切了几道水题复习

标签：acm

1. poj 1088

#include<cstdio>
#include<cmath>
#include<stdlib.h>
#include<map>
#include<set>
#include<time.h>
#include<vector>
#include<queue>
#include<string>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define eps 1e-8
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define LL long long
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
typedef pair<int , int> pii;

#define maxn 100 + 10

int d[maxn][maxn];
int a[maxn][maxn];
int go[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};

int R, C;
int ans;

int dfs(int x, int y)
{
    if(d[x][y] > 0) return d[x][y];

    int res = 0;
    for(int i = 0; i < 4; i++)
    {
        int dx = x + go[i][0];
        int dy = y + go[i][1];
        if(dx < 1 || dy < 1 || dx > R || dy > C || a[dx][dy] >= a[x][y]) continue;
        int t = dfs(dx, dy);
        res = max(res, t);
    }
    res++;
    ans = max(ans, res);
    return d[x][y] = res;
}

int main()
{
    while(~scanf("%d%d", &R, &C))
    {
        ans = -INF;
        for(int i = 1; i <= R; i++)
        for(int j = 1; j <= C; j++)
        scanf("%d", &a[i][j]);

        memset(d, -1, sizeof d);
        for(int i = 1; i <= R; i++)
        for(int j = 1; j <= C; j++)
        if(d[i][j] < 0)
        dfs(i, j);

        printf("%d\n", ans);
    }
    return 0;
}

#include<cstdio>
#include<cmath>
#include<stdlib.h>
#include<map>
#include<set>
#include<time.h>
#include<vector>
#include<queue>
#include<string>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define eps 1e-8
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define LL long long
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
typedef pair<int , int> pii;
#define maxn 500000 + 10

int a[maxn];
int n;

int lowbit(int x)
{
    return x & (-x);
}

void modify(int x,int add)//一维
{
    while(x<=n)
    {
        a[x]+=add;
        x+=lowbit(x);
    }
}

int get_sum(int x)
{
    int ret=0;
    while(x > 0)
    {
        ret+=a[x];
        x-=lowbit(x);
    }
    return ret;
}

map<int, int> M;
set<int> S;


struct B
{
    int x, pos;
}b[maxn];

int cmp(B t1, B t2)
{
    return t1.x < t2.x ;
}

int bb[maxn];

int main()
{
    while(~scanf("%d", &n) && n)
    {
        for(int i = 1; i <= n; i++)
        {
            scanf("%d", &b[i].x);
            b[i].pos = i;
        }

        sort(b + 1, b + 1 + n, cmp);

        for(int i = 1; i <= n; i++)
        bb[b[i].pos] = i;

        memset(a, 0, sizeof a);
        LL ans = 0;
        for(int i = 1 ; i <= n; i++)
        {
            modify(bb[i], 1);
            ans += i - get_sum(bb[i]);
        }
        printf("%I64d\n", ans);
    }
    return 0;
}

/*

5
9
1
0
5
4
3
1
2
3
0

*/

3. hdu

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>

using namespace std;
#define maxn 400000 + 10

int L[maxn], R[maxn];
int pos[maxn];
int a[maxn];
int n;

void get_left()
{
    int head = 1, tail = 0;
    for(int i = 1; i <= n; i++)
    {
        while(head <= tail && a[pos[tail]] >= a[i])
        tail--;
        L[i] = i - pos[tail] - 1;
        pos[++tail] = i;
    }
}

void get_right()
{
    int head = 1, tail = 0;
    pos[tail] = n + 1;
    for(int i = n; i >= 1; i--)
    {
        while(head <= tail && a[pos[tail]] >= a[i])
        tail--;
        R[i] = pos[tail] - i - 1;
        pos[++tail] = i;
    }
}

int main()
{
    while(~scanf("%d", &n))
    {
        memset(L, 0, sizeof L);
        memset(R, 0, sizeof R);
        for(int i = 1; i <= n; i++)
        scanf("%d", a + i);
        get_left();
        get_right();


        for(int i = 1; i <= n; i++)
        {
            printf("%d * %d\n", L[i], R[i]);
        }
        long long  ans = -1;
        for(int i = 1; i <= n; i++)
        ans = max(ans, ((long long)L[i] + R[i] + 1) * a[i]);
        printf("%I64d\n", ans);
    }
    return 0;
}

/*

6
5 8 4 4 8 4

*/

4.hdu 1541

#include<cstdio>
#include<cmath>
#include<stdlib.h>
#include<map>
#include<set>
#include<time.h>
#include<vector>
#include<queue>
#include<string>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define eps 1e-8
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define LL long long
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
typedef pair<int , int> pii;
#define N 32010
#define maxn 15010

int n;
int a[N];
int ans[maxn];

int lowbit(int x)
{
    return x & (-x);
}

void modify(int x, int add)
{
    while(x <= N)
    {
        a[x] += add;
        x += lowbit(x);
    }
}

int get_sum(int x)
{
    int ret = 0;
    while(x > 0)
    {
        ret += a[x];
        x -= lowbit(x);
    }
    return ret;
}

int main()
{
    while(~scanf("%d", &n))
    {
        int x, y;
        memset(ans, 0, sizeof ans);
        memset(a, 0, sizeof a);
        for(int i = 0; i < n; i++)
        {
            scanf("%d%d", &x, &y);
            x++;
            ans[get_sum(x)]++;
            modify(x, 1);
        }
        for(int i = 0; i < n; i++)
        printf("%d\n", ans[i]);
    }
    return 0;
}

切了几道水题复习

标签：acm

原文地址：http://blog.csdn.net/dojintian/article/details/47401973