題干：

A group of thieves is approaching a museum in the country of zjsxzy,now they are in city A,and the museum is in city B,where keeps many broken legs of zjsxzy.Luckily,GW learned the conspiracy when he is watching stars and told it to zjsxzy.?
Zjsxzy decided to caught these thieves,and he let the police to do this,the police try to catch them on their way from A to B. Although the thieves might travel this way by more than one group, zjsxzy's excellent police has already gather the statistics that the cost needed on each road to guard it.?
Now ,zjsxzy's conutry can be described as a N*N matrix A,Aij indicates the city(i,j) have bidirectionals road to city(i+1,j) and city(i,j+1),gurad anyone of them costs Aij.?
Now give you the map,help zjsxzy to calculate the minimium cost.We assume thieves may travel in any way,and we will catch all passing thieves on a road if we guard it.?

Input

The first line is an integer T,followed by T test cases.?
In each test case,the first line contains a number N(1<N<=400).?
The following N lines,each line is N numbers,the jth number of the ith line is Aij.?
The city A is always located on (1,1) and the city B is always located on (n,n).?
Of course,the city (i,j) at the last row or last line won't have road to (i,j+1) or (i+1,j).?

Output

For each case,print a line with a number indicating the minimium cost to arrest all thieves.

Sample Input

1 3 10 5 5 6 6 20 4 7 9

Sample Output

18

Hint

The map is like this:

題目大意：

有一個n*n個點的格子。輸入n*n條邊，每個邊權代表(i,j) -> (i,j+1) 和 (i,j) -> (i+1,j)這兩條邊的邊權。求左上角到右下角的最小割、

解題報告：

? 直接轉對偶圖跑最短路。但是建圖的時候細節要注意一下。別建多了邊，寫錯了符號啥的。

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 = 4e5 + 5; const int INF = 0x3f3f3f3f; struct Edge {int u,v,w;int ne; } e[MAX<<2]; int n; int head[MAX],tot; void add(int u,int v,int w) {//加雙向邊 e[++tot].v = v; e[tot].w = w; e[tot].ne = head[u]; head[u] = tot;e[++tot].v = u; e[tot].w = w; e[tot].ne = head[v]; head[v] = tot; } int ID(int x,int y) {return (x-1)*n + y; } int dis[MAX],vis[MAX]; struct Point {int pos,dis;Point(int pos=0,int dis=0):pos(pos),dis(dis){}bool operator < (const Point & b) const {return dis > b.dis;} }; int Dij(int st,int ed) {priority_queue<Point> pq;pq.push(Point(st,0));dis[st]=0;while(pq.size()) {Point cur = pq.top();pq.pop();if(vis[cur.pos]) continue;vis[cur.pos] = 1;for(int i = head[cur.pos]; ~i; i = e[i].ne) {int v = e[i].v;if(dis[v] > dis[cur.pos] + e[i].w) {dis[v] = dis[cur.pos] + e[i].w;pq.push(Point(v,dis[v]));}} }return dis[ed]; } int main() {int t;cin>>t;while(t--) {tot=0;scanf("%d",&n);for(int i = 0; i<=n*n + 2; i++) head[i] = -1,dis[i] = INF,vis[i] = 0;int st = n*n+1,ed=n*n+2;for(int x,i = 1; i<=n; i++) {for(int j = 1; j<=n; j++) {scanf("%d",&x);if( (i==1&&j!=n)|| (j==n&&i!=n) ) add(ID(i,j-(j==n)),ed,x);//和匯點 if( (i==n&&j!=n)|| (j==1&&i!=n) ) add(ID(i-(i==n),j),st,x);//和起點 if(i != 1 && i != n && j != n) add(ID(i-1,j),ID(i,j),x);//水平行之間if(j != 1 && j != n && i != n) add(ID(i,j-1),ID(i,j),x);//豎直列之間 }}printf("%d\n",Dij(st,ed)); }return 0 ; }

總結：

? 建圖的過程中雖然非常的顯而易見，但是還是出現了細節問題。比如剛開始直接寫的和源點和匯點的連邊，直接就是i==1||j==n代表和匯點相連，顯然這樣寫會有重邊，不光如此，第一行最右側那一條邊在連邊的時候，要注意是ID(i,j-1)和ed相連。

還有判斷水平邊(以水平邊為例)的時候，不僅要有i!=1&&i!=n，還要有j!=n這一個條件！總之別落下了判斷條件就行。

總結

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