LightOJ - 1047 Neighbor House(DP)

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题目大意：给你N个相邻的点，现在要求你将这些点染色(红，蓝，绿)，使得任意相邻的点的颜色不一样，且染色的费用达到最小

解题思路：用dp[i][j]表示前i个点都染色了，且第i个点染成j的颜色需要的最小费用
则dp[i][j] = min(dp[i][k]) + val[i][j]

#include <cstdio>#include <algorithm>#include <cstring>using namespace std;const int N = 30;int mat[N][3];int dp[N][3];int n, cas = 1;void init() {    scanf("%d", &n);    for (int i = 0; i < n; i++)        for (int j = 0; j < 3; j++)            scanf("%d", &mat[i][j]);}void solve() {    memset(dp, 0x3f, sizeof(dp));    dp[0][0] = mat[0][0];    dp[0][1] = mat[0][1];    dp[0][2] = mat[0][2];    for (int i = 1; i < n; i++)         for (int j = 0; j < 3; j++)             for (int k = 0; k < 3; k++)                if (j != k) dp[i][j] = min(dp[i][j], dp[i - 1][k] + mat[i][j]);    printf("Case %d: %d\n", cas++, min(min(dp[n - 1][0], dp[n - 1][1]), dp[n - 1][2]));}int main() {    int test;    scanf("%d", &test);    while (test--) {        init();        solve();    }    return 0;}

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