BZOJ 1003 [ZJOI2006]物流运输trans ★(Dijkstra + DP)

题目链接

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

思路

先Dijkstra暴力求出i..j天内不变换路线的最少花费，然后dp[i] = min(cost[1..i], dp[j]+cost[j+1][i]+K).

总结：

1. BZOJ 题目质量果然高啊……做一道一次升华……

2.自己太弱……

代码

[cpp]
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <string>
#include <cstring>
#include <vector>
#include <queue>
#define MID(x,y) ((x+y)/2)
#define MEM(a,b) memset(a,b,sizeof(a))
using namespace std;

struct edge{
int v, w;
edge(int _v, int _w){
v = _v, w = _w;
}
};
typedef vector <edge> VI;
priority_queue <pair<int, int>, vector<pair<int, int> >, greater<pair<int, int> > > PQ;
const int maxn = 25;
VI adj[maxn];
int n, m, K, e, d;
int cost[105][105], dist[maxn], flag[25][105], dp[105];
bool vis[maxn];

void dij(int l, int r){
while(!PQ.empty()) PQ.pop();
MEM(vis, 0);
for (int i = 2; i <= m; i ++) dist[i] = 0x3fffffff;
for (int i = 1; i <= m; i ++)
for (int j = l; j <= r; j ++)
if (flag[i][j]){
vis[i] = 1;
break;
}
dist[1] = 0;
PQ.push(make_pair(0, 1));
while(!PQ.empty()){
int u = PQ.top().second;
PQ.pop();
for (int i = 0; i < (int)adj[u].size(); i ++){
int v = adj[u][i].v;
if (!vis[v] && dist[v] > dist[u] + adj[u][i].w){
dist[v] = dist[u] + adj[u][i].w;
PQ.push(make_pair(dist[v], v));
}
}
}
if (dist[m] != 0x3fffffff)
cost[l][r] = (r - l + 1) * dist[m];
else
cost[l][r] = 0x3fffffff;
//printf("l = %d r = %d cost = %d\n", l, r, cost[l][r]);
}

int main(){
//freopen("test.in", "r", stdin);
//freopen("test.out", "w", stdout);
scanf("%d %d %d %d", &n, &m, &K, &e);
for (int i = 0; i <= m; i ++) adj[i].clear();
for (int i = 0; i < e; i ++){
int u, v, w;
scanf("%d %d %d", &u, &v, &w);
adj[u].push_back(edge(v, w));
adj[v].push_back(edge(u, w));
}
scanf("%d", &d);
MEM(flag, 0);
for (int i = 0; i < d; i ++){
int p, l, r;
scanf("%d %d %d", &p, &l, &r);
for(int j = l; j <= r; j ++)
flag[p][j] = 1;
}
for (int i = 1; i <= n; i ++){
for (int j = i; j <= n; j ++){
dij(i, j);
}
}
for (int i = 1; i <= n; i ++) dp[i] = 0x3fffffff;
dp[0] = 0;
for (int i = 1; i <= n; i ++){
dp[i] = cost[1][i];
for (int j = 0; j < i; j ++){
if (cost[j+1][i] == 0x3fffffff) continue;
dp[i] = min(dp[i], dp[j] + cost[j+1][i] + K);
}
}
printf("%d\n", dp[n]);
return 0;
}
[/cpp]