HDU 4334——Trouble——————【贪心&水题】

Trouble

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5388 Accepted Submission(s): 1494


[align=left]Problem Description[/align]
Hassan is in trouble. His mathematics teacher has given him a very difficult problem called 5-sum. Please help him.
The 5-sum problem is defined as follows: Given 5 sets S_1,...,S_5 of n integer numbers each, is there a_1 in S_1,...,a_5 in S_5 such that a_1+...+a_5=0?

[align=left]Input[/align]
First line of input contains a single integer N (1≤N≤50). N test-cases follow. First line of each test-case contains a single integer n (1<=n<=200). 5 lines follow each containing n integer numbers in range [-10^15, 1 0^15]. I-th line denotes set S_i for 1<=i<=5.

[align=left]Output[/align]
For each test-case output "Yes" (without quotes) if there are a_1 in S_1,...,a_5 in S_5 such that a_1+...+a_5=0, otherwise output "No".

[align=left]Sample Input[/align]

2

2

1 -1

1 -1

1 -1

1 -1

1 -1

3

1 2 3

-1 -2 -3

4 5 6

-1 3 2
-4 -10 -1

[align=left]Sample Output[/align]

No

Yes

题目大意：给你5行每行n个数，问你是否可以在每行中取出一个数，让a1+a2+a3+a4+a5=0。
解题思路：开始写了搜索，但是一直超时。然而超时好像是必然的，没办法剪枝。其实可以将前1、2行合成一行a，将3、4行合成一行b，从小到大排序。枚举剩下的最后一行c，枚举a，逆序枚举b。贪心思想在如果b的当前值跟a和c的和小于0，那么可以直接让a的指针加一，如果大于0，可以让b的指针减一，如果逆序遍历完b或正序遍历完a以后还没找到，那么就继续枚举c。

#include<bits/stdc++.h>
using namespace std;
typedef long long INT;
INT Map[6][210];
INT a[42000];
INT b[42000];
INT c[210];
int main(){
int t,n;
scanf("%d",&t);
while(t--){
scanf("%d",&n);
for(int i=1;i<=5;i++){
for(int j=0;j<n;j++){
scanf("%lld",&Map[i][j]);
}
}
int na=0,nb=0,nc=0;
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
a[na++]=Map[1][i]+Map[2][j];//合并1、2行
b[nb++]=Map[3][i]+Map[4][j];//合并3、4行
}
}
int nn=n*n;
sort(a,a+nn);   //递增排序
sort(b,b+nn);   //递增排序
na=1,nb=1,nc=1;
for(int i=1;i<nn;i++){  //去重
if(a[i]!=a[i-1]){
a[na++]=a[i];
}
}
for(int i=1;i<nn;i++){  //去重
if(b[i]!=b[i-1]){
b[nb++]=b[i];
}
}
c[0]=Map[5][0];
for(int i=1;i<n;i++){   //去重
if(Map[5][i]!=Map[5][i-1]){
c[nc++]=Map[5][i];
}
}
int flag=0;
for(int i=0;i<nc&&(!flag);i++){  //n的复杂度
for(int j=0,k=nb-1;k>=0&&j<na; ){//n*n的复杂度
INT cc=c[i]+a[j]+b[k];
if(cc<0){       //降低复杂度。模拟跳出一层循环
j++;
}else if(cc==0){
flag=1;
break;
}else{
k--;
}
}
}
if(flag){
puts("Yes");
}else {
puts("No");
}
}
return 0;
}