Topcoder 720 div2-C RainbowGraph

题目描述

https://community.topcoder.com/stat?c=problem_statement&pm=14667

解题思路

查看其它同学的解题思路时发现日本的同学把这种dp叫做bitDp，看不懂日语…，感觉跟状态压缩是一个意思。

首先求出颜色相同的团内一共有多少种走法，进而求出在不同的团下共有多少种走法。
在求不同团共有多少种走法的时候，如果仅考虑当路径从该团走出时是哪个点记录状态，会比枚举加和所有的入口、出口点循环少。所以需要先计算出来。

状态一定要放在循环最外层保证在内部循环时改变后的状态可以再后续遍历到。

代码

import java.util.ArrayList;/** * Created by liang on 2017/11/20. */public class RainbowGraph {    private final long mod = 1000_000_007;    public  int countWays(int[] color, int[] a, int[] b){        ArrayList<ArrayList<Integer>> c = new ArrayList<ArrayList<Integer>>();        long[][][] dp = new long[1<<10][10][10];        int al = 0;        int[][] map = new int[color.length][color.length];        for (int i = 0; i < 10; i++) {            c.add(new ArrayList<>());        }        for (int i = 0; i < color.length; i++) {            c.get(color[i]).add(i);        }        for (int i = 0; i < a.length; i++) {            map[a[i]][b[i]] = 1;            map[b[i]][a[i]] = 1;        }        for (int i = 0; i < 10; i++) {            ArrayList<Integer> tmp = c.get(i);            if (tmp.size() == 0) continue;            for (int j = 0; j < tmp.size(); j++) {                long[][] cnt = new long[1<<10][tmp.size()];                cnt[1<<j][j] = 1;                for (int k = 0; k < 1<<tmp.size(); k++) {                    for (int l = 0; l < tmp.size(); l++) {                        if (cnt[k][l]==0) continue;                        for (int m = 0; m < tmp.size(); m++) {                            if( (k>>m & 1) == 0 && map[tmp.get(l)][tmp.get(m)] == 1) {                                cnt[k | 1<<m ][m] = (cnt[k | 1<<m][m] + cnt[k][l])%this.mod;                            }                        }                    }                }                for (int k = 0; k < tmp.size(); k++) {                    dp[i][j][k] = cnt[(1<<tmp.size()) - 1][k];                }            }        }        long[][] res = new long[1<<10][100];        for (int i = 0; i < 10; i++) {            ArrayList<Integer> tmp = c.get(i);            for (int j = 0; j < tmp.size(); j++) {                for (int k = 0; k < tmp.size(); k++) {                    res[1<<i][tmp.get(j)] = (res[1<<i][tmp.get(j)] + dp[i][j][k])%this.mod;                }            }        }        for (int j = 1; j < 1 << 10; j++) {            for (int i = 1; i < color.length; i++) {                if (res[j][i] == 0) {                    continue;                }                for (int k = 0; k < 10; k++) {                    if ((j >> k & 1) == 0) {                        ArrayList<Integer> tmp = c.get(k);                        for (int l = 0; l < tmp.size(); l++) {                            if (map[i][tmp.get(l)] == 1) {                                for (int m = 0; m < tmp.size(); m++) {                                    res[j | 1 << k][tmp.get(m)] = (res[j | 1 << k][tmp.get(m)] + res[j][i]*dp[k][l][m]) % this.mod;                                }                            }                        }                    }                }            }        }        for (int i = 0; i < 10; i++) {            if(c.get(i).size()!=0){                al |= 1<<i;            }        }        long ans = 0;        for (int i = 0; i < color.length; i++) {            ans = (ans + res[al][i])%this.mod;        }        return (int)ans;    }    public static void main(String[] args) {        RainbowGraph r = new RainbowGraph();        int[] color = new int[]{0,0,0,1,1,1,2,2,2};        int[] a = new int[]{0,1,2,3,4,5,6,7,8,0,3,6};        int[] b = new int[]{1,2,0,4,5,3,7,8,6,3,6,0};        int ans = r.countWays(color, a, b);        System.out.println(ans);    }}