題干：

The empire is under attack again. The general of empire is planning to defend his castle. The land can be seen as?NN?towns and?MM?roads, and each road has the same length and connects two towns. The town numbered?11?is where general's castle is located, and the town numbered?NN?is where the enemies are staying. The general supposes that the enemies would choose a shortest path. He knows his army is not ready to fight and he needs more time. Consequently he decides to put some barricades on some roads to slow down his enemies. Now, he asks you to find a way to set these barricades to make sure the enemies would meet at least one of them. Moreover, the barricade on the?ii-th road requires?wiwi?units of wood. Because of lacking resources, you need to use as less wood as possible.

Input

The first line of input contains an integer?tt, then?tt?test cases follow.?
For each test case, in the first line there are two integers?N(N≤1000)N(N≤1000)?and?M(M≤10000)M(M≤10000).?
The?ii-the line of the next?MM?lines describes the?ii-th edge with three integers?u,vu,vand?ww?where?0≤w≤10000≤w≤1000?denoting an edge between?uu?and?vv?of barricade cost?ww.

Output

For each test cases, output the minimum wood cost.

Sample Input

1 4 4 1 2 1 2 4 2 3 1 3 4 3 4

Sample Output

4

題目大意：

我在1號點，敵人在n號點，敵人會走n到1的最短路徑進攻你了，現在你需要在路徑上放置障礙，但是每條邊上放置障礙有一個花費，求能阻擋所有敵人放置最少花費的障礙。

解題報告：

因為邊的長度都是1，所以直接bfs求最短路，然后建出最短路子圖，然后求最小割就行了。題目思路很清晰，算是一道比較簡單的網絡流。

AC代碼：

#include<cstring> #include<cstdio> #include<algorithm> #include<iostream> #include<queue> using namespace std; const int MAX = 2e5 +5; int n; int tot,TOT; struct Edge {int to,ne,w; } e[100005 * 2]; struct EE {int fr,to,ne,w; }E[100005 +5]; int head[10005],HEAD[10005]; int st,ed; int dis[10050],q[10005];//一共多少個點跑bfs，dis數組和q數組就開多大。 void add(int u,int v,int w) {e[++tot].to=v;e[tot].w=w;e[tot].ne=head[u];head[u]=tot; } void addE(int u,int v,int w) {E[++TOT].to=v;E[TOT].fr = u;E[TOT].w=w;E[TOT].ne=HEAD[u];HEAD[u]=TOT; } bool bfs(int st,int ed) {memset(dis,-1,sizeof(dis));int front=0,tail=0;q[tail++]=st;dis[st]=0;while(front<tail) {int cur = q[front];if(cur == ed) return 1;front++;for(int i = head[cur]; i!=-1; i = e[i].ne) {if(e[i].w&&dis[e[i].to]<0) {q[tail++]=e[i].to;dis[e[i].to]=dis[cur]+1;}}}if(dis[ed]==-1) return 0;return 1; } int dfs(int cur,int limit) {//limit為源點到這個點的路徑上的最小邊權 if(limit==0||cur==ed) return limit;int w,flow=0;for(int i = head[cur]; i!=-1; i = e[i].ne) { if(e[i].w&&dis[e[i].to]==dis[cur]+1) {w=dfs(e[i].to,min(limit,e[i].w));e[i].w-=w;e[i^1].w+=w;flow+=w;limit-=w;if(limit==0) break;}}if(!flow) dis[cur]=-1;return flow; } int dinic() {int ans = 0;while(bfs(st,ed)) ans+=dfs(st,0x7fffffff);return ans; } int DIS[MAX]; int vis[MAX]; struct Point {int pos,step;Point(){}Point(int pos,int step):pos(pos),step(step){} }; int flag[MAX]; void bfs() {for(int i = 1; i<=n; i++) vis[i] = 0,DIS[i] = -1;queue<Point> q;q.push(Point(1,0));vis[1] = 1;while(!q.empty()) {Point cur = q.front();q.pop(); // if(vis[cur.pos] == 1) continue;for(int i = HEAD[cur.pos]; ~i; i = E[i].ne) {int v = E[i].to;if(vis[v] == 1) {if(cur.step+1 == DIS[v]) {flag[i] = 1;}}else {DIS[v] = cur.step+1;vis[v] = 1;flag[i] = 1;q.push(Point(v,DIS[v]));}}} } int main() {int t,m;cin>>t;while(t--) {scanf("%d%d",&n,&m);//inittot=1,TOT=0;for(int i = 0; i<=2*m; i++) flag[i] = 0;for(int i = 1; i<=n; i++) head[i] = HEAD[i] = -1; for(int u,v,w,i = 1; i<=m; i++) {scanf("%d%d%d",&u,&v,&w);addE(u,v,w);addE(v,u,w);} bfs();for(int i = 1; i<=TOT; i++) {if(flag[i] == 1) {add(E[i].fr,E[i].to,E[i].w);add(E[i].to,E[i].fr,0);}}st=1,ed=n;printf("%d\n",dinic()); }return 0; } /* 3 4 4 1 2 1 2 4 2 3 1 3 4 3 46 7 1 2 1 2 6 7 1 3 2 2 4 5 4 6 5 1 5 4 5 6 34 4 1 2 1 2 4 2 3 1 3 4 3 4*/

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