题意 你要从第1个城市到第N个城市去 有m条路 每条路用a, b, c, p, r 表示 你从第a个城市到第b个城市时 若之前经过或现在位于第c个城市 过路费就是p元 否则就是r元 求你到达第N个城市最少用多少过路费
由于最多只有10个城市 10条路 这个题就变得很简单了 直接暴力dfs就行 可以用状态压缩来存储已经走过了哪些城市 由于最多只有10条路 从某个城市出发要一条 回这个城市也要一条 所以一个城市最多经过5次 这个可以作为dfs的结束条件
#include
#include
#include
using namespace std; const int N = 11, INF = 2333333; int n, m, ans, vis[N]; struct road{ int a, b, c, p, r; } rd[N]; //当前所在城市, 到过哪些城市, 当前已经用了多少过路费 void dfs(int p, int s, int v) { if(vis[p] > 5) return; s = s | 1 << (p - 1); if(p == n) { ans = min(ans, v); return; } for(int i = 0; i < m; ++i) { if(rd[i].a != p) continue; ++vis[rd[i].b]; if(s & 1 << (rd[i].c - 1)) //到过城市c dfs(rd[i].b, s, v + rd[i].p); else dfs(rd[i].b, s, v + rd[i].r); --vis[rd[i].b]; //回溯 } } int main() { while(~scanf(%d%d, &n, &m)) { for(int i = 0; i < m; ++i) scanf(%d%d%d%d%d, &rd[i].a, &rd[i].b, &rd[i].c, &rd[i].p, &rd[i].r); ans = INF; memset(vis, 0, sizeof(vis)); dfs(1, 0, 0); if(ans == INF) puts(impossible); else printf(%d , ans); } return 0; }
Paid Roads
Description A network of m roads connects N cities (numbered from 1 to N). There may be more than one road connecting one city with another. Some of the roads are paid. There are two ways to pay for travel on a paid road i from city ai to city bi: in advance, in a city ci (which may or may not be the same as ai);after the travel, in the city bi. The payment is Pi in the first case and Ri in the second case. Write a program to find a minimal-cost route from the city 1 to the city N. Input The first line of the input contains the values of N and m. Each of the following m lines describes one road by specifying the values of ai, bi, ci, Pi, Ri (1 ≤ i ≤ m). Adjacent values on the same line are separated by one or more spaces. All values are integers, 1 ≤ m, N ≤ 10, 0 ≤ Pi , Ri ≤ 100, Pi ≤ Ri (1 ≤ i ≤ m). Output The first and only line of the file must contain the minimal possible cost of a trip from the city 1 to the city N. If the trip is not possible for any reason, the line must contain the word ‘impossible’. Sample Input
4 5
1 2 1 10 10
2 3 1 30 50
3 4 3 80 80
2 1 2 10 10
1 3 2 10 50 Sample Output
110 Source Northeastern Europe 2002, Western Subregion |