HDU 1003 Max Sum

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1.题目描述：点击打开链接

2.解题思路：本题是经典的最大连续和问题，效率高的解决方法有两种：（1）利用公式：A[i]+A[i+1]+...+A[j]=S[j]-S[i-1];（2）利用动态规划；（3）边读入边计算（精简版的动态规划）。对于（1），二重循环还是会TLE的，应该事先计算好0~j之间的最小值及其位置。对于（2）定义d(i)表示以i结尾的最大连续字串的和，再用一个辅助变量保存开始值s，状态转移方程如下：

d(i)=d(i-1)+t[i](d[i-1]+t[i]>=t[i])

d(i)=t[i](d(i-1)+t[i]<t[i])

两种方法的时间复杂度均为O(N)。

3.代码：

（方法一）

#define _CRT_SECURE_NO_WARNINGS #include<iostream>#include<algorithm>#include<string>#include<sstream>#include<set>#include<vector>#include<stack>#include<map>#include<queue>#include<deque>#include<cstdlib>#include<cstdio>#include<cstring>#include<cmath>#include<ctime>#include<functional>using namespace std;typedef long long LL;#define INF 100000000+10const int maxn = 100000 + 10;int A[maxn];LL S[maxn];LL min_a[maxn];int pos_a[maxn];int n;int main(){//freopen("test.txt", "r", stdin);int T;int rnd = 0;cin >> T;while (T--){memset(A, 0, sizeof(A));memset(min_a, 0, sizeof(min_a));memset(S, 0, sizeof(S));memset(pos_a, 0, sizeof(pos_a));int start = 1;int end = 1;cin >> n;for (int i = 1; i <= n; i++)cin >> A[i];S[0] = 0;LL max_x = A[1];for (int i = 1; i <= n; i++){S[i] = S[i - 1] + A[i];}pos_a[1] = 0;min_a[1] = 0;for (int i = 2; i <= n; i++)//计算0到i-1之间的最小值及其位置{min_a[i] = min_a[i - 1];pos_a[i] = pos_a[i - 1];if (S[i - 1] < min_a[i]){min_a[i] = S[i - 1];pos_a[i] = i - 1;}}for (int j = 2; j <= n; j++)if (max_x<S[j] - min_a[j]){max_x = S[j] - min_a[j];end = j;start = pos_a[j]+1;}printf("Case %d:\n", ++rnd);printf("%lld %d %d\n", max_x, start, end);if (T)cout << endl;}return 0;}

（方法二）

#include<cstring>#include<algorithm>#include<iostream>#include<cstdio>using namespace std;int t[100010];int dp[100010];int main(){    int n;    cin >> n;    int Kase=1;    while(n--)    {        int a;        cin >> a;        int s=1;        int e=1;        int as=1;        int ae=1;        int Ans;        memset(dp,0,sizeof(dp));        cin >> t[1];        dp[1]=t[1];        Ans=dp[1];        for(int i=2;i<=a;i++)        {            cin >> t[i];            if(dp[i-1]+t[i]>=t[i])            {                e=i;//记录结尾                dp[i]=dp[i-1]+t[i];            }            else            {                s=i;//开头                e=i;//结尾                dp[i]=t[i];            }            if(dp[i]>Ans)//更新最大值和端点            {                Ans=dp[i];                as=s;                ae=e;            }        }        printf("Case %d:\n",Kase++);        printf("%d %d %d\n",Ans,as,ae);        if(n) cout << endl;    }    return 0;}

（方发二精简版）

#define _CRT_SECURE_NO_WARNINGS #include<iostream>#include<algorithm>#include<string>#include<sstream>#include<set>#include<vector>#include<stack>#include<map>#include<queue>#include<deque>#include<cstdlib>#include<cstdio>#include<cstring>#include<cmath>#include<ctime>#include<functional>using namespace std;#define INF 0xfffffffint main(){freopen("test.txt", "r", stdin);int T;int rnd = 0;cin >> T;while (T--){int sum = -INF;int Max = -INF;int n, x;int l, L, R;cin >> n;for (int i = 0; i < n; i++){cin >> x;if (sum + x < x)sum = x, l = i;else sum += x;if (Max < sum)Max = sum, L = l, R = i;}printf("Case %d:\n", ++rnd);printf("%d %d %d\n", Max, L + 1, R + 1);if (T)cout << endl;}return 0;}

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