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Trucking
Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 796 Accepted Submission(s): 268
Problem Description
A certain local trucking company would like to transport some goods on a cargo truck from one place to another. It is desirable to transport as much goods as possible each trip. Unfortunately, one cannot always use the roads in the shortest route: some roads may have obstacles (e.g. bridge overpass, tunnels) which limit heights of the goods transported. Therefore, the company would like to transport as much as possible each trip, and then choose the shortest route that can be used to transport that amount.
For the given cargo truck, maximizing the height of the goods transported is equivalent to maximizing the amount of goods transported. For safety reasons, there is a certain height limit for the cargo truck which cannot be exceeded.
For the given cargo truck, maximizing the height of the goods transported is equivalent to maximizing the amount of goods transported. For safety reasons, there is a certain height limit for the cargo truck which cannot be exceeded.
Input
The input consists of a number of cases. Each case starts with two integers, separated by a space, on a line. These two integers are the number of cities (C) and the number of roads (R). There are at most 1000 cities, numbered from 1. This is followed by R lines each containing the city numbers of the cities connected by that road, the maximum height allowed on that road, and the length of that road. The maximum height for each road is a positive integer, except that a height of -1 indicates that there is no height limit on that road. The length of each road is a positive integer at most 1000. Every road can be travelled in both directions, and there is at most one road connecting each distinct pair of cities. Finally, the last line of each case consists of the start and end city numbers, as well as the height limit (a positive integer) of the cargo truck. The input terminates when C = R = 0.
Output
For each case, print the case number followed by the maximum height of the cargo truck allowed and the length of the shortest route. Use the format as shown in the sample output. If it is not possible to reach the end city from the start city, print "cannot reach destination" after the case number. Print a blank line between the output of the cases.
Sample Input
5 6 1 2 7 5 1 3 4 2 2 4 -1 10 2 5 2 4 3 4 10 1 4 5 8 5 1 5 10 5 6 1 2 7 5 1 3 4 2 2 4 -1 10 2 5 2 4 3 4 10 1 4 5 8 5 1 5 4 3 1 1 2 -1 100 1 3 10 0 0
Sample Output
Case 1: maximum height = 7 length of shortest route = 20 Case 2: maximum height = 4 length of shortest route = 8 Case 3: cannot reach destination
Source
Recommend
呢个最短路首先要用一次DP求出最高限度,最高限度有D似最大流。
我用左两次最短路,第一次求用Dijkstra最高限度,第二次用SPFA求最短路。
其实求最短路我就系改左Dijkstra内核既松弛技术, 改成求最大流增广路既一种DP。
下面贴代码:
4234500 | 2011-07-21 10:46:35 | Accepted | 2962 | 250MS | 8092K | 2613 B | C++ | 10SGetEternal{(。)(。)}! |
#include<iostream> #include<queue> using namespace std; #define INFI 0x7fffffff #define MAXI 1001 struct node { int h , w; } g[MAXI][MAXI]; int v , h , n , dp[MAXI]; bool vis[MAXI]; int Mem_DFS(int s , int minh) { int i , temph; if (dp[s] != 0) return dp[s]; if (s == v) return minh; vis[s] = 1; for (i = 0; i < n; i++) if (!vis[i] && !(g[s][i].h == -1 && g[s][i].w == INFI)) { if (g[s][i].h < minh) temph = Mem_DFS(i , g[s][i].h); else temph = Mem_DFS(i , minh); if (dp[s] < temph) dp[s] = temph; } vis[s] = 0; return dp[s]; } void Dijkstra(int u , int minh) { int i , j , tmax , temph; dp[u] = minh; for (i = 0; i < n; i++) { tmax = -11; for (j = 0; j < n; j++) if (!vis[j] && dp[j] > tmax) { tmax = dp[j]; u = j; } vis[u] = 1; for (j = 0; j < n; j++) if (!vis[j] && g[u][j].h != -1) { if (g[u][j].h < dp[u]) temph = g[u][j].h; else temph = dp[u]; if (temph > dp[j]) dp[j] = temph; } } } void SPFA(int u) { int i; queue<int> Q; Q.push(u); vis[u] = 1; dp[u] = 0; while (!Q.empty()) { u = Q.front(); Q.pop(); vis[u] = 0; for (i = 0; i < n; i++) if (i != u && g[u][i].h >= h && dp[i] - dp[u] > g[u][i].w) { dp[i] = dp[u] + g[u][i].w; if (!vis[i]) { Q.push(i); vis[i] = 1; } } } } int main() { int u , w , m , Case = 0; bool ok; scanf("%d%d" , &n , &m); while (n || m) { ok = 1; for (u = 0; u < n; u++) for (v = 0; v < n; v++) { g[u][v].w = INFI; g[u][v].h = -1; } while (m--) { scanf("%d%d%d%d" , &u , &v , &h , &w); u--; v--; if (h == -1) h = INFI; if (g[u][v].h == h) { if (g[u][v].w > w) g[u][v].w = g[v][u].w = w; } else if (g[u][v].h < h) { g[u][v].h = g[v][u].h = h; g[u][v].w = g[v][u].w = w; } } for (u = 0; u < n; u++) { dp[u] = 0; vis[u] = 0; } scanf("%d%d%d" , &u , &v , &h); u--; v--; //dp[v] = Mem_DFS(u , h); Dijkstra(u , h); if (dp[v] != 0) { h = dp[v]; for (w = 0; w < n; w++) { dp[w] = INFI; vis[w] = 0; } SPFA(u); if (dp[v] != INFI) ok = 1; } else ok = 0; printf("Case %d:\n" , ++Case); if (ok) { printf("maximum height = %d\n" , h); printf("length of shortest route = %d\n" , dp[v]); } else printf("cannot reach destination\n"); scanf("%d%d" , &n , &m); if (n || m) putchar('\n'); } return 0; }
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HDU 1075 What Are You Talking About
2011-08-04 11:00 831What Are You Talking About Tim ... -
最潮最短路算法:SPFA
2011-08-14 11:32 580首先理解单源最短路一滴概念(以下概念不作证明,证明自行bai ... -
HDU 1058 Humble Numbers
2011-08-02 15:55 1168Humble Numbers Time Limit: 200 ... -
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2011-08-02 16:13 762find your present (2) Time Lim ... -
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2011-08-02 21:00 994Train Problem I Time Limit: 20 ... -
2142 HDU box
2011-08-02 21:21 733box Time Limit: 3000/1000 MS ( ... -
HDU 2151 Worm
2011-08-01 20:48 784Worm Time Limit: 1000/1000 MS ... -
HDU 2722 Here We Go(relians) Again
2011-08-02 00:06 968Here We Go(relians) Again Time ... -
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2011-07-31 21:47 983Stars Time Limit: 1000MS ... -
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2011-07-31 21:26 860Long Long Message Time Li ... -
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2011-07-31 21:08 1013分拆素数和 Time Limit: 1000/1000 MS ... -
ZOJ 3512 Financial Fraud .
2011-07-31 20:49 1226Financial Fraud Time Limit: 3 ... -
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2011-07-31 20:47 1069Tell me the area Time Limit: 3 ... -
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2011-07-31 20:45 569find the safest road Time Limi ... -
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2011-07-31 20:19 756Surround the Trees Time Limit: ... -
HDU 1234 开门人和关门人 .
2011-07-31 20:17 639开门人和关门人 Time Limit: 2000/1000 ...
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