Time Limit: 1000MS | Memory Limit: 30000K | |
Total Submissions: 15180 | Accepted: 5585 |
Description
The tests which the contractor completed were computer simulations of battlefield and hostile attack conditions. Since they were only preliminary, the simulations tested only the CATCHER's vertical movement capability. In each simulation, the CATCHER was fired at a sequence of offensive missiles which were incoming at fixed time intervals. The only information available to the CATCHER for each incoming missile was its height at the point it could be intercepted and where it appeared in the sequence of missiles. Each incoming missile for a test run is represented in the sequence only once.
The result of each test is reported as the sequence of incoming missiles and the total number of those missiles that are intercepted by the CATCHER in that test.
The General Accounting Office wants to be sure that the simulation test results submitted by the military contractor are attainable, given the constraints of the CATCHER. You must write a program that takes input data representing the pattern of incoming missiles for several different tests and outputs the maximum numbers of missiles that the CATCHER can intercept for those tests. For any incoming missile in a test, the CATCHER is able to intercept it if and only if it satisfies one of these two conditions:
The incoming missile is the first missile to be intercepted in this test.
-or-
The missile was fired after the last missile that was intercepted and it is not higher than the last missile which was intercepted.
Input
Output
Note: The number of missiles for any given test is not limited. If your solution is based on an inefficient algorithm, it may not execute in the allotted time.
Sample Input
389 207 155 300 299 170 158 65 -1 23 34 21 -1 -1
Sample Output
Test #1: maximum possible interceptions: 6 Test #2: maximum possible interceptions: 2
Source
題目大意:
求最長不升子序列長度
反省&感想:
WA了很多炮,最後發現是因爲,dp[n-1]不一定就是答案,因爲dp[n-1]可能會是初始化的1(當a[n-1]大於其他所有a時)
加了一個循環遍歷所有dp求最大值就過了
自己還是不夠仔細.....一定要在腦中將dp過程理清楚,保證正確才行
下面是ac代碼:
#include <iostream>
#include <string.h>
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <memory.h>
#include <string>
#include <vector>
#include <list>
#include <map>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <numeric>
#include <functional>
#define maxn 6000005
using namespace std;
int a[maxn];
int dp[maxn];
int main()
{
int cas=1;
while(1){
scanf("%d",a);
if(a[0]==-1) return 0;
int t=0;
for(int i=1;a[i-1]!=-1;i+=1,t+=1){
scanf("%d",&a[i]);
}
for(int i=0;i<t;i+=1) dp[i]=1;
int ans=-1;
dp[0]=1;
for(int i=0;i<t;i+=1){
for(int j=0;j<i;j+=1){
if(a[i]<=a[j]){
dp[i]=max(dp[i],dp[j]+1);
}
}
ans=max(ans,dp[i]);
}
printf("Test #%d:\n maximum possible interceptions: %d\n\n",cas++,ans);
}
return 0;
}