hdu 2830 Matrix Swapping II dp 動態規劃

標籤：hdu   2830   matrix swapping ii   dp   動態規劃   

//這題和之前的2870題意相似，不過多了可以任意交換兩列的功能//先開始沒看見能交換兩列。。。結果裸的範例都沒過，最後想了想//任意交換兩列，那我們就可以直接對第i行第j列得到的cnt[i][j]//這一行驚醒排序就可以了，其中cnt[i][j]表示前i行第j列有多少//個相同的1，進行降序排序得到a[j]，那麼a[j]*j的最大值就是我//們所要的答案，因為j前面的肯定是高於j的。//注釋部分是我之前的2870的裸的代碼，具體可以看我的部落格dp46道專題 //哎，繼續練吧。。。#include <algorithm>#include <bitset>#include <cassert>#include <cctype>#include <cfloat>#include <climits>#include <cmath>#include <complex>#include <cstdio>#include <cstdlib>#include <cstring>#include <ctime>#include <deque>#include <functional>#include <iostream>#include <list>#include <map>#include <numeric>#include <queue>#include <set>#include <stack>#include <vector>#define ceil(a,b) (((a)+(b)-1)/(b))#define endl '\n'#define gcd __gcd#define highBit(x) (1ULL<<(63-__builtin_clzll(x)))#define popCount __builtin_popcountlltypedef long long ll;using namespace std;const int MOD = 1000000007;const long double PI = acos(-1.L);template<class T> inline T lcm(const T& a, const T& b) { return a/gcd(a, b)*b; }template<class T> inline T lowBit(const T& x) { return x&-x; }template<class T> inline T maximize(T& a, const T& b) { return a=a<b?b:a; }template<class T> inline T minimize(T& a, const T& b) { return a=a<b?a:b; }const int maxn = 1024;char mp[maxn][maxn];int cnt[maxn][maxn];int l[maxn];int r[maxn];int n,m;int ans;int a[maxn];bool cmp(int a,int b){return a>b;}void init(){for (int i=1;i<=n;i++)scanf("%s",mp[i]+1);//for (int i=1;i<=n;i++)//printf("%s\n",mp[i]+1);memset(cnt,0,sizeof(cnt));ans = 0;}void solve(){for (int i=1;i<=n;i++){for (int j=1;j<=m;j++){l[j] = r[j] = j;if (mp[i][j]=='1')cnt[i][j] = cnt[i-1][j]+1;else cnt[i][j] = 0;}for (int j=1;j<=m;j++)a[j] = cnt[i][j];sort(a+1,a+m+1,cmp);//for (int j=2;j<=m;j++){//if (!cnt[i][j])continue;//while(cnt[i][l[j]-1]>=cnt[i][j])//l[j] = l[l[j]-1];//}//for (int j=m-1;j>=1;j--){//if (!cnt[i][j])continue;//while(cnt[i][r[j]+1]>=cnt[i][j])//r[j] = r[r[j]+1];//}for (int j=1;j<=m;j++){//if (!cnt[i][j])continue;ans = max(ans,a[j]*j);}}printf("%d\n",ans);}int main() {    //freopen("G:\\Code\\1.txt","r",stdin);while(scanf("%d%d",&n,&m)!=EOF){init();solve();}return 0;}

hdu 2830 Matrix Swapping II dp 動態規劃