referred to as the maximal sub-rectangle.
As an example, the maximal sub-rectangle of the array:
0 -2 -7 0
9 2 -6 2
-4 1 -4 1
-1 8 0 -2
is in the lower left corner:
9 2
-4 1
-1 8
and has a sum of 15.
As an example, the maximal sub-rectangle of the array:
0 -2 -7 0
9 2 -6 2
-4 1 -4 1
-1 8 0 -2
is in the lower left corner:
9 2
-4 1
-1 8
and has a sum of 15.
Input
The input consists of an N * N array of integers. The input begins with a single positive integer N on a line by itself, indicating the size of the square two-dimensional array. This is followed by N^2 integers separated by whitespace (spaces and newlines). These are the N^2 integers of the array, presented in row-major order. That is, all numbers in the first row, left to right, then all numbers in the second row, left to right, etc. N may be as large as 100. The numbers in the array will be in the range [-127,127].
Output
Output the sum of the maximal sub-rectangle.
Sample Input
4 0 -2 -7 0 9 2 -6 2 -4 1 -4 1 -1 8 0 -2
Sample Output
15
题意：给出一个矩阵，求和最大的子矩阵，输出和。
思路：对某两行，列于列相加，组成新的数组，求最大连续和。
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define N 105
int a[N][N];
int sum[N];
int main()
{
//freopen("in.txt", "r", stdin);
int n;
while(~scanf("%d", &n))
{
int i, j, k, p;
for(i=0; i<n; i++){
for(j=0; j<n; j++){
scanf("%d", &a[i][j]);
}
}
int max_s = -(1<<30), tmp;
for(i=0; i<n; i++){
memset(sum, 0, sizeof(sum));
for(j=i; j<n; j++){
tmp = 0;
for(k=0; k<n; k++){
//for(p=i; p<=j; p++)
// tmp += a[p][k];
sum[k] += a[j][k];
tmp += sum[k];
if(tmp > max_s) max_s = tmp;
if(tmp < 0) tmp = 0;
}
}
}
printf("%d\n", max_s);
}
return 0;
}
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