bzoj1003 [ZJOI2006]物流运输

传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=1003

【题解】

瞎预处理瞎[i,j]天的最短路

dp处理即可。f[i]=min(f[j]+dis[i,j,n]*(i-j))+K

# include <queue>
# include <stdio.h>
# include <string.h>
# include <iostream>
# include <algorithm>
// # include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int M = 2e3 + 10, N = 30;
const int mod = 1e9+7;

# define RG register
# define ST static

int n, day, K, m; 
int head[N], nxt[M], to[M], w[M], tot=0;
inline void add(int u, int v, int _w) {
    ++tot; nxt[tot] = head[u]; head[u] = tot; to[tot] = v;
    w[tot] = _w;
}
inline void adde(int u, int v, int _w) {
    add(u, v, _w);
    add(v, u, _w);
}

int pn;
struct pa {
    int p, a, b;
    pa() {}
    pa(int p, int a, int b) : p(p), a(a), b(b) {}
}p[M];

bool cannot[N];

ll dis[110][110][N];
bool vis[N];
ll f[110];
queue<int> q;

inline void spfa(int si, int sj) {
    ll *d = dis[si][sj];
    memset(vis, 0, sizeof vis);
    while(!q.empty()) q.pop();
    for (int i=1; i<=n; ++i) d[i] = 1e9;
    q.push(1); d[1] = 0;
    while(!q.empty()) {
        int top = q.front(); q.pop(); vis[top] = 0;
        for (int i=head[top]; i; i=nxt[i]) {
            if(d[to[i]] > d[top] + w[i] && !cannot[to[i]]) {
                d[to[i]] = d[top] + w[i];
                if(!vis[to[i]]) {
                    q.push(to[i]);
                    vis[to[i]] = 1;
                }
            }
        }
    }
}
    

int main() {
    cin >> day >> n >> K >> m;
    for (int i=1, u, v, l; i<=m; ++i) {
        scanf("%d%d%d", &u, &v, &l);
        adde(u, v, l);
    }
    cin >> pn;
    for (int i=1; i<=pn; ++i) 
        scanf("%d%d%d", &p[i].p, &p[i].a, &p[i].b);
    
    for (int i=1; i<=day; ++i)
        for (int j=i; j<=day; ++j) {
            for (int k=1; k<=n; ++k) cannot[k] = 0;
            for (int k=1; k<=pn; ++k) {
                if(p[k].a > j || p[k].b < i) continue;
                cannot[p[k].p] = 1;
            }
            spfa(i, j);
        }
    f[0] = 0;
    for (int i=1; i<=day; ++i) f[i] = 1e9;
    for (int i=1; i<=day; ++i) 
        for (int j=0; j<=i-1; ++j) 
            f[i] = min(f[i], f[j] + dis[j+1][i][n]*(i-j) + K);
    printf("%lld\n", f[day] - K);
    return 0;
}