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POJ 3123

陆公子521 2022-12-07 阅读 139


​​https://cn.vjudge.net/problem/POJ-3123​​

题意:

n 个城市,m 条路,给定八个点(也就是四对),使每队点连通且总权和最小。

分析:

dp[i][j] 表示 i 状态下以 j 为起点的最小总权和。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <bitset>
#include <map>
typedef long long int ll;
const int MOD = (int)1e9 + 7;
const int INF = 99999999;
using namespace std;

int n, m;
map<string, int> city;
int dp1[1 << 8][35];
int dp2[1 << 8];
int d[35][35];
int vis[35];

bool check(int s)
{
for (int i = 0; i < 4; i++)
{
if (((s & 3) != 3) && ((s & 3) != 0))
return false;

s >>= 2;
}

return true;
}

int main()
{
while (~scanf("%d%d", &n, &m) && (n || m))
{
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
d[i][j] = (i == j) ? 0 : INF;

string s1, s2;
int w;
for (int i = 0; i < n; i++)
{
cin >> s1;
city[s1] = i;
}
for (int i = 0; i < m; i++)
{
cin >> s1 >> s2 >> w;
int u = city[s1];
int v = city[s2];

if (w < d[u][v])
d[u][v] = d[v][u] = w;
}

// Floyd
for (int k = 0; k < n; k++)
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
d[i][j] = min(d[i][j], d[i][k] + d[k][j]);

for (int i = 0; i < 8; i++)
{
cin >> s1;

for (int j = 0; j < n; j++)
dp1[1 << i][j] = d[j][city[s1]];
}

for (int i = 0; i < (1 << 8); i++)
{
if (!(i & (i - 1)))
continue;

for (int j = 0; j < n; j++)
{
dp1[i][j] = INF;

for (int sub = (i - 1) & i; sub != 0; sub = (sub - 1) & i)
dp1[i][j] = min(dp1[i][j], dp1[sub][j] + dp1[i - sub][j]);
}

memset(vis, 0, sizeof(vis));

int min_w, min_i;
for (int j = 0; j < n; j++)
{
min_w = INF;

for (int k = 0; k < n; k++)
{
if (dp1[i][k] < min_w && !vis[k])
{
min_w = dp1[i][k];
min_i = k;
}
}

vis[min_i] = 1;

for (int k = 0; k < n; k++)
dp1[i][min_i] = min(dp1[i][min_i], dp1[i][k] + d[k][min_i]);
}
}

for (int i = 0; i < (1 << 8); i++)
{
dp2[i] = INF;

for (int j = 0; j < n; j++)
dp2[i] = min(dp2[i], dp1[i][j]);
}

for (int i = 0; i < (1 << 8); i++)
{
if (check(i))
{
for (int j = i; j != 0; j = (j - 1) & i)
{
if (check(j))
dp2[i] = min(dp2[i], dp2[j] + dp2[i - j]);
}
}
}

printf("%d\n", dp2[(1 << 8) - 1]);
}
return 0;
}


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