2014 ACM/ICPC 亚洲区 北京站

题目链接  2014北京区域赛

Problem A

Problem B

直接DFS+剪枝

剪枝条件：当前剩余的方块数量cnt < 2 * max{a[i]} - 1，则停止往下搜。

因为这样搜下去接下来肯定会出现相邻方块颜色相同的情况。

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b)	for (int i(a); i >= (b); --i)
#define MP		make_pair
#define fi		first
#define se		second


typedef long long LL;

const int N = 205;

int f[N][N];
int a[N], b[N];
int n;
int T;
int ca = 0;


int dp(int l, int r){
	if (l > r) return 0;
	if (~f[l][r]) return f[l][r];
	int &ret = f[l][r];

	ret = 1 << 30;
	rep(i, l, r) ret = min(ret, dp(l, i - 1) + a[i] + b[l - 1] + b[r + 1] + dp(i + 1, r));
	return ret;
}


int main(){

	scanf("%d", &T);
	while (T--){
		memset(f, -1, sizeof f);
		scanf("%d", &n);
		memset(a, 0, sizeof a);
		memset(b, 0, sizeof b);
		rep(i, 1, n) scanf("%d", a + i);
		rep(i, 1, n) scanf("%d", b + i);
		printf("Case #%d: %d\n", ++ca, dp(1, n));
	}

	return 0;
}

 

Problem C

Problem D

考虑区间DP

$f[i][j] = min(f[i][k-1] + a[k] + b[k - 1] + b[k + 1] + f[k + 1][j])$

边界条件处理并不麻烦，

记忆化搜索就可以了。

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b)	for (int i(a); i >= (b); --i)
#define MP		make_pair
#define fi		first
#define se		second


typedef long long LL;

const int N = 205;

int f[N][N];
int a[N], b[N];
int n;
int T;
int ca = 0;


int dp(int l, int r){
	if (l > r) return 0;
	if (~f[l][r]) return f[l][r];
	int &ret = f[l][r];

	ret = 1 << 30;
	rep(i, l, r) ret = min(ret, dp(l, i - 1) + a[i] + b[l - 1] + b[r + 1] + dp(i + 1, r));
	return ret;
}


int main(){

	scanf("%d", &T);
	while (T--){
		memset(f, -1, sizeof f);
		scanf("%d", &n);
		memset(a, 0, sizeof a);
		memset(b, 0, sizeof b);
		rep(i, 1, n) scanf("%d", a + i);
		rep(i, 1, n) scanf("%d", b + i);
		printf("Case #%d: %d\n", ++ca, dp(1, n));
	}

	return 0;
}

 

 

Problem E

Problem F

Problem G

Problem H

Problem I

Problem J

Problem K