POJ 3411 Paid Roads

DescriptionA network of m roads connects N cities (numbered from 1 toN). 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 roadi from city ai to city bi:in advance, in a city ci (which may or may not be the same asai);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.InputThe 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 ofai, 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).OutputThe first and only line of the file must contain the minimal possible cost of a trip from the city 1 to the cityN. 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

解题思路：一开始误以为每条边只能走一次，然后果断用了状压DP来搞这题，结果交的时候WA到死，无奈去POJ看discuss，看有个人的一组数据中表示同一条边可能经过多次，那我的状压DP很显然是错误的。后来网上有人总结每个节点的次数最多能能访问的次数，因此我们完全可以采用dfs来进行搜索即可，给定每个节点限制访问次数为3次，然后在所有可能的路径中选取最小值即可，只到每个节点最多能够访问的次数为3次，那这道题就水到家了，但自己就是想不到，努力刷体提高自己的水平吧。

参考博客：http://blog.csdn.net/lyy289065406/article/details/6689310

#include <ctime>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <string>#include <vector>#include <deque>#include <list>#include <stack>#include <queue>#include <map>#include <set>#include <numeric>#include <utility>#include <algorithm>#include <functional>using namespace std;const int inf = 0x3f3f3f3f;const int maxn = 110;struct Edge {    int a, b, c, p, r, next;    Edge() { }    Edge(int t_a, int t_b, int t_c, int t_p, int t_r, int t_next) : a(t_a), b(t_b), c(t_c), p(t_p), r(t_r), next(t_next) { }}edges[maxn];int head[20];int edge_sum;void init_graph() {    edge_sum = 0;    memset(head, -1, sizeof(head));}void add_edge(int a, int b, int c, int p, int r) {    edges[edge_sum].a = a;    edges[edge_sum].b = b;    edges[edge_sum].c = c;    edges[edge_sum].p = p;    edges[edge_sum].r = r;    edges[edge_sum].next = head[a];    head[a] = edge_sum++;}int vis[20];int ans;int n, m;void dfs(int u, int cost) {    if(ans <= cost) return ;    if(u == n) {        ans = min(ans, cost);        return ;    }    for(int i = head[u]; i != -1; i = edges[i].next) {        int v = edges[i].b;        if(vis[v] < 3) {            vis[v]++;            int t = edges[i].r;            if(vis[edges[i].c]) {                t = min(t, edges[i].p);            }            dfs(v, cost + t);            vis[v]--;        }    }    return ;}int main() {    //freopen("aa.in", "r", stdin);    int a, b, c, p, r;    while(scanf("%d %d", &n, &m) != EOF) {    init_graph();    for(int i = 0; i < m; ++i) {        scanf("%d %d %d %d %d", &a, &b, &c, &p, &r);        add_edge(a, b, c, p, r);    }    ans = inf;    dfs(1, 0);    if(ans < inf) {        printf("%d\n", ans);    } else {        printf("impossible\n");    }    }    return 0;}/*struct Edge {    int a, b, c, p, r;    Edge() { }    Edge(int t_a, int t_b, int t_c, int t_p, int t_r) : a(t_a), b(t_b), c(t_c), p(t_p), r(t_r) { }}E[20];int dp[(1<<10)+10][15];int vh[(1<<10)+10];int n, m;int main() {    freopen("aa.in", "r", stdin);    while(scanf("%d %d", &n, &m) != EOF) {    for(int i = 0; i < m; ++i) {        scanf("%d %d %d %d %d", &E[i].a, &E[i].b, &E[i].c, &E[i].p, &E[i].r);        E[i].a--;        E[i].b--;        E[i].c--;    }    memset(vh, 0, sizeof(vh));    for(int i = 0; i < (1<<m); ++i) {        for(int j = 0; j < m; ++j) {            if(i&(1<<j)) {                vh[i] |= (1<<E[j].a);                vh[i] |= (1<<E[j].b);            }        }    }    memset(dp, inf, sizeof(dp));    dp[0][0] = 0;    for(int i = 1; i < (1<<m); ++i) {        for(int j = 0; j < n; ++j) {            if(!(vh[i]&(1<<j))) continue;            for(int k = 0; k < m; ++k) {                if(E[k].b == j && (i&(1<<k)) && dp[i^(1<<k)][E[k].a] < inf) {                    dp[i][j] = min(dp[i][j], dp[i^(1<<k)][E[k].a] + E[k].r);                    if(vh[i^(1<<k)]&(1<<E[k].c)) {                        dp[i][j] = min(dp[i][j], dp[i^(1<<k)][E[k].a] + E[k].p);                    }                }            }        }    }    int ans = inf;    for(int i = 0; i <(1<<m); ++i) {        ans = min(ans, dp[i][n-1]);    }    if(ans < inf) {        printf("%d\n", ans);    } else {        printf("impossible\n");    }    }    return 0;}*/