DP入门练习

T1

题目:codevs4815江哥的dp题a

codevs4815

一个简单的DP,注意开long long(不然会全WA),以及初始条件(这题有负数,所以要把f设成极小值.还要保证转移正确).

#include <iostream>
#include <cstdio>
#include <cstring>
#define ll long long
const int maxN = 1000 + 7;
using namespace std;

ll f[maxN][maxN][2],w[maxN];

int main() {
	ll n,k;
	memset(f,-63,sizeof(f));
	scanf("%lld%lld",&n,&k);
	for(int i = 1;i <= n;++ i)
		scanf("%lld",&w[i]);
	for(int i = 0;i <= n;++ i) f[i][0][0] = 0;
	for(int i = 1;i <= n;++ i) {
		for(int j = 1;j <= k;++ j) {
			f[i][j][0] = max(f[i - 1][j][0],f[i - 1][j][1]);
			f[i][j][1] = max(f[i - 1][j - 1][0] + w[i],f[i][j][1]);
		}
	}

	printf("%lld",max(f[n][k][0],f[n][k][1]));
}

T2

题目:codevs1695 windows 2013

codevs1695

简单的DP,和上一个题差不多.注意题目中的开头先用

#include <iostream>
#include <cstring>
#include <cstdio>
#define ll long long
const int maxN = 100 + 7;
using namespace std;

ll f[maxN][2],a[maxN],b[maxN],c[maxN],n;//f[i][0]当前用勺   f[i][1]当前用筷

int main() {
	memset(f,0x3f,sizeof(f));
	scanf("%lld",&n);
	for(int i = 1;i <= n;++ i)
		scanf("%lld%lld%lld",&a[i],&b[i],&c[i]);//勺子筷子i道菜a_i,b_i。交换c_i
	f[1][1] = b[1];
	f[1][0] = c[1] + a[1];
	for(int i = 2;i <= n;++ i) {
		f[i][0] = min(f[i - 1][1] + c[i] + a[i],f[i - 1][0] + a[i]);
		f[i][1] = min(f[i - 1][0] + c[i] + b[i],f[i - 1][1] + b[i]);
	}
	printf("%lld",min(f[n][0],f[n][1]));
	return 0;
}

luogu2066 机器分配

题目链接:luogu2066

设状态f[i][j]表示第i个公司分配j台的最佳答案.那么就由dp转移方程.

dp[i][j] =dp[i - 1][j - k] + w[i][k];

输出路径,用一个数组存下路径就OK了.这里要注意的是按字典序进行输出,.

#include <iostream>
#include <cstdio>
#include <queue>
#define ll long long
const int maxN = 20;
const int maxM = 20;
using namespace std;

queue<ll>q;
ll n,m;
ll w[maxN][maxM],f[maxN][maxM],path[maxN][maxM][maxN];

int main() {
	scanf("%lld%lld",&n,&m);
	for(int i = 1;i <= n;++ i ){
		for(int j = 1;j <= m;++ j) {
			scanf("%lld",&w[i][j]);
		}
	}
	for(int i = 1;i <= n;++ i ) {
		for(int j = 0;j <=m ;++ j ) {
			for(int k = 0;k <= j;++ k) {
				if(f[i - 1][k] + w[i][j - k] > f[i][j]) {
					f[i][j] = f[i - 1][k] + w[i][j - k];
					for(int l = 1;l < i;l ++ )path[i][j][l] = path[i - 1][k][l];
					path[i][j][i] = j - k;
				}
			}
		}
	}
	printf("%d\n",f[n][m]);
	for(int i=1;i<=n;i++) cout<<i<<" "<<path[n][m][i]<<endl;
	return 0;
}

luogu 1564膜拜

luogu1564

先求出i - j人数差.设f[i]表示前i人最少的分配的数,然后就有dp方程

f[i] = min(f[i],f[j] + 1)(j满足一定的条件)

#include<cstdio>
#include<algorithm>
using namespace std;

int n,m,a[2501],b[2501],f[2501];

int main()
{
    int i;
    scanf("%d%d",&n,&m);
    for(i=1,x;i<=n;i++)
    {
        scanf("%d",&x);
        if(x == 1)
          a[i] = a[i - 1] + 1,b[i] = b[i - 1];
        else
          a[i] = a[i - 1],b[i] = b[i - 1] + 1;
    }
    for(i = 1;i <= n;i ++)
    {
        f[i] = i;
        for(j = i;j >= 0;j --)
          if(a[i] - a[j] == i - j || b[i] - b[j] == i - j || abs(b[i] - b[j] - a[i] + a[j]) <= m)
            f[i] = min(f[i],f[j]+1);
    }
    printf("%d",f[n]);
    return 0;
}

luogu1115最大字段和

luogu1115

f[i]表示第i个位置的最大字段和.然后得转移方程:f[i] = max(f[i - 1] + a[i],a[i])

用一个变量不断地记录maxn

#include <iostream>
#include <cstdio>
#include <cstring>
const int maxN = 200000 + 7;
using namespace std;

int f[maxN],maxn = -0x7fffffff,a[maxN];

int main() {
	int n;
	scanf("%d",&n);
	for(int i = 1;i <= n;++ i )
		scanf("%d",&a[i]);
	for(int i = 1;i <= n;++ i) {
		f[i] = max(f[i - 1] + a[i],a[i]);
		maxn = max(f[i],maxn);
	}
	printf("%d",maxn);
	return 0;
}

poj2479 -- Maximum sum

poj2479

正反两边最大子段和,然后枚举断点即可.

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxN = 50000 + 7;

int f[maxN],a[maxN],dp[maxN],qwq[maxN],qaq[maxN];

int main() {
	int T,n;
	scanf("%d",&T);
	while(T --) {
		int Ans = -0x7fffffff;
		scanf("%d",&n);
		memset(f,0,sizeof(f));
		memset(dp,0,sizeof(dp));
		memset(qwq,0,sizeof(qwq));
		memset(qaq,0,sizeof(qaq));
		for(int i = 1;i <= n;++ i)
			scanf("%d",&a[i]);
		for(int i = 1;i <= n;++ i)
			f[i] = max(f[i - 1] + a[i],a[i]),qwq[i] = max(f[i],qwq[i - 1]);
		for(int j = n;j >= 1; -- j)
			dp[j] = max(dp[j + 1] + a[j],a[j]),qaq[j] = max(dp[j],qaq [j - 1]);
		for(int i = 1;i <= n;++ i) {
			Ans = max(Ans,qwq[i] + qaq[i + 1]);
		}
		printf("%d\n",Ans);
	}
}