LightOJ 1047 - Neighbor House(DP)

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题意：n个房子涂色，相邻须不同色。涂色有不同花费，求最小花费。

思路：dp[i][now][last]表示到第i个房子，现在涂now色，上一次涂last色的最小花费。

代码略长，但意思比较清楚。

#include <algorithm>#include <iostream>#include <sstream>#include <cstring>#include <cstdio>#include <vector>#include <string>#include <queue>#include <stack>#include <cmath>#include <set>#include <map>using namespace std;typedef long long LL;#define mem(a, n) memset(a, n, sizeof(a))#define ALL(v) v.begin(), v.end()#define si(a) scanf("%d", &a)#define sii(a, b) scanf("%d%d", &a, &b)#define siii(a, b, c) scanf("%d%d%d", &a, &b, &c)#define pb push_back#define eps 1e-8const int inf = 0x3f3f3f3f, N = 20 + 5, MOD = 1e9 + 7;int T, cas = 0;int n, m;int r[N], g[N], b[N];int dp[N][3][3];  // dp[i][now][last]int main(){#ifdef LOCAL    freopen("/Users/apple/input.txt", "r", stdin);//freopen("/Users/apple/out.txt", "w", stdout);#endif    si(T);    while(T --) {    si(n);    mem(dp, 0x3f);    r[0] = g[0] = b[0] = 0;    mem(dp, 0x3f);    mem(dp[0], 0);    for(int i = 1; i <= n; i ++) siii(r[i], g[i], b[i]);    for(int i = 1; i <= n; i ++) {    dp[i][0][1] = min(dp[i-1][1][0], dp[i-1][1][2]) + r[i];    dp[i][0][2] = min(dp[i-1][2][0], dp[i-1][2][1]) + r[i];    dp[i][1][0] = min(dp[i-1][0][1], dp[i-1][0][2]) + g[i];    dp[i][1][2] = min(dp[i-1][2][0], dp[i-1][2][1]) + g[i];    dp[i][2][0] = min(dp[i-1][0][1], dp[i-1][0][2]) + b[i];    dp[i][2][1] = min(dp[i-1][1][0], dp[i-1][1][2]) + b[i];    }    int ans = inf;    for(int i = 0; i < 3; i ++) for(int j = 0; j < 3; j ++) ans = min(ans, dp[n][i][j]);    printf("Case %d: %d\n", ++ cas, ans);    }        return 0;}

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