HDU-4571 travel in time Dynamic Planning

Meaning: Given N points, each point has a stay time, and the satisfaction of the stay. There are m sides, each representing the time required for moving between the two points, now, how much satisfaction can be obtained from a specified point of s to an EIP within the specified t time? The satisfaction of the points that require staying increases sequentially.

Solution: although this is a DP, you should pay special attention to the processing of the two endpoints, because both endpoints can choose to pass by, rather than stay, so you need to virtualize the two endpoints, the two endpoints indicate the starting point and the ending point respectively. The distance between the virtual endpoint and other nodes is also caused by the unidirectional EDGE connection, and the bidirectional edge times out. At the beginning, I decided not to process the end point, but only to determine whether the end point is the end point when I found this node, and to update the end point in two ways, one is access, the other is pass by, and it turns out that it will not work after the event. Therefore, if a passing update enters the queue, it will be determined by the next update as the meaning of the stay, there will be a constraint on increasing satisfaction, and if you do not take the initiative to join the team, it will also be due to the previous update.

The Code is as follows:

#include <cstdlib>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <queue>using namespace std;int N, M, T, S, E, S2, E2;int ct[105];int sa[105];int mp[105][105];int f[105][305];const int INF = 0x3f3f3f3f;void floyd() {    for (int k = 0; k < N; ++k) {        for (int i = 0; i < N; ++i) {            if (mp[i][k] == INF || i == k) continue;            for (int j = 0; j < N; ++j) {                if (mp[k][j] == INF || k == j) continue;                if (mp[i][j] > mp[i][k] + mp[k][j]) {                    mp[i][j] = mp[i][k] + mp[k][j];                }            }        }    }}char vis[105][305];struct Node {    int x, t;}info;int spfa() {    memset(f, 0, sizeof (f));    memset(vis, 0, sizeof (vis));    f[S2][0] = 0;    vis[S2][0] = 1;    info.x = S2, info.t = 0;    queue<Node>q;    q.push(info);    while (!q.empty()) {        Node cur = q.front();        q.pop();        vis[cur.x][cur.t] = 0;        int u = cur.x, v;        for (int v = 0; v < N; ++v) {            if (mp[u][v] == INF || v == u) continue;            int tt = cur.t + ct[v] + mp[u][v];            if (tt <= T && (sa[u] < sa[v] || v == E2)) {                if (f[v][tt] < f[u][cur.t] + sa[v]) {                    f[v][tt] = f[u][cur.t] + sa[v];                    if (!vis[v][tt]) {                        vis[v][tt] = 1;                        info.x = v, info.t = tt;                        q.push(info);                    }                }            }        }    }    int ret = 0;    for (int i = 0; i <= T; ++i) {        ret = max(ret, max(f[E][i], f[E2][i]));    }    return ret;}int main() {    int W;    scanf("%d", &W);    for (int ca = 1; ca <= W; ++ca) {        scanf("%d%d%d%d%d", &N, &M, &T, &S, &E);        memset(mp, 0x3f, sizeof (mp));        for (int i = 0; i < N; ++i) {            scanf("%d", &ct[i]);        }         for (int i = 0; i < N; ++i) {            scanf("%d", &sa[i]);        }        S2 = N, sa[S2] = -1;        E2 = N+1, sa[E2] = 0, ct[E2] = 0;        int a, b, c;        for (int i = 0; i < M; ++i) {            scanf("%d %d %d", &a, &b, &c);            mp[a][b] = mp[b][a] = min(c, mp[a][b]);        }        floyd();        for (int i = 0; i < N; ++i) {            mp[S2][i] = mp[S][i];            mp[i][E2] = mp[i][E];        }        mp[S2][S] = 0;        mp[E][E2] = 0;        N += 2;        printf("Case #%d:\n%d\n", ca, spfa());    }    return 0;}