[BZOJ 1003] 物流运输

最新推荐文章于 2018-11-11 18:44:24 发布

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最新推荐文章于 2018-11-11 18:44:24 发布

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原文链接：http://www.cnblogs.com/miecoku/p/9494747.html

本文提供了一个BZOJ1003问题的详细解决方案，采用C++实现，涉及图遍历、最短路径算法等核心概念。通过递归深度优先搜索(DFS)寻找所有可能的状态并计算最小代价。

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#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int N = 1 << 21;

struct Edge {
    int v, d;
    Edge *next;
    Edge (int v, int d, Edge *next) : v(v), d(d), next(next) {}
    Edge () {}
}*head[101];
int n, m, cnt, e, K, tp[N], F[N], dis[N];
void dfs(int u, int p) {
    if( u == m) {
        if( !F[p]) tp[++ cnt] = p; F[p] = 1; return ;
    }
    for (Edge *now = head[u]; now; now = now->next) if( dis[p|(1<<now->v-1)] > dis[p]+now->d) 
    dis[p|(1<<now->v-1)] = dis[p]+now->d, dfs(now->v, p|(1<<now->v-1));
}
int ok[101], G[30][102], f[101];
int main() {
#ifndef ONLINE_JUDGE
    freopen("bzoj1003.in", "r", stdin);
#endif
    scanf("%d%d%d%d", &n, &m, &K, &e); 
    for ( int i = 1; i <= n; ++ i) ok[i] = (1 << m) - 1;
    for ( int i = 1, a, b, c; i <= e; ++ i) {
        scanf("%d%d%d", &a, &b, &c);
        head[a] = new Edge(b, c, head[a]); head[b] = new Edge(a, c, head[b]);
    }
    memset(dis, 0x3f, sizeof(dis)); dis[1] = 0; dfs(1, 1);
    scanf("%d", &e);
    for ( int i = 1, a, b, c; i <= e; ++ i) 
        scanf("%d%d%d", &c, &a, &b), G[c][a]++, G[c][b+1]--;
    for ( int i = 2; i <= m; ++ i) 
        for ( int k = 1; k <= n; ++ k) 
            G[i][k] += G[i][k-1], ok[k] ^= ((G[i][k] != 0) << i-1);

    memset(f, 0x3f, sizeof(f)); f[0] = -K;
//    for ( int i = 1; i <= n; ++ i) printf("%d ", ok[i]); puts(""); // de bug
//    for ( int i = 1; i <= cnt; ++ i) printf("%d ", tp[i]); puts(""); // de bug
//    printf("%d\n", cnt);
    for ( int i = 1, l; i <= n; ++ i) 
        for ( int k = 1; k <= cnt; ++ k) {
            l = i;
            for ( ; l && ((ok[l]&tp[k]) == tp[k]); -- l) f[i] = min(f[i], f[l-1]+dis[tp[k]]*(i-l+1)+K);
        }
    printf("%d\n", f[n]); 
    return 0;
}