[POJ 2135] Farm Tour【费用流】
Problem:
| Time Limit: 1000MS | Memory Limit: 65536K |
Description
When FJ's friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000) fields numbered 1..N, the first of which contains his house and the Nth of which contains the big barn. A total M (1 <= M <= 10000) paths that connect the fields in various ways. Each path connects two different fields and has a nonzero length smaller than 35,000.
To show off his farm in the best way, he walks a tour that starts at his house, potentially travels through some fields, and ends at the barn. Later, he returns (potentially through some fields) back to his house again.
He wants his tour to be as short as possible, however he doesn't want to walk on any given path more than once. Calculate the shortest tour possible. FJ is sure that some tour exists for any given farm.
Input
* Line 1: Two space-separated integers: N and M.
* Lines 2..M+1: Three space-separated integers that define a path: The starting field, the end field, and the path's length.
Output
Sample Input
4 5 1 2 1 2 3 1 3 4 1 1 3 2 2 4 2
Sample Output
6
Source
Solution:
这题数据有毒。。。N 的范围必须开到 2000。。
题目本身是一道很好的最小费用最大流模板题，求的是从 1 -> N 再从 N -> 1 的不重复路径最短路。
由于每条路径只能走一次，所以每条边的容量为 1。而需要求的是最短路，所以将路径长度作为费用。
注意本题的边均为无向边，当成两条有向边处理即可，这样的话总边数要开到 4 * M。
最小费用最大流可以用 SPFA 版的 Edmonds-Karp 算法求解。
Code: O(VkE2), k为节点平均入队次数 [1624K, 32MS]
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
int N, M;
struct Edge{
int from, np, cap, cost;
Edge *nxt, *rev;
};
struct Graph{
Edge *V[2005], E[40010];
int tope;
inline void addedge(int u, int v, int cap, int cost = 0){
E[++tope].from = u, E[tope].np = v, E[tope].cap = cap, E[tope].cost = cost;
E[tope].nxt = V[u], E[tope].rev = &E[tope + 1], V[u] = &E[tope];
E[++tope].from = v, E[tope].np = u, E[tope].cap = 0, E[tope].cost = -cost;
E[tope].nxt = V[v], E[tope].rev = &E[tope - 1], V[v] = &E[tope];
}
} G;
Edge *prev[2005];
int dis[2005];
int q[2005], fr, re;
bool inq[2005];
inline int MCMF(){
int mincost = 0;
do{
fr = re = 0, q[re++] = 0;
memset(inq, 0, sizeof(inq)), inq[0] = 1;
memset(prev, 0, sizeof(prev));
memset(dis, 0x3f, sizeof(dis)), dis[0] = 0;
while(fr != re){
int u = q[fr++]; inq[u] = 0;
for(register Edge *ne = G.V[u]; ne; ne = ne->nxt)
if(ne->np && ne->cap && dis[u] + ne->cost < dis[ne->np]){
dis[ne->np] = dis[u] + ne->cost;
prev[ne->np] = ne;
if(!inq[ne->np]) q[re++] = ne->np, inq[ne->np] = 1;
}
}
if(prev[2001]){
int dflow = 0x3f3f3f3f, dcost = 0;
for(register Edge *ne = prev[2001]; ne; ne = prev[ne->from]) dflow = min(dflow, ne->cap);
for(register Edge *ne = prev[2001]; ne; ne = prev[ne->from])
ne->cap -= dflow, ne->rev->cap += dflow, dcost += ne->cost * dflow;
mincost += dcost;
}
} while(prev[2001]);
return mincost;
}
int main(){
scanf("%d%d", &N, &M);
G.addedge(0, 1, 2), G.addedge(N, 2001, 2);
for(register int i = 1; i <= M; i++){
int st, en, len;
scanf("%d%d%d", &st, &en, &len);
G.addedge(st, en, 1, len), G.addedge(en, st, 1, len);
}
int mincost = MCMF();
printf("%d\n", mincost);
return 0;
}
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