題干：

There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a village C such that there is a road between A and C, and C and B are connected.?

We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum.?

Input

The first line is an integer N (3 <= N <= 100), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 1000]) between village i and village j.?

Then there is an integer Q (0 <= Q <= N * (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1 <= a < b <= N), which means the road between village a and village b has been built.?

Output

You should output a line contains an integer, which is the length of all the roads to be built such that all the villages are connected, and this value is minimum.?

Sample Input

3 0 990 692 990 0 179 692 179 0 1 1 2

Sample Output

179

題目大意：

? 大概意思就是給你一個(gè)圖，然后告訴你每?jī)蓚€(gè)點(diǎn)之間的距離，然后給你一個(gè)q，下面q行，每行兩個(gè)數(shù)代表這兩個(gè)點(diǎn)是連通的，問(wèn)你為了讓整個(gè)圖是個(gè)連通圖，最短還需要修多長(zhǎng)的路。

解題報(bào)告：

? 最小生成樹(shù)裸題，，復(fù)習(xí)一下，，不解釋了、、最后就是，加不加那個(gè)cnt計(jì)數(shù)對(duì)這道題都無(wú)所謂，因?yàn)閿?shù)據(jù)量太小了，都是31ms。下面兩個(gè)代碼都貼上。

AC代碼：（加cnt計(jì)數(shù)）

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define ll long long #define pb push_back #define pm make_pair #define fi first #define se second using namespace std; const int MAX = 2e5 + 5; int n,tot; int f[205]; struct Edge {int u,v;int w;Edge(){}Edge(int u,int v,int w):u(u),v(v),w(w){} } e[MAX<<1]; bool cmp(Edge a,Edge b) {return a.w < b.w; } int getf(int v) {return f[v] == v ? v : f[v] = getf(f[v]); } void merge(int u,int v) {int t1 = getf(u);int t2 = getf(v);if(t1!=t2) f[t2]=t1; } int main() {while(~scanf("%d",&n)) {tot=0;for(int i = 1; i<=n; i++) f[i] = i;for(int i = 1,w; i<=n; i++) {for(int j = 1; j<=n; j++) {scanf("%d",&w);if(i==j) continue;e[++tot] = Edge(i,j,w);e[++tot] = Edge(j,i,w);}} int q,cnt = 0;scanf("%d",&q);while(q--) {int u,v;scanf("%d%d",&u,&v);if(getf(u)!=getf(v)) merge(u,v),cnt++;}sort(e+1,e+tot+1,cmp);ll ans = 0;for(int i = 1; i<=tot; i++) {int u = e[i].u,v = e[i].v;if(getf(u) == getf(v)) continue;merge(u,v);cnt++;ans += 1LL * e[i].w;if(cnt == n-1) break;}printf("%lld\n",ans);}return 0 ;}

AC代碼2：（不加cnt計(jì)數(shù)）

//不帶cnt計(jì)數(shù) 版本 #include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define ll long long #define pb push_back #define pm make_pair #define fi first #define se second using namespace std; const int MAX = 2e5 + 5; int n,tot; int f[205]; struct Edge {int u,v;int w;Edge(){}Edge(int u,int v,int w):u(u),v(v),w(w){} } e[MAX<<1]; bool cmp(Edge a,Edge b) {return a.w < b.w; } int getf(int v) {return f[v] == v ? v : f[v] = getf(f[v]); } void merge(int u,int v) {int t1 = getf(u);int t2 = getf(v);if(t1!=t2) f[t2]=t1; } int main() {while(~scanf("%d",&n)) {tot=0;for(int i = 1; i<=n; i++) f[i] = i;for(int i = 1,w; i<=n; i++) {for(int j = 1; j<=n; j++) {scanf("%d",&w);if(i==j) continue;e[++tot] = Edge(i,j,w);e[++tot] = Edge(j,i,w);}} int q;scanf("%d",&q);while(q--) {int u,v;scanf("%d%d",&u,&v);merge(u,v);}sort(e+1,e+tot+1,cmp);ll ans = 0;for(int i = 1; i<=tot; i++) {int u = e[i].u,v = e[i].v;if(getf(u) == getf(v)) continue;merge(u,v);ans += 1LL * e[i].w;}printf("%lld\n",ans);}return 0 ;}

?

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