HDU 3339 In Action【最短路+01背包模板/主要是建模看谁是容量、价值】
Since 1945, when the first nuclear bomb was exploded by the Manhattan Project team in the US, the number of nuclear weapons have soared across the globe.
Nowadays,the crazy boy in FZU named AekdyCoin possesses some nuclear weapons and wanna destroy our world. Fortunately, our mysterious spy-net has gotten his plan. Now, we need to stop it.
But the arduous task is obviously not easy. First of all, we know that the operating system of the nuclear weapon consists of some connected electric stations, which forms a huge and complex electric network. Every electric station has its power value. To start the nuclear weapon, it must cost half of the electric network's power. So first of all, we need to make more than half of the power diasbled. Our tanks are ready for our action in the base(ID is 0), and we must drive them on the road. As for a electric station, we control them if and only if our tanks stop there. 1 unit distance costs 1 unit oil. And we have enough tanks to use.
Now our commander wants to know the minimal oil cost in this action.
InputThe first line of the input contains a single integer T, specifying the number of testcase in the file.
For each case, first line is the integer n(1<= n<= 100), m(1<= m<= 10000), specifying the number of the stations(the IDs are 1,2,3...n), and the number of the roads between the station(bi-direction).
Then m lines follow, each line is interger st(0<= st<= n), ed(0<= ed<= n), dis(0<= dis<= 100), specifying the start point, end point, and the distance between.
Then n lines follow, each line is a interger pow(1<= pow<= 100), specifying the electric station's power by ID order.OutputThe minimal oil cost in this action.
If not exist print "impossible"(without quotes).Sample Input
2 2 3 0 2 9 2 1 3 1 0 2 1 3 2 1 2 1 3 1 3
Sample Output
5 impossible
#include<cstdio> #include<string> #include<cstdlib> #include<cmath> #include<iostream> #include<cstring> #include<set> #include<queue> #include<algorithm> #include<vector> #include<map> #include<cctype> #include<stack> #include<sstream> #include<list> #include<assert.h> #include<bitset> #include<numeric> #define debug() puts("++++") #define gcd(a,b) __gcd(a,b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define fi first #define se second #define pb push_back #define sqr(x) ((x)*(x)) #define ms(a,b) memset(a,b,sizeof(a)) #define sz size() #define be begin() #define pu push_up #define pd push_down #define cl clear() #define lowbit(x) -x&x #define all 1,n,1 #define rep(i,x,n) for(int i=(x); i<(n); i++) #define in freopen("in.in","r",stdin) #define out freopen("out.out","w",stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int,int> P; const int INF = 0x3f3f3f3f; const LL LNF = 1e18; const int maxn = 50000+20; const int maxm = 1e6 + 10; const double PI = acos(-1.0); const double eps = 1e-8; const int dx[] = {-1,1,0,0,1,1,-1,-1}; const int dy[] = {0,0,1,-1,1,-1,1,-1}; int dir[4][2] = {{0,1},{0,-1},{-1,0},{1,0}}; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; int t,n,m,u,v,w,x,y,p[maxn],tot,sum; int vis[maxn],dp[maxn]; int dis[maxn]; struct node { int u,v,w,nxt; }e[maxn<<2]; int head[maxn]; void init() { tot=0; ms(head,-1); ms(dp,INF); ms(dis,INF); ms(vis,0); ms(e,INF); } void add(int u,int v,int w) { e[tot].v=v; e[tot].w=w; e[tot].nxt=head[u]; head[u]=tot++; } void spfa() { queue<int>q; for(int i=1;i<=n;i++) vis[i]=0,dis[i]=INF; vis[0]=1,dis[0]=0; q.push(0); while(!q.empty()) { int u = q.front(); q.pop(); vis[u]=0; for(int i=head[u]; i!=-1; i=e[i].nxt) { int v = e[i].v; if(dis[v] > dis[u]+e[i].w) { dis[v] = dis[u]+e[i].w; if(!vis[v]) { vis[v]=1; q.push(v); } } } } } int main() { scanf("%d",&t); while(t--) { scanf("%d%d",&n,&m); init(); for(int i=0;i<m;i++) { scanf("%d%d%d",&u,&v,&w); add(u,v,w); add(v,u,w); } spfa(); sum=0; for(int i=1;i<=n;i++) { scanf("%d",&p[i]); //电量 sum += p[i]; //总发电量 } dp[0]=0; for(int i=1; i<=n; i++) //个数 { for(int j=sum; j>=p[i] ;j--) { dp[j] = min(dp[j], dp[j-p[i]]+dis[i]); } } int ans=INF; for(int i=sum/2+1; i<=sum; i++) ans=min(ans,dp[i]); if(ans>=INF) cout<<"impossible"<<endl; else cout<<ans<<endl; } } /* 【题意】 题意是，有若干发电站，有无数的坦克，每个发电站有一个发电量， 每个坦克能到达的发电站的发电量要达到总的发电量的一半以上，这个发射装置才能被摧毁，否则不能。 但是到达每个发电站都有一段距离，坦克的耗油量 = 这段距离，现在要求最少耗油量来摧毁这个发射装置。 【类型】 最短路+01背包 【分析】 首先是求，基地0能到达的每个发电站的最短路 然后用01背包求出消耗的最少油量。要是所能到达的发电站产生的电量都没有超过总产电量的一半，则无法摧毁。 以总路径为背包，点的能量值为价值，对每个点只有去和不去两种状态，用01背包解决。 【时间复杂度&&优化】 【trick】 在t组数据中sum忘记清零了！ 【数据】 spfa:218MS dij:187MS */
#include<cstdio> #include<string> #include<cstdlib> #include<cmath> #include<iostream> #include<cstring> #include<set> #include<queue> #include<algorithm> #include<vector> #include<map> #include<cctype> #include<stack> #include<sstream> #include<list> #include<assert.h> #include<bitset> #include<numeric> #define debug() puts("++++") #define gcd(a,b) __gcd(a,b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define fi first #define se second #define pb push_back #define sqr(x) ((x)*(x)) #define ms(a,b) memset(a,b,sizeof(a)) #define sz size() #define be begin() #define pu push_up #define pd push_down #define cl clear() #define lowbit(x) -x&x #define all 1,n,1 #define rep(i,x,n) for(int i=(x); i<(n); i++) #define in freopen("in.in","r",stdin) #define out freopen("out.out","w",stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int,int> P; const int INF = 0x3f3f3f3f; const LL LNF = 1e18; const int maxn = 50000+20; const int maxm = 1e6 + 10; const double PI = acos(-1.0); const double eps = 1e-8; const int dx[] = {-1,1,0,0,1,1,-1,-1}; const int dy[] = {0,0,1,-1,1,-1,1,-1}; int dir[4][2] = {{0,1},{0,-1},{-1,0},{1,0}}; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; int t,n,m,u,v,w,x,y,p[maxn],tot,sum; int vis[maxn],dp[maxn]; int dis[maxn]; struct node { int u,v,w,nxt; }e[maxn<<2]; struct cmp { bool operator()(int a,int b) { return dis[a]>dis[b]; } }; int head[maxn]; void init() { tot=0; ms(head,-1); ms(dp,INF); ms(dis,INF); ms(vis,0); ms(e,INF); } void add(int u,int v,int w) { e[tot].v=v; e[tot].w=w; e[tot].nxt=head[u]; head[u]=tot++; } void spfa() { priority_queue<int,vector<int>,cmp >q; for(int i=1;i<=n;i++) dis[i]=INF; //vis[0]=1; dis[0]=0; q.push(0); while(!q.empty()) { int u = q.top(); q.pop(); //vis[u]=0; for(int i=head[u]; i!=-1; i=e[i].nxt) { int v = e[i].v; if(dis[v] > dis[u]+e[i].w) { dis[v] = dis[u]+e[i].w; q.push(v); } } } } int main() { scanf("%d",&t); while(t--) { scanf("%d%d",&n,&m); init(); for(int i=0;i<m;i++) { scanf("%d%d%d",&u,&v,&w); add(u,v,w); add(v,u,w); } spfa(); sum=0; for(int i=1;i<=n;i++) { scanf("%d",&p[i]); //电量 sum += p[i]; //总的发电量 } dp[0]=0; for(int i=1; i<=n; i++) //个数 { for(int j=sum; j>=p[i] ;j--) //电量 { dp[j] = min(dp[j], dp[j-p[i]]+dis[i]); } } int ans=INF; for(int i=sum/2+1; i<=sum; i++) ans=min(ans,dp[i]); if(ans>=INF) cout<<"impossible"<<endl; else cout<<ans<<endl; } }
