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;
}