• Codeforces 1340C Nastya and Unexpected Guest(01bfs)


    http://codeforces.com/contest/1340/problem/C

    我完了,连普及的题都不会做。

    (f[i][j])表示在(j)时刻到(i),最少多少个红绿灯回合。

    发现i只用往i-1和i+1走,设(t=|x1-x2|),则(j+t<g)(j+t=g)(此时回合数+1)

    那么相当于有一些边权=(0),有一些边权(=1),01bfs即可。

    从网上学到01bfs原来可以用deque写。

    Code:

    #include<bits/stdc++.h>
    #define fo(i, x, y) for(int i = x, _b = y; i <= _b; i ++)
    #define ff(i, x, y) for(int i = x, _b = y; i <  _b; i ++)
    #define fd(i, x, y) for(int i = x, _b = y; i >= _b; i --)
    #define ll long long
    #define pp printf
    #define hh pp("
    ")
    using namespace std;
     
    const int N = 10005;
     
    const int M = 1005;
     
    int n, m, a[N], g, r;
    int dis[N][M], bz[N][M];
     
    struct P {
    	int x, y;
    	P(int _x = 0, int _y = 0) {
    		x = _x, y = _y;
    	}
    };
     
    int ans;
     
    deque<P> q;
     
    int main() {
    	scanf("%d %d", &n, &m);
    	fo(i, 1, m) scanf("%d", &a[i]);
    	scanf("%d %d", &g, &r);
    	sort(a + 1, a + m + 1);
    	m = unique(a + 1, a + m + 1) - (a + 1);
    	bz[1][0] = 1; q.push_front(P(1, 0));
    	ans = 1e9;
    	while(!q.empty()) {
    		P b = q.front(); q.pop_front();
    		int x = b.x, y = b.y;
    		if(y == 0 && g >= n - a[x]) {
    			ans = min(ans, dis[x][y] * (g + r) + n - a[x]);
    		}
    		if(x > 1) {
    			int v = b.y + a[x] - a[x - 1];
    			if(v < g) {
    				if(!bz[x - 1][v]) {
    					bz[x - 1][v] = 1;
    					dis[x - 1][v] = dis[x][y];
    					q.push_front(P(x - 1, v));
    				}
    			} else
    			if(v == g) {
    				if(!bz[x - 1][0]) {
    					bz[x - 1][0] = 1;
    					dis[x - 1][0] = dis[x][y] + 1;
    					q.push_back(P(x - 1, 0));
    				}
    			}
    		}
    		if(x < m) {
    			int v = b.y + a[x + 1] - a[x];
    			if(v < g) {
    				if(!bz[x + 1][v]) {
    					bz[x + 1][v] = 1;
    					dis[x + 1][v] = dis[x][y];
    					q.push_front(P(x + 1, v));
    				}
    			} else
    			if(v == g) {
    				if(!bz[x + 1][0]) {
    					bz[x + 1][0] = 1;
    					dis[x + 1][0] = dis[x][y] + 1;
    					q.push_back(P(x + 1, 0));
    				}
    			}
    		}
    	}
    	pp("%d
    ", ans == 1e9 ? -1 : ans);
    }
    
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  • 原文地址:https://www.cnblogs.com/coldchair/p/12773496.html
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