uva 10604

状态压缩  奇怪的是A与B混合 和 B与A 混合得到的热量可能不同

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <set>
#include <stack>
#include <vector>
#include <sstream>
#include <cstring>
#include <string>
#include <map>
#include <queue>
#include <algorithm>
#include <iostream>
#define FFI freopen("in.txt", "r", stdin)
#define maxn 1000100
#define INF 0x3f3f3f3f
#define inf 10000000
#define MOD 1000000007
#define ULL unsigned long long
#define LL long long
#define _setm(houge) memset(houge, INF, sizeof(houge))
#define _setf(houge) memset(houge, -1, sizeof(houge))
#define _clear(houge) memset(houge, 0, sizeof(houge))
using namespace std;

int g[10][10], r[10][10];
int dp[1500][10];
int a[20], n, k;

int main() {
	// FFI;
	int t;
	scanf("%d", &t);
	while(t --) {
		scanf("%d", &n);
		for(int i = 1; i <= n; ++ i) {
			for(int j = 1; j <= n; ++ j) {
				scanf("%d%d", &g[i][j], &r[i][j]);
			}
		}
		scanf("%d", &k);
		for(int i = 0; i < k; ++ i) {
			scanf("%d", &a[i]);
		}
		_setm(dp);
		for(int i = 0; i < k; ++ i) {
			dp[1<<i][a[i]] = 0;
		}
		int tot = (1<<k);
		for(int s = 0; s < tot; ++ s) {
			for(int i = 1; i <= n; ++ i) {
				if(dp[s][i] == INF) continue;
				for(int s2 = 0; s2 < tot; ++ s2) {
					if(s&s2) continue;
					for(int j = 1; j <= n; ++ j) {
						if(dp[s2][j] == INF) continue;
						int S = s|s2, now = g[i][j], v = r[i][j];
						int now2 = g[j][i], v2 = r[j][i];
						dp[S][now] = min(dp[S][now], dp[s][i]+dp[s2][j]+v);
						dp[S][now2] = min(dp[S][now2], dp[s][i]+dp[s2][j]+v2);
					}
				}
			}
		}
		int ans = INF;
		for(int i = 1; i <= n; ++ i) {
			ans = min(ans, dp[tot-1][i]);
		}
		printf("%d\n", ans);
		char a[2];
		scanf("%s", a);
	}
	return 0;
}

  

uva 10604

时间: 2024-10-11 07:04:04

uva 10604的相关文章

小白书关于动态规划

10192 最长公共子序列 http://uva.onlinejudge.org/index.php?option=com_onlinejudge& Itemid=8&page=show_problem&category=114&problem=1133&mosmsg= Submission+received+with+ID+13297616 */ #include <cstdio> #include <string.h> #include&

UVA 562 Dividing coins --01背包的变形

01背包的变形. 先算出硬币面值的总和,然后此题变成求背包容量为V=sum/2时,能装的最多的硬币,然后将剩余的面值和它相减取一个绝对值就是最小的差值. 代码: #include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; #define N 50007 int c[102],d

UVA 10341 Solve It

Problem F Solve It Input: standard input Output: standard output Time Limit: 1 second Memory Limit: 32 MB Solve the equation: p*e-x + q*sin(x) + r*cos(x) + s*tan(x) + t*x2 + u = 0 where 0 <= x <= 1. Input Input consists of multiple test cases and te

UVA 11014 - Make a Crystal(容斥原理)

UVA 11014 - Make a Crystal 题目链接 题意:给定一个NxNxN的正方体,求出最多能选几个整数点.使得随意两点PQ不会使PQO共线. 思路:利用容斥原理,设f(k)为点(x, y, z)三点都为k的倍数的点的个数(要扣掉一个原点O).那么全部点就是f(1),之后要去除掉共线的,就是扣掉f(2), f(3), f(5)..f(n).n为素数.由于这些素数中包括了合数的情况,而且这些点必定与f(1)除去这些点以外的点共线,所以扣掉.可是扣掉后会扣掉一些反复的.比方f(6)在f

[UVa] Palindromes(401)

UVA - 401 Palindromes Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description A regular palindrome is a string of numbers or letters that is the same forward as backward. For example, the string "ABCDED

uva 401.Palindromes

题目链接:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=342 题目意思:给出一段字符串(大写字母+数字组成).判断是否为回文串 or 镜像串 or 回文镜像串 or 什么都不是.每个字母的镜像表格如下 Character Reverse Character Reverse Character Reverse A A M M Y Y B

[2016-02-19][UVA][129][Krypton Factor]

UVA - 129 Krypton Factor Time Limit: 3000MS Memory Limit: Unknown 64bit IO Format: %lld & %llu Submit Status Description You have been employed by the organisers of a Super Krypton Factor Contest in which contestants have very high mental and physica

[2016-02-03][UVA][514][Rails]

时间:2016-02-03 22:24:52 星期三 题目编号:UVA 514 题目大意:给定若干的火车(编号1-n),按1-n的顺序进入车站, 给出火车出站的顺序,问是否有可能存在 分析:    FIFO,用栈模拟一遍即可, 方法:    根据输入的顺序,从1-n开始,当前操作的为i 如果i是当前对应的编号,那么直接跳过(进入B) 如果不是,根据当前需求的编号,小于i,就从栈顶弹出一个元素, 看这个元素是否是需求的,是则继续.否则NO 1 2 3 4 5 6 7 8 9 10 11 12 13

uva 11584 Partitioning by Palindromes 线性dp

// uva 11584 Partitioning by Palindromes 线性dp // // 题目意思是将一个字符串划分成尽量少的回文串 // // f[i]表示前i个字符能化成最少的回文串的数目 // // f[i] = min(f[i],f[j-1] + 1(j到i是回文串)) // // 这道题还是挺简单的,继续练 #include <algorithm> #include <bitset> #include <cassert> #include <