題干：

The bovine population boom down on the farm has caused serious congestion on the cow trails leading to the barn. Farmer John has decided to conduct a study to find the bottlenecks in order to relieve the 'traffic jams' at milking time.

The pasture contains a network of?M?(1 ≤?M?≤ 50,000) one-way trails, each of which connects exactly two different intersections from the set of?N?(1 ≤?N?≤ 5,000) intersections conveniently numbered 1..N; the barn is at intersection number?N. Each trail connects one intersection point to another intersection point with a higher number. As a result, there are no cycles and, as they say on the farm, all trails lead to the barn. A pair of intersection points might be connected by more than one trail.

During milking time rush hour, the cows start from their respective grazing locations and head to the barn. The grazing locations are exactly those intersection points with no trails connecting into them. Each cow traverses a 'path', which is defined as a sequence of trails from a grazing location to the barn.

Help FJ finding the busiest trail(s) by computing the largest number of possible paths that contain any one trail. The answer is guaranteed to fit in a signed 32-bit integer.

Input

Line 1: Two space-separated integers:?N?and?M.?
Lines 2..?M+1: Two space-separated intersection points.

Output

Line 1: The maximum number of paths passing through any one trail.

Sample Input

7 7 1 3 3 4 3 5 4 6 2 3 5 6 6 7

Sample Output

4

Hint

Here are the four possible paths that lead to the barn:?
1 3 4 6 7?
1 3 5 6 7?
2 3 4 6 7?
2 3 5 6 7

題目大意：

n個點(diǎn)m條有向邊，求在入度為零的點(diǎn)到n號點(diǎn)的所有路徑中，哪條邊被這些路徑覆蓋的次數(shù)最多，求這個次數(shù)。題目約定每一條邊總是由標(biāo)號小的連向標(biāo)號大的。

解題報告：

? 本來想dfs，發(fā)現(xiàn)不用，，，因?yàn)槭菢?biāo)號小的連向標(biāo)號大的，所以直接掃一遍就行了。但是這題不能用vector存邊，因?yàn)槟阈枰肋@一條邊的編號，所以老老實(shí)實(shí)寫結(jié)構(gòu)體。。。正著掃一遍維護(hù)經(jīng)過每一條邊的次數(shù)和每個頂點(diǎn)的出現(xiàn)次數(shù)，同樣的倒著掃一遍。最后枚舉每一條邊，根據(jù)乘法原理答案就是兩者乘積，維護(hù)最大值就行了。這題不能用網(wǎng)絡(luò)流那種^1的建邊方法，因?yàn)樾枰婪聪蜻叺膆ead[u]，所以需要單開數(shù)組。

注意可能有重邊，比如：

3 3
1 2
1 2
2 3

這組數(shù)據(jù)就輸出2.

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define F first #define S second #define ll long long #define pb push_back #define pm make_pair using namespace std; typedef pair<int,int> PII; const int MAX = 5000 + 5; vector<int> vv[MAX],pp[MAX]; int in[MAX],out[MAX],vis[MAX],tot1,tot2,dpV[MAX],dpP[MAX]; int head[MAX],HEAD[MAX]; struct Edge {int to,ne,sum; } e[MAX*10],E[MAX*10]; void add(int u,int v) {e[++tot1].to = v;e[tot1].ne = head[u];head[u] = tot1;E[++tot2].to = u;E[tot2].ne = HEAD[v];HEAD[v] = tot2; } int main() {int n,m;cin>>n>>m;memset(head,-1,sizeof head);memset(HEAD,-1,sizeof HEAD);for(int a,b,i = 1; i<=m; i++) {scanf("%d%d",&a,&b);add(a,b);in[b]++;out[a]++;}for(int i = 1; i<=n; i++) {if(in[i] == 0) dpV[i] = 1;if(out[i] == 0) dpP[i] = 1;}for(int cur = 1; cur<=n; cur++) {for(int i = head[cur]; ~i; i = e[i].ne) {e[i].sum = dpV[cur];dpV[e[i].to] += e[i].sum;}}for(int cur = n; cur>=1; cur--) {for(int i = HEAD[cur]; ~i; i = E[i].ne) {E[i].sum = dpP[cur];dpP[E[i].to] += E[i].sum;}}int ans = 0 ;for(int i = 1; i<=tot1; i++) {ans = max(ans,e[i].sum * E[i].sum);}printf("%d\n",ans);return 0 ; }

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