hdoj 2437 Jerboas

类型：拓扑排序

题目：http://acm.hdu.edu.cn/showproblem.php?pid=2437

来源：2008 Asia Chengdu Regional Contest Online

思路：题目是求到某个满足条件的节点的最短路且路径长度为k的倍数。使用二维数组dist[i][j]记录当从源点到当前节点i且满足路径长度 % k = j时的最短路径长度

则最终结果为dist[i][0]，i为P节点且长度最小。

因为该图为有向无环图，所以从i点到j点路径唯一，v节点的值只能由其前驱更新。所以可以以拓扑序列的顺序更新图中节点的dist值。

若dist[u][i]已知，则可以用dist[u][i] + w 更新f[v][(i + c) % k]的最优值。根据拓扑顺序的性质，知道递推式满足无后效性。

初值dist[s][0] = 0。

// hdoj 2437 Jerboas// re ac 109MS 4464K#include <iostream>#include <string>#include <queue>#include <stack>#include <algorithm>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>using namespace std;#define FOR(i,a,b) for(i = (a); i < (b); ++i)#define FORE(i,a,b) for(i = (a); i <= (b); ++i)#define FORD(i,a,b) for(i = (a); i > (b); --i)#define FORDE(i,a,b) for(i = (a); i >= (b); --i)#define CLR(a,b) memset(a,b,sizeof(a))#define PB(x) push_back(x)const int MAXN = 1010;const int MAXM = 20010;const int hash_size = 25000002;const int INF = 0x7f7f7f7f;char str[MAXN];int cnt, cas = 1;int n, m, s, k;int head[MAXN], ru[MAXN], dist[MAXN][MAXN];struct edge {    int v, w, nxt;}e[MAXM];void addedge(int u, int v, int w) {    e[cnt].v = v;    e[cnt].w = w;    e[cnt].nxt = head[u];    head[u] = cnt++;}void topo() {    int i, j, u;    dist[s][0] = 0;    while(1) {        FORE(u, 1, n)            if(ru[u] == 0)                break;        if(u > n)            break;        for(i = head[u]; i != -1; i = e[i].nxt) {            int v = e[i].v;            --ru[v];            FOR(j, 0, k) {                if(dist[u][j] == -1)                    continue;                int len = dist[u][j] + e[i].w;                int rem = len % k;                if(dist[v][rem] == -1 || dist[v][rem] > len)                    dist[v][rem] = len;            }        }        ru[u] = -1;    }}void solve() {    int i;    int enddy = -1, enddz = -1;    topo();    FORE(i, 1, n) {        int tmpy = dist[i][0];        if(str[i] == 'P' && tmpy != -1 && (enddy == -1 || enddz > tmpy))            enddy = i, enddz = tmpy;    }    if(enddy == -1)        printf("Case %d: -1 -1\n", cas++);    else        printf("Case %d: %d %d\n", cas++, enddz, enddy);}void init() {    int i, u, v, w;    CLR(head, -1);    CLR(ru, 0);    CLR(dist, -1);    cnt = 0;    scanf("%d %d %d %d", &n, &m, &s, &k);    scanf("%s", str + 1);    FORE(i, 1, m) {        scanf("%d %d %d", &u, &v, &w);        addedge(u, v, w);        ++ru[v];    }}int main() {    int t;    scanf("%d", &t);    while(t--) {        init();        solve();    }    return 0;}