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n个点m条路, 每条路有权值, 给出起点和终点, 求一条路使得权值最小。可以使路过的路中, k条路的权值忽略。
其实就是多一维, 具体看代码
#include<bits/stdc++.h> using namespace std; #define pb(x) push_back(x) #define ll long long #define mk(x, y) make_pair(x, y) #define lson l, m, rt<<1 #define mem(a) memset(a, 0, sizeof(a)) #define rson m+1, r, rt<<1|1 #define mem1(a) memset(a, -1, sizeof(a)) #define mem2(a) memset(a, 0x3f, sizeof(a)) #define rep(i, a, n) for(int i = a; i<n; i++) #define ull unsigned long long typedef pair<int, int> pll; const double PI = acos(-1.0); const double eps = 1e-8; const int mod = 1e9+7; const int inf = 1061109567; const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} }; const int maxn = 5e4+5; int dis[12][60005], in[12][60005], num, head[maxn*2]; int n, m, k, s, t; struct edge { int to, nextt, w; }e[maxn*2]; struct node { int u, x; node(){} node(int u, int x):u(u),x(x){} }; void init() { mem1(head); num = 0; } void add(int u, int v, int w) { e[num].to = v; e[num].nextt = head[u]; e[num].w = w; head[u] = num++; } queue <node> q; int dij() { mem2(dis); q.push(node(s, 0)); dis[0][s] = 0; in[0][s] = 1; while(!q.empty()) { node tmp = q.front(); q.pop(); int u = tmp.u, x = tmp.x; in[x][u] = 0; for(int i = head[u]; ~i; i = e[i].nextt) { int v = e[i].to; if(dis[x][v]>dis[x][u]+e[i].w) { dis[x][v] = dis[x][u]+e[i].w; if(!in[x][v]) { in[x][v] = 1; q.push(node(v, x)); } } } if(x<k) { x++; for(int i = head[u]; ~i; i = e[i].nextt) { int v = e[i].to; if(dis[x][v]>dis[x-1][u]) { dis[x][v] = dis[x-1][u]; if(!in[x][v]) { in[x][v] = 1; q.push(node(v, x)); } } } } } int ans = inf; for(int i = 0; i<=k; i++) ans = min(ans, dis[i][t]); return ans; } int main() { int x, y, z; cin>>n>>m>>k>>s>>t; init(); while(m--) { scanf("%d%d%d", &x, &y, &z); add(x, y, z); add(y, x, z); } int ans = dij(); cout<<ans<<endl; return 0; }
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原文地址:http://www.cnblogs.com/yohaha/p/5068070.html