BZOJ 1003

1003: [ZJOI2006]物流运输

DP+SPFA
注意边是双向的，以及n,m不能搞反

BZOJ 1003

/**************************************************************
    Problem: 1003
    User: Dream_Tonight
    Language: C++
    Result: Accepted
    Time:80 ms
    Memory:1392 kb
****************************************************************/
 
#include <algorithm>
#include <iostream>
#include <cstring>
#include <vector>
#include <string>
#include <cstdio>
#include <cmath>
#include <stack>
#include <queue>
#include <list>
#include <map>
#include <set>
#include <iostream>
#include <cstdio>
 
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
#define mp make_pair
#define pb push_back
const int maxn = 1e5 + 7;
const int INF = 1e9;
 
struct node {
    int t;
    int v;
 
    node(int a, int b) {
        t = a;
        v = b;
    }
};
 
ll v[105], d[105];
vector<node> ed[1005];
ll f[105], dp[105][105];
vector<int> st[105];
 
long long spfa(int l, int r, int m) {
    queue<int> q;
    memset(v, 0, sizeof(v));
    for (int i = 1; i <= m; i++) d[i] = INF;
    for (int i = l; i <= r; i++) {
        for (int j = 0; j < st[i].size(); j++) {
            v[st[i][j]] = 1;
        }
    }
    q.push(1);
    d[1] = 0;
    while (!q.empty()) {
        int n = q.front();
        q.pop();
        for (int i = 0; i < ed[n].size(); i++) {
            int tt = ed[n][i].t;
            int vv = ed[n][i].v;
            if (v[tt] == 0 && d[tt] >= d[n] + vv) {
                d[tt] = d[n] + vv;
                q.push(tt);
            }
        }
    }
    return d[m];
}
 
int main() {
    int n, m, k, e, s, t, p, d;
    scanf("%d%d%d%d", &n, &m, &k, &e);
    for (int i = 0; i < e; i++) {
        scanf("%d%d%d", &s, &t, &p);
        ed[s].pb(node(t, p));
        ed[t].pb(node(s, p));
    }
    scanf("%d", &d);
    for (int i = 0; i < d; i++) {
        scanf("%d%d%d", &p, &s, &t);
        for (int j = s; j <= t; j++) {
            st[j].push_back(p);
        }
    }
    for (int i = 1; i <= n; i++) {
        for (int j = i; j <= n; j++) {
            dp[i][j] = spfa(i, j, m);
        }
    }
    f[0] = 0;
    for (int i = 1; i <= n; i++) {
        f[i] = 1e18;
        for (int j = 0; j < i; j++) {
            f[i] = min(f[i], f[j] + dp[j + 1][i] * (i - j) + k);
        }
    }
    printf("%lld", f[n] - k);
    return 0;
}