• HDU 4360 As long as Binbin loves Sangsang spfa


    题意:

    给定n个点m条边的无向图

    每次必须沿着LOVE走,到终点时必须是完整的LOVE,且至少走出一个LOVE,

    问这样情况下最短路是多少,在一样短情况下最多的LOVE个数是多少。

    有自环。

    #include <cstdio>
    #include <cstring>
    #include <cmath>
    #include <queue>
    #include <iostream>
    #include <algorithm>
    using namespace std;
    typedef __int64 ll;
    const ll Inf = 4611686018427387904LL;
    const int N = 1314 + 100;
    const int E = 13520 * 2 + 100;
    const int M = N * 4 + 100;
    struct Edge {
    	ll len;
    	int v, f, nex;
    	Edge() {
    	}
    	Edge(int _v, int _f, ll _len, int _nex) {
    		v = _v;
    		f = _f;
    		len = _len;
    		nex = _nex;
    	}
    };
    struct node{
    	int to, f;
    	node(int b=0,int d=0):to(b),f(d){}
    };
    Edge eg[E];
    ll dis[N][4], tim[N][4];
    bool vis[N][4];
    int T, n, g[N], idx;
    
    int re(char c) {
    	if (c == 'L')
    		return 0;
    	else if (c == 'O')
    		return 1;
    	else if (c == 'V')
    		return 2;
    	else
    		return 3;
    }
    void addedge(int u, int v, ll len, int f) {
    	eg[idx] = Edge(v, f, len, g[u]);
    	g[u] = idx++;
    }
    void spfa() {
    	memset(vis, 0, sizeof vis);
    	for (int i = 0; i < n; ++i)
    		for (int j = 0; j < 4; ++j) {
    			dis[i][j] = Inf;
    			tim[i][j] = 0;
    	}
    	queue<node>q;
    	q.push(node(0,3));
    	dis[0][3] = 0;
    	tim[0][3] = 0;
    	while(!q.empty()){
    		node u = q.front(); q.pop(); vis[u.to][u.f] = 0;
    		for(int i = g[u.to]; ~i; i = eg[i].nex){
    			int y = eg[i].v, f = eg[i].f;
    			if(f != (u.f+1)%4)continue;
    			bool yes = false;
    			if(dis[y][f] > dis[u.to][u.f]+eg[i].len)
    			{
    				dis[y][f] = dis[u.to][u.f]+eg[i].len;
    				tim[y][f] = tim[u.to][u.f];
    				if(f == 3)
    					tim[y][f]++;
    				yes = true;
    			}
    			else if(dis[y][f] == dis[u.to][u.f]+eg[i].len) {
    				ll tmp = tim[u.to][u.f];
    				if(f == 3)
    					tmp++;
    				if(tmp > tim[y][f])
    					tim[y][f] = tmp, yes = true;
    			}
    			else if(tim[y][f]==0) {
    				ll tmp = tim[u.to][u.f];
    				if(f == 3)
    					tmp++;
    				if(tmp > tim[y][f])
    					dis[y][f] = dis[u.to][u.f]+eg[i].len, tim[y][f] = tmp, yes = true;
    			}
    			if(yes && vis[y][f] == 0)
    				vis[y][f] = 1, q.push(node(y, f));
    		}
    	}
    }
    void work() {
    	int m, u, v; ll len;
    	char s[5];
    	memset(g, -1, sizeof g);
    	idx = 0;
    
    	scanf("%d %d", &n, &m);
    	while (m -- > 0) {
    		scanf("%d%d%I64d%s", &u, &v, &len, s);
    		-- u; -- v;
    		addedge(u, v, len, re(s[0]));
    		addedge(v, u, len, re(s[0]));
    	}
    	spfa();
    	ll ansdis = dis[n - 1][3], ansnum = tim[n - 1][3];
    	printf("Case %d: ", ++T);
    	if (ansdis == Inf || ansnum == 0) {
    		puts("Binbin you disappoint Sangsang again, damn it!");
    	} else {
    printf("Cute Sangsang, Binbin will come with a donkey after travelling %I64d meters and finding %I64d LOVE strings at last.
    ", ansdis, ansnum);
    	}
    }
    int main() {
    	int cas;
    	T = 0;
    	scanf("%d", &cas);
    	while (cas -- > 0)
    		work();
    	return 0;
    }
    /*
    99
    4 4
    1 2 1 L
    2 4 1 O
    4 1 1 V
    1 4 1 E
    
    1 4
    1 1 1 L
    1 1 1 O
    1 1 1 V
    1 1 1 E
    
    1 0
    
    
    */


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  • 原文地址:https://www.cnblogs.com/hrhguanli/p/4049984.html
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