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HDU 4114 Disney's FastPass (状压DP)
题意:给定 n 个区域,然后给定两个区域经过的时间,然后你有 k 个景点,然后给定个每个景点的区域和有票没票的等待时间,从哪些区域能够得到票,问你从景点1开始,最后到景点1,而且要经过看完这k个景点。
析:一个状压DP,dp[s1][s2][i] 表示已经访问了 s1 中的景点,拥有 s2 的票,当前在 i 区域,然后两种转移一种是去下一个景点,另一种是去下一个区域拿票。当时输入,写错了,卡了好长时间。。。。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #include <list> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1e5 + 10; const int mod = 1000; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } int d[55][55]; int dp[1<<8][1<<8][55]; int ft[10], t[10], p[10]; int st[55]; int main(){ int T; cin >> T; for(int kase = 1; kase <= T; ++kase){ int K; scanf("%d %d %d", &n, &m, &K); memset(d, INF, sizeof d); for(int i = 0; i < m; ++i){ int u, v, val; scanf("%d %d %d", &u, &v, &val); --u, --v; d[u][v] = d[v][u] = val; } for(int i = 0; i < n; ++i) d[i][i] = 0; memset(st, 0, sizeof st); for(int i = 0; i < K; ++i){ int tt; scanf("%d %d %d %d", p+i, t+i, ft+i, &tt); --p[i]; for(int j = 0; j < tt; ++j){ int x; scanf("%d", &x); st[x-1] |= 1<<i; } } for(int k = 0; k < n; ++k) for(int i = 0; i < n; ++i) for(int j = 0; j < n; ++j) if(d[i][k] != INF && d[k][j] != INF) d[i][j] = min(d[i][j], d[i][k] + d[k][j]); memset(dp, INF, sizeof dp); dp[0][st[0]][0] = 0; int all = 1<<K; for(int i = 0; i < all; ++i) for(int j = 0; j < all; ++j){ for(int k = 0; k < n; ++k){ if(dp[i][j][k] == INF) continue; for(int l = 0; l < K; ++l){ if(i&(1<<l)) continue; int &tmp = dp[i|(1<<l)][j|st[p[l]]][p[l]]; if(j&(1<<l)) tmp = min(tmp, dp[i][j][k] + d[k][p[l]] + ft[l]); else tmp = min(tmp, dp[i][j][k] + d[k][p[l]] + t[l]); } for(int l = 0; l < n; ++l) dp[i][j|st[l]][l] = min(dp[i][j|st[l]][l], dp[i][j][k] + d[k][l]); } } int ans = dp[all-1][0][0]; for(int i = 0; i < n; ++i) for(int j = 0; j < all; ++j) ans = min(dp[all-1][j][i] + d[i][0], ans); printf("Case #%d: %d\n", kase, ans); } return 0; }
HDU 4114 Disney's FastPass (状压DP)
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