标签:turn http alt pair ini define max return set
题解:状压dp,dp[ s ][ i ]表示当前已经游玩的点(R中的点),且终点是 i 的最小花费。预处理出任意两点之间的最短距离。总的来说,面向数据编程。
。。。。。。。心里有句mmp。调试了一个小时,始终0%,所有数据怒开long long,然后过了,,,,,
#include<bits/stdc++.h> #define ll long long #define P pair<int,int> #define pb push_back #define lson root << 1 #define INF (int)2e9 + 7 #define maxn (int)1e4 + 7 #define rson root << 1 | 1 #define LINF (unsigned long long int)1e18 #define mem(arry, in) memset(arry, in, sizeof(arry)) using namespace std; ll n, m, R; ll d[204][204], dp[1 << 16][16], r[20]; void Inite(){ for(int i = 1; i <= 204; i++){ for(int j = 1; j <= 204; j++) d[i][j] = (i == j ? 0 : INF); } for(int i = 0; i < (1 << 16); i++){ for(int j = 0; j < 16; j++) dp[i][j] = INF; } } void Floyd(){ for(int k = 1; k <= n; k++){ for(int i = 1; i <= n; i++){ for(int j = 1; j <= n; j++){ d[i][j] = min(d[i][j], d[i][k] + d[k][j]); } } } } void Solve(){ for(int i = 0; i < R; i++) dp[1 << i][i] = 0; for(int s = 0; s < (1 << R); s++) { for(int u = 0; u < R; u++) if(s & (1 << u)) { for(int v = 0; v < R; v++) if(!(s & (1 << v))) { dp[s | (1 << v)][v] = min(dp[s | (1 << v)][v], dp[s][u] + d[r[u]][r[v]]); } } } res = dp[(1 << R) - 1][0]; for(int i = 1; i < R; i++) res = min(res, dp[(1 << R) - 1][i]); cout << res << endl; } int main() { cin >> n >> m >> R; for(int i = 0; i < R; i++) cin >> r[i]; Inite(); for(int i = 1; i <= m; i++) { int u, v, w; cin >> u >> v >> w; d[u][v] = d[v][u] = w; } Floyd(); Solve(); return 0; }
标签:turn http alt pair ini define max return set
原文地址:https://www.cnblogs.com/zgglj-com/p/9193817.html