暑假集训 || 区间DP

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

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

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 = ;
const int INF = 1e9+;
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-].y-pos[r].y;
    return ans;
}
int main()
{
    int n;
    while(~scanf("%d", &n))
    {
        for(int i = ; i <= n; i++) scanf("%d %d", &pos[i].x, &pos[i].y);
        for(int i = ; i <= n; i++)
            for(int j = ; j <= n; j++)
            {
                if(i == j) f[i][j] = , s[i][j] = i;
                else f[i][j] = INF;
            }
        for(int len = ; len <= n; len++)
        {
            for(int l = ; l <= n-len+; l++)
            {
                int r = l+len-;
                for(int k = s[l][r-]; k <= s[l+][r]; k++)
                {
                    int ans =  f[l][k-] + f[k][r] + cal(l, k, r);
                    if(ans <= f[l][r])
                    {
                        f[l][r] = ans;
                        s[l][r] = k;
                    }
                }
            }
        }
        printf("%d\n", f[][n]);
    }
    return ;
}

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

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 = ;
int a[SZ], f[SZ][SZ];
int main()
{
    int T, tt = ;
    scanf("%d", &T);
    while(T--)
    {
        int n;
        scanf("%d", &n);
        for(int i = ; i <= n; i++)
            scanf("%d", &a[i]);
        memset(f, , sizeof(f));
        for(int i = ; i <= n; i++) f[i][i] = ;
        for(int len = ; len <= n; len++)
            for(int l = ; l <= n; l++)
            {
                int r = l+len;
                if(r > n) break;
                if(a[l] == a[r]) f[l][r] = f[l][r-];
                else f[l][r] = f[l][r-] + ;
                for(int k = l+; k < r-; k++)
                    if(a[l] == a[k])
                        f[l][r] = min(f[l][r], f[l][k-] + f[k][r]);
            }
        printf("Case %d: %d\n", ++tt, f[][n]);
    }
    return ;
}