[BZOJ1003][ZJOI2006]物流运输（最短路+DP）

可以得到，如果时间在一个区间[l,r]内的运输路线一样，那么这一部分中，一定不能经过的点就是第l天到第r天的所有停用的码头的并集。费用为不经过不能经过的点的1到m的最短路乘以r−l+1。所以这时候可以建立DP模型，f[i]表示到第i天的最小费用，那么有：
f[i]=minij=1(f[j−1]+dis[j,i]∗(i−j+1)+K∗(j>1))
其中dis[j,i]表示在不能经过第j天到第i天内被停用过的码头的情况下，1到m的最短路，f[0]=0。最后结果就是f[n]。
据说有状压的做法。
代码：

#include <cmath>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;inline int read() {    int res = 0; bool bo = 0; char c;    while (((c = getchar()) < '0' || c > '9') && c != '-');    if (c == '-') bo = 1; else res = c - 48;    while ((c = getchar()) >= '0' && c <= '9')        res = (res << 3) + (res << 1) + (c - 48);    return bo ? ~res + 1 : res;}const int N = 3e4 + 5, M = 605, INF = 0x3f3f3f3f;int n, m, K, E, q, ecnt, nxt[N], adj[N], go[N], dis[N], val[N], is[M][M],len, que[N], f[N];bool can[N], vis[N];void add_edge(int u, int v, int w) {    nxt[++ecnt] = adj[u]; adj[u] = ecnt; go[ecnt] = v; val[ecnt] = w;    nxt[++ecnt] = adj[v]; adj[v] = ecnt; go[ecnt] = u; val[ecnt] = w;}int spfa() {    int i; for (i = 1; i <= m; i++) dis[i] = INF;    dis[que[len = 1] = 1] = 0;    for (i = 1; i <= len; i++) {        int u = que[i]; vis[u] = 0;        for (int e = adj[u], v; e; e = nxt[e])            if (can[v = go[e]] && dis[u] + val[e] < dis[v]) {                dis[v] = dis[u] + val[e];                if (!vis[v]) vis[que[++len] = v] = 1;            }    }    return dis[m];}int main() {    int i, j, k, x, y, z; n = read(); m = read(); K = read(); E = read();    for (i = 1; i <= E; i++) x = read(), y = read(), z = read(),        add_edge(x, y, z); q = read();    while (q--) {        x = read(); y = read(); z = read();        is[x][y]++; is[x][z + 1]--;    }    for (i = 1; i <= m; i++) for (j = 2; j <= n; j++) is[i][j] += is[i][j - 1];    for (i = 1; i <= n; i++) {        f[i] = INF; for (j = 1; j <= m; j++) can[j] = 1;        for (j = i; j; j--) {            for (k = 1; k <= m; k++) if (is[k][j]) can[k] = 0;            int val = spfa(); if (val >= INF) continue;            f[i] = min(f[i], f[j - 1] + val * (i - j + 1) + K * (j > 1));        }    }    cout << f[n] << endl;    return 0;}