BZOJ 1003: [ZJOI2006]物流运输trans SPFA+DP

原题链接：http://www.lydsy.com/JudgeOnline/problem.php?id=1003

题解：

dp就好，令dp[i]表示第i天的答案，那么dp[i]=min{Cost(1,i),Cost(j+1,i)+dp[j]+K}，其中Cost(i,j)表示从i到j都用同一种方案。这种dp和划分问题很类似。

代码：

#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<cstdio>
#include<climits>
#define INF 0x3f3f3f3f
#define MAX_M 220
#define MAX_N 1110
using namespace std;

int N,M,K,E;

int dp[MAX_N];
int d[MAX_M];

queue<int> que;
bool inQue[MAX_M];

struct edge {
public:
    int to, cost;

    edge(int t, int c) : to(t), cost(c) { }

    edge() { }
};

vector<edge> G[MAX_M];
bool used[MAX_M];

vector<int> days[MAX_N];

int spfa() {
    while (que.size())que.pop();
    memset(inQue, 0, sizeof(inQue));
    fill(d, d + M + 1, INF);
    que.push(1);
    inQue[1] = 1;
    d[1] = 0;
    while (que.size()) {
        int u = que.front();
        que.pop();
        inQue[u]=0;
        for (int i = 0; i < G[u].size(); i++) {
            int v = G[u][i].to, c = G[u][i].cost;
            if (d[v] > d[u] + c && (!used[v])) {
                d[v] = d[u] + c;
                if (!inQue[v]) {
                    inQue[v] = 1;
                    que.push(v);
                }
            }
        }
    }
    return d[M];
}

int COST(int i,int j) {
    memset(used, 0, sizeof(used));
    for (int k = i; k <= j; k++)
        for (int t = 0; t < days[k].size(); t++)
            used[days[k][t]] = 1;
    int tmp=spfa();
    if(tmp==INF)return INF;
    return (j - i + 1) * tmp;
}

int main() {
    //freopen("1003.in", "r", stdin);
    //freopen("1003.out", "w", stdout);
    scanf("%d%d%d%d", &N, &M, &K, &E);
    for (int i = 0; i < E; i++) {
        int u, v, c;
        scanf("%d%d%d", &u, &v, &c);
        G[u].push_back(edge(v, c));
        G[v].push_back(edge(u, c));
    }
    int D;
    scanf("%d",&D);
    while (D--) {
        int P, a, b;
        scanf("%d%d%d", &P, &a, &b);
        for (int i = a; i <= b; i++)days[i].push_back(P);
    }
    for (int i = 1; i <= N; i++) {
        dp[i] = COST(1,i);
        for (int j = 1; j < i; j++)
            dp[i] = min(dp[i], COST(j + 1, i) + dp[j] + K);
    }
    printf("%d
", dp[N]);
    return 0;
}