Codeforces 711c 简单dp

题目：http://codeforces.com/problemset/problem/711/C

题意：

有n棵树，m（1-m）种颜色，要求划分成k组，每组是连续的同一种颜色的树。
刚开始树有的已经涂上色了，有的没涂色用0表示。第i棵树涂上第j种颜色需要颜料w[i][j]。已经涂色的不能再涂，问最少需要多少颜料？

分析：

这题是比较简单的dp
d[i][j][k]表示第i个树涂上第j种颜色形成k组所花费的最少颜料
如果当前树已经有了颜色，那么d[i][c[i]][k]=min{ d[i-1][c[i]][k], d[i-1][j!=c[i]][k-1] }
否则d[i][j][k]=min{ d[i-1][j][k], d[i-1][x!=j][k-1] }+w[i][j]
其实想法就是如果当前涂色个上一个涂色一样，那么就d[i][j][k]由d[i-1][j][k]得到，如果不一样，由d[i-1][x!=j][k-1]得到
时间复杂度O(n^4)，优化一下可以O(n^3)

#include<bits/stdc++.h>
using namespace std;
#define rep(i,s,t) for(int i=(s);i<(t);i++)
#define per(i,t,s) for(int i=(t);i>=(s);i--)
#define fi first
#define se second
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int>pii;
const ll INF = 1e15 + 9;
const int N = 100 + 9;
ll d[N][N][N], w[N][N];
int c[N], n, m, K;
int main() {
    //freopen ("f.txt", "r", stdin);

    scanf ("%d%d%d", &n, &m, &K);
    rep (i, 1, n + 1) scanf ("%d", &c[i]);
    rep (i, 1, n + 1) rep (j, 1, m + 1) scanf ("%I64d", &w[i][j]);
    rep (i, 0, n + 1) rep (j, 0, m + 1) rep (k, 0, K + 1) d[i][j][k] = INF;

    d[0][0][0] = 0;
    rep (i, 1, n + 1) {
        if (c[i] != 0) {
            rep (k, 1, K + 1) {
                d[i][c[i]][k] = min (d[i][c[i]][k], d[i - 1][c[i]][k]);
                ll minn = INF;
                rep (z, 0, m + 1) if (z != c[i]) minn = min (minn, d[i - 1][z][k - 1]);
                d[i][c[i]][k] = min (d[i][c[i]][k], minn);
            }
        } else
            rep (k, 1, K + 1) {
            rep (j, 1, m + 1) {
                d[i][j][k] = min (d[i][j][k], d[i - 1][j][k] + w[i][j]);
                ll minn = INF;
                rep (z, 0, m + 1) if (z != j) minn = min (minn, d[i - 1][z][k - 1]);
                d[i][j][k] = min (d[i][j][k], minn + w[i][j]);
            }
        }
    }
    ll ans = INF;
    rep (i, 1, m + 1) ans = min (ans, d[n][i][K]);
    if (ans == INF) printf ("-1\n");
    else printf ("%I64d\n", ans);
    return 0;
}