BZOJ 2200 道路与航线 (算竞进阶习题)

dijkstra + 拓扑排序

这道题有负权边,但是卡了spfa,所以我们应该观察题目性质。
负权边一定是单向的,且不构成环,那么我们考虑先将正权边连上。然后dfs一次找到所有正权边构成的联通块,将他们看成点,那么负权边和这些点就构成了一张DAG。
对于DAG,我们可以拓扑排序,在排序的过程中,我们把入度为0的联通块里的所有点松弛一次,如果访问到联通块外的点,就让其入度减1,然后重复在拓扑排序中跑最短路的过程即可得到答案。

输出答案的时候注意一个坑,因为存在负权边,当有的点不能从起点达到的时候,INF可能被负权边松弛。。

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
using namespace std;
typedef long long ll;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int X = 0, w = 0; char ch = 0;
    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
    return w ? -X : X;
}
inline int gcd(int a, int b){ return a % b ? gcd(b, a % b) : b; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
template<typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template<typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template<typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
    return ans;
}

const int N = 40005;
int t, r, p, s, cnt, tot, head[N], c[N], d[N], dist[N];
bool vis[N];
struct Edge { int v, next, w; } edge[N<<5];

void addEdge(int a, int b, int w){
    edge[cnt].v = b, edge[cnt].w = w, edge[cnt].next = head[a], head[a] = cnt ++;
}

void dfs(int s){
    vis[s] = true;
    c[s] = tot;
    for(int i = head[s]; i != -1; i = edge[i].next){
        int u = edge[i].v;
        if(vis[u]) continue;
        dfs(u);
    }
}

void topSort(){
    full(dist, INF);
    queue<int> q;
    for(int i = 1; i <= tot; i ++){
        if(!d[i]) q.push(i);
    }
    dist[s] = 0;
    while(!q.empty()){
        full(vis, false);
        int cur = q.front(); q.pop();
        priority_queue< pair<int, int>, vector< pair<int, int> >, greater< pair<int, int> > > pq;
        for(int i = 1; i <= t; i ++){
            if(c[i] == cur) pq.push(make_pair(dist[i], i));
        }
        while(!pq.empty()){
            int f = pq.top().second, dis = pq.top().first; pq.pop();
            if(vis[f]) continue;
            vis[f] = true;
            for(int i = head[f]; i != -1; i = edge[i].next){
                int u = edge[i].v;
                if(dist[u] > dis + edge[i].w){
                    dist[u] = dis + edge[i].w;
                    if(c[u] == c[f]) pq.push(make_pair(dist[u], u));
                }
                if(c[u] != c[f]){
                    if(!--d[c[u]]) q.push(c[u]);
                }
            }
        }
    }
}

int main(){

    full(head, -1);
    t = read(), r = read(), p = read(), s = read();
    for(int i = 0; i < r; i ++){
        int u = read(), v = read(), c = read();
        addEdge(u, v, c), addEdge(v, u, c);
    }
    for(int i = 1; i <= t; i ++){
        if(!vis[i]) ++tot, dfs(i);
    }
    for(int i = 0; i < p; i ++){
        int u = read(), v = read(), c = read();
        addEdge(u, v, c);
    }
    for(int cur = 1; cur <= t; cur ++){
        for(int i = head[cur]; i != -1; i = edge[i].next){
            int u = edge[i].v;
            if(c[u] != c[cur]) d[c[u]] ++;
        }
    }
    topSort();
    for(int i = 1; i <= t; i ++){
        printf(dist[i] >= 100000000 ? "NO PATH\n" : "%d\n", dist[i]);
    }
    return 0;
}
posted @ 2019-04-18 21:49  清楚少女ひなこ  阅读(204)  评论(0编辑  收藏  举报