vijos 1038 添加括号

石子合并类似的问题,要求找出括号的添加方法，最小中间和之和，各个中间和.

设dp[i][j]为i到j之间最小中间和之和

则dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j] + sum[i][j] | i<= k < j) sum[i][j]为i到j的和

base case:dp[i][i] = 0

#include <cstdio>
#include <climits>
#include <memory.h>
using namespace std;
const int MAX = 23;

int dp[MAX][MAX];
int pos[MAX][MAX];
int val[MAX], sum[MAX][MAX];

void print_path(int i, int j){
	if(i == j){
		printf("%d", val[i]);
		return;
	}
	printf("(");
	print_path(i, pos[i][j]);
	printf("+");
	print_path(pos[i][j] + 1, j);
	printf(")");
}

int print_intermedia(int x, int y){
	if(x == y)return val[x];
	int v = print_intermedia(x, pos[x][y]) + print_intermedia(pos[x][y] + 1, y);
	printf("%d ", v);
	return v;
}

int main(int argc, char const *argv[]){
	int N;
	scanf("%d", &N);
	for(int i = 1; i <= N; ++i){
		scanf("%d", &val[i]);
	}
	memset(sum, -1, sizeof(sum));

	for(int i = 1; i <= N; ++i){
		dp[i][i] = 0;
		pos[i][i] = i;
		sum[i][i]= val[i];
		for(int j = i + 1; j <= N; ++j){
			sum[i][j] = sum[i][j - 1] + val[j];
		}
	}
	
	for(int j = 1; j < N; ++j){
		for(int i = 1; i + j <= N; ++i){
			int opt_v = INT_MAX, opt_p = 0;
			for(int k = i + j - 1; k >= i; --k){
				if(opt_v > dp[i][k] + dp[k + 1][i + j] + sum[i][i + j]){
					opt_v = dp[i][k] + dp[k + 1][i + j] + sum[i][i + j];
					opt_p = k;
				}
			}
			pos[i][i + j] = opt_p;
			dp[i][i + j] = opt_v;
		}
	}
	print_path(1, N);
	printf("\n");
	printf("%d\n", dp[1][N]);
	print_intermedia(1, N);
	printf("\n");

	return 0;	
}

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vijos 1038 添加括号

原文：http://blog.csdn.net/zxjcarrot/article/details/21888315