LightOJ 1047 - Neighbor House （dp）

题意:

n<=20个房子涂3色,每个房子涂不同颜色代价不用,相邻房子颜色不能相同,求涂完的最小代价

分析:

煞笔题,dp[i][j]:=前i个房子涂完,且第i个房子颜色为j的最小代价
ans=min{dp[n][i],i∈[0,3]}

代码:

////  Created by TaoSama on 2015-11-13//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int n, a[25][3], dp[25][3];int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    int t; scanf("%d", &t);    int kase = 0;    while(t--) {        scanf("%d", &n);        for(int i = 1; i <= n; ++i)            for(int j = 0; j < 3; ++j)                scanf("%d", &a[i][j]);        memset(dp, 0x3f, sizeof dp);        int ans = INF;        for(int i = 0; i < 3; ++i) dp[1][i] = a[1][i];        for(int i = 2; i <= n; ++i) {            for(int j = 0; j < 3; ++j) {                for(int k = 0; k < 3; ++k) {                    if(j == k) continue;                    dp[i][j] = min(dp[i][j], dp[i - 1][k] + a[i][j]);                }                if(i == n) ans = min(ans, dp[i][j]);            }        }        printf("Case %d: %d\n", ++kase, ans);    }    return 0;}