洛谷P3953 逛公园(dp 拓扑排序)

题意

题目链接

sol

去年考noip的时候我好像连最短路计数都不会啊qwq。。

首先不难想到一个思路，\(f[i][j]\)表示到第\(i\)个节点，与最短路之差长度为\(j\)的路径的方案数

首先把每个节点的最短路求出来

转移的时候按拓扑序(也就是按距离从小到大排序)转移一下

然而有\(0\)边的时候会挂掉，原因是会有dis相同的时候，这时候单按dis排序会无法判断转移方向

一种方案是直接把所有\(0\)边加入到新图中，拓扑排序一遍。得到第二关键字

同时判断一下\(0\)环

#include<bits/stdc++.h>
#define pair pair<int, int>
#define mp make_pair
#define fi first
#define se second 
using namespace std;
const int maxn = 100001, inf = 1e9 + 7;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int t, n, m, k, mod, f[maxn][51], dis1[maxn], disn[maxn], vis[maxn], inder[maxn], id[maxn];
vector<pair> v[maxn], rv[maxn];
vector<pair<pair<int, int>, int> > p;
vector<int> e[maxn];
void init() {
    memset(f, 0, sizeof(f));
    memset(inder, 0, sizeof(inder));
    p.clear();
    f[1][0] = 1;
    n = read(); m = read(); k = read(); mod = read();
    for(int i = 1; i <= n; i++) v[i].clear(), e[i].clear();
    for(int i = 1; i <= m; i++) {
        int x = read(), y = read(), z = read();
        v[x].push_back(mp(y, z)); rv[y].push_back(mp(x, z));
        if(!z) e[x].push_back(y), inder[y]++;
    }
}
void add(int &x, int y) {
    if(x + y < 0) x = (x + y + mod);
    else x = (x + y >= mod ? x + y - mod : x + y);
}
int topsort() {
    int cnt = 0; queue<int> q;
    for(int i = 1; i <= n; i++) if(!inder[i]) id[i] = ++cnt, q.push(i);
    while(!q.empty()) {
        int p = q.front(); q.pop();
        for(int i = 0, to; i < e[p].size(); i++) 
            if(!(--inder[to = e[p][i]])) q.push(to), id[to] = ++cnt;
    }
    for(int i = 1; i <= n; i++) if(inder[i] && (dis1[i] + disn[i] <= dis1[n] + k)) return -1;
    return 0;
}
void dij(int bg, int *dis) {
    priority_queue<pair> q; q.push(mp(0, bg)); 
    memset(vis, 0, sizeof(vis));
    for(int i = 1; i <= n; i++) dis[i] = inf + 1; dis[bg] = 0;
    while(!q.empty()) {
        if(vis[q.top().se]) {q.pop(); continue;}
        int p = q.top().se; q.pop(); vis[p] = 1;
        if(bg == 1) {
            for(int i = 0, to; i < v[p].size(); i++) 
                if(dis[to = v[p][i].fi] > dis[p] + v[p][i].se) 
                    dis[to] = dis[p] + v[p][i].se, q.push(mp(-dis[to], to));            
        } else {
            for(int i = 0, to; i < rv[p].size(); i++) 
                if(dis[to = rv[p][i].fi] > dis[p] + rv[p][i].se) 
                    dis[to] = dis[p] + rv[p][i].se, q.push(mp(-dis[to], to));           
        }
    }   
}
int solve() {
    dij(1, dis1);
    dij(n, disn);
    if(topsort() == -1) return -1;
    for(int i = 1; i <= n; i++) p.push_back(mp(mp(dis1[i], id[i]), i));
    sort(p.begin(), p.end());
    f[1][0] = 1;
    for(int j = 0; j <= k; j++) {
        for(int i = 0; i < n; i++) {
            int x = p[i].se;
            for(int k = 0; k < v[x].size(); k++) {
                int to = v[x][k].fi, w = v[x][k].se, ps = dis1[x] + j + w - dis1[to];
                if(ps <= k) add(f[to][ps], f[x][j]);
            }
        }
    }
    int ans = 0;
    for(int i = 0; i <= k; i++) add(ans, f[n][i]);
    return ans;
}
int main() {
    t = read(); 
    while(t--) {
        init();
        printf("%d\n", solve());            
    }
    return 0;
}
/*
*/