uvalive 4256(dp)

标签：uva   dp   

题意：有从1到n的数字组成一个无向连通图，给出了连通情况，然后给出一个数字序列，问这个序列要求相邻的点要么相等要么在图中是直接连通的，问最少修改序列中的几个点可以让序列满足要求。

题解：f[i][j]表示前i个数组组成的序列以数字j结尾的最少修改点，那么f[i][j] = min{f[i][j]，f[i - 1][k] + (d[i] != j)}，此时j==k或g[j][k] == 1。最后f[len][k]所有数字过一遍选出最大值就可以了。

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int N = 205;
const int INF = 0x3f3f3f3f;
int n, m, g[N][N], d[N], f[N][N];

int main() {
	int t;
	scanf("%d", &t);
	while (t--) {
		scanf("%d%d", &n, &m);
		memset(g, 0, sizeof(g));
		int a, b;
		for (int i = 0; i < m; i++) {
			scanf("%d%d", &a, &b);
			g[a][b] = g[b][a] = 1;
		}
		scanf("%d", &m);
		for (int i = 1; i <= m; i++)
			scanf("%d", &d[i]);
		memset(f, INF, sizeof(f));
		for (int i = 1; i <= n; i++)
			f[1][i] = (i != d[1]);
		for (int i = 2; i <= m; i++)
			for (int j = 1; j <= n; j++)
				for (int k = 1; k <= n; k++)
					if (j == k || g[j][k])
						f[i][j] = min(f[i][j], f[i - 1][k] + (j != d[i]));
		int res = INF;
		for (int i = 1; i <= n; i++)
			res = min(res, f[m][i]);
		printf("%d\n", res);
	}
	return 0;
}

uvalive 4256(dp)

标签：uva   dp   

原文地址：http://blog.csdn.net/hyczms/article/details/45035409