暑假集训 || 区间DP

最新推荐文章于 2025-07-13 21:50:15 发布

转载最新推荐文章于 2025-07-13 21:50:15 发布 · 101 阅读

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原文链接：http://www.cnblogs.com/pinkglightning/p/9508378.html

区间DP

经典石子合并问题V1 复杂度 On3

int a[SZ], sum[SZ], f[SZ][SZ];
int main()
{
    int n;
    scanf("%d", &n);
    for(int i = 1; i <= n; i++)
    {
        scanf("%d", &a[i]);
        sum[i] = sum[i-1] + a[i];
    }
    for(int len = 2; len <= n; len++)
    {
        for(int l = 1; l <= n-len+1; l++)
        {
            int r = l+len-1;
            int ans = INF;
            for(int k = l; k < r; k++)
                ans = min(ans, f[l][k] + f[k+1][r] + sum[r] - sum[l-1]);
            f[l][r] = ans;
        }
    }
    printf("%d\n", f[1][n]);
    return 0;
}

View Code

V2 复杂度 On2

环形问题可以在后面再接一段数组

int main()
{
    int n;
    scanf("%d", &n);
    for(int i = 1; i <= n; i++)
    {
        scanf("%d", &a[i]);
        sum[i] = sum[i-1] + a[i];
        a[i+n] = a[i];
    }
    for(int i = n+1; i <= 2*n; i++) sum[i] = sum[i-1] + a[i];
    for(int i = 0; i <= 2*n; i++)
        for(int j = 0; j <= 2*n; j++)
        {
            if(i == j) f[i][j] = 0, s[i][j] = i;
            else f[i][j] = INF;
        }
    for(int len = 2; len <= n; len++)
    {
        for(int l = 1; l <= 2*n-len+1; l++)
        {
            int r = l+len-1;
            for(int k = s[l][r-1]; k <= s[l+1][r]; k++)
            {
                int ans =  f[l][k] + f[k+1][r] + sum[r] - sum[l-1];
                if(ans < f[l][r])
                {
                    f[l][r] = ans;
                    s[l][r] = k;
                }
            }
        }
    }
    int res = INF;
    for(int i = 1; i <= n; i++)
        res = min(res, f[i][i+n-1]);
    printf("%d\n", res);
    return 0;
}

View Code

V3 复杂度 Onlogn

HDU 3516

给一堆点，在平面内选择一个位置做根，只能向右和向上连向点，问最小的连线总长度

竟然。。是区间DP问题。。。还要四边形优化

发现把两段合并好的树l~k-1 k~r 再合并在一起需要花费cal

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
const int SZ = 2100;
const int INF = 1e9+10;
int a[SZ], sum[SZ], f[SZ][SZ], s[SZ][SZ];
struct node
{
    int x, y;
}pos[SZ];
int cal(int l, int k, int r)
{
    int ans = pos[k].x-pos[l].x + pos[k-1].y-pos[r].y;
    return ans;
}
int main()
{
    int n;
    while(~scanf("%d", &n))
    {
        for(int i = 1; i <= n; i++) scanf("%d %d", &pos[i].x, &pos[i].y);
        for(int i = 0; i <= n; i++)
            for(int j = 0; j <= n; j++)
            {
                if(i == j) f[i][j] = 0, s[i][j] = i;
                else f[i][j] = INF;
            }
        for(int len = 2; len <= n; len++)
        {
            for(int l = 1; l <= n-len+1; l++)
            {
                int r = l+len-1;
                for(int k = s[l][r-1]; k <= s[l+1][r]; k++)
                {
                    int ans =  f[l][k-1] + f[k][r] + cal(l, k, r);
                    if(ans <= f[l][r])
                    {
                        f[l][r] = ans;
                        s[l][r] = k;
                    }
                }
            }
        }
        printf("%d\n", f[1][n]);
    }
    return 0;
}

View Code

POJ 2955 括号匹配

为什么忘性这么大。。

int f[SZ][SZ], s[SZ][SZ];
int main()
{
    char s[111];
    while(1)
    {
        scanf(" %s", s+1);
        if(s[1] == 'e') break;
        int n = strlen(s)-1;
        memset(f, 0, sizeof(f));
        for(int len = 2; len <= n; len++)
            for(int l = 1; l <= n-len+1; l++)
            {
                int r = l+len-1;
                if((s[l] == '(' && s[r] == ')') || (s[l] == '[' && s[r] == ']')) f[l][r] = f[l+1][r-1]+2;
                for(int k = l; k < r; k++)
                    f[l][r] = max(f[l][r], f[l][k] + f[k+1][r]);
            }
        printf("%d\n", f[1][n]);
    }
    return 0;
}

View Code

LightOJ 1422

题意：每一场Party都要穿相应的衣服，一次可以套着穿多件，如果某两场Party衣服一样，同一件就可以接着用，脱下过的衣服就不能再穿了，现在要求最少穿的衣服。

思路： f[i][j] 表示 i-j天里需要的衣服数，考虑区间dp

在l - r天内，若a[l] == a[r] 则第r天可以不穿，f[l][r] = f[l][r-1]

否则第r天要穿，f[l][r] = f[l][r-1]+1

然后枚举l - r之间的k，若a[l] == a[k] 则第k天可以不换，f[l][r] = min(f[l][r], f[l][k-1] + f[k][r])

区间DP的边界问题真的是个玄学orz

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;
const int SZ = 150;
int a[SZ], f[SZ][SZ];
int main()
{
    int T, tt = 0;
    scanf("%d", &T);
    while(T--)
    {
        int n;
        scanf("%d", &n);
        for(int i = 1; i <= n; i++)
            scanf("%d", &a[i]);
        memset(f, 63, sizeof(f));
        for(int i = 1; i <= n; i++) f[i][i] = 1;
        for(int len = 1; len <= n; len++)
            for(int l = 1; l <= n; l++)
            {
                int r = l+len;
                if(r > n) break;
                if(a[l] == a[r]) f[l][r] = f[l][r-1];
                else f[l][r] = f[l][r-1] + 1;
                for(int k = l+1; k < r-1; k++)
                    if(a[l] == a[k])
                        f[l][r] = min(f[l][r], f[l][k-1] + f[k][r]);
            }
        printf("Case %d: %d\n", ++tt, f[1][n]);
    }
    return 0;
}

View Code

转载于:https://www.cnblogs.com/pinkglightning/p/9508378.html