HDU 1422 annular maximal nonnegative sub-segment

Test instructions

• A number of columns that form a ring to find the maximum length of the sub-segments where each step is greater than or equal to 0

Solution:

• Similar rings are used two arrays to achieve the way, but look at others code found not necessary, open more space, directly to the subscript to take the surplus operation can achieve the ideal effect.
• I am the starting point of the enumeration ring (from 0 to n-1), and then each starting point is a sequence of n, and the maximum substring length is calculated with a method similar to the maximal continuous substring, so that the complexity is o ( < Span style= "Position:absolute; Clip:rect (2.029em 1000.003em 2.776em-0.477em); Top: -2.611em; Left:0.003em; " >n 2 )
• There is a better way of thinking: you put this twice times the length of the sequence (0 to 2n-1), to find the longest to meet the conditions of the sub-segment (note that the length of the sub-paragraph to exit, or else error), this idea does not control a sequence length of n constraints, Actually only this entire sequence satisfies the condition is the length is n the error and calculates the result of 2n, here the preceding one achieves N to exit can be specially solved.

My Code:

#include <queue>#include <set>#include <map>#include <stack>#include <string>#include <vector>#include <algorithm>#include <cstdio>#include <cstring>#include <iostream>using namespace STD;//HDU 1422//problems//typedef Long Long intLL;Const intM =1000009, INF =0x3fffffff;intN, A[m *2], Dp[m], key, ans;intMainvoid) { while(~scanf("%d", &n)) {memset(DP,0,sizeof(DP)); Ans = key = a[0] =0; dp[0] =1; for(inti =1; I <= N; i++) {intx, y;scanf("%d%d", &x, &y);        A[i] = a[i + N] = x-y; } for(inti =1; I <= N; i++) {if(Dp[i-1] >0) {Key-= a[i-1]; Dp[i] = dp[i-1] -1; }if(Key <0) { for(intj = i + Dp[i]-1; ;                    j--) {Key-= a[j]; dp[i]--;if(Key >=0) Break; }            }Else{ for(intj = i + dp[i]; J < i + N; J + +) {if(Key + A[j] <0) Break;                    Key + = A[j];                dp[i]++; }            }if(Dp[i] > ans) ans = dp[i];if(ans = = N) Break; }printf("%d\n", ans); }return 0;}

Better code:

#include <stdio.h>intMain () {inti,n,a,b,sum,cnt,max,ar[100005]; while(SCANF ("%d", &n)!=eof) { for(i=0; i<n;i++) {scanf ("%d%d", &a,&b);        Ar[i]=a-b; } sum=cnt=0; max=-1; for(i=0;i<2*n-1; i++) {Sum+=ar[i%n];if(sum>=0) {cnt++;if(max<cnt) max=cnt;if(cnt==n) Break; }ElseSum=cnt=0; }printf("%d\ n", Max); }return 0;}

HDU 1422 annular maximal nonnegative sub-segment