洛谷 #1850. 换教室

题意

有n个时间段，每个时间段有2节课，最多可以换m次课，概率给出

把校园抽象成无向图，求耗费最小体力值

题解

这是一道期望Dp,f[i][j][k]表示考虑到第i个教室，换了j次，k为0/1，表示这次是否换

分四种情况，具体见代码

调试记录

double的输入提示符是%lf，不是%llf

$害老子调了一个小时$

#include <cstdio>
#include <algorithm>
#include <cstring>
#define maxn 2005
#define INF 0x3f3f3f3f

using namespace std;

int n, m, v, e;
int c[maxn], d[maxn], dis[305][305];
double f[maxn][maxn][2];
double k[maxn];

void Floyd(){
	for (int i = 1; i <= v; i++)
		for (int j = 1; j < i; j++)
			dis[j][i] = dis[i][j] = INF;
	for (int x, y, w, i = 1; i <= e; i++){
		scanf("%d%d%d", &x, &y, &w);
		dis[x][y] = dis[y][x] = min(dis[x][y], w);
	}
	for (int k = 1; k <= v; k++){
		for (int i = 1; i <= v; i++){
			for (int j = 1; j < i; j++)
				dis[j][i] = dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]);
		}
	}
}

int main(){
	scanf("%d%d%d%d", &n, &m, &v, &e);
	//memset(f, 0x3f, sizeof f);
	
	for (int i = 1; i <= n; i++) scanf("%d", &c[i]);
	for (int i = 1; i <= n; i++) scanf("%d", &d[i]);
	for (int i = 1; i <= n; i++) scanf("%lf", &k[i]);
	
	Floyd();
	
	for (int i = 1; i <= n; i++)
		for (int j = 0; j <= m; j++)
			f[i][j][0] = f[i][j][1] = INF;
	
	f[1][0][0] = 0, f[1][1][1] = 0;
	
	for (int i = 2; i <= n; i++){
		double w1 = dis[c[i - 1]][c[i]]; /*两次都不换*/ 
		double w2 = dis[d[i - 1]][c[i]] * k[i - 1] + dis[c[i - 1]][c[i]] * (1.0 - k[i - 1]);/*前一次换，后一次不换*/ 
		double w3 = dis[c[i - 1]][d[i]] * k[i] + dis[c[i - 1]][c[i]] * (1.0 - k[i]); /*前一次不换，后一次换*/
		double w4 = dis[d[i - 1]][d[i]] * k[i - 1] * k[i] //都成功 
				  + dis[d[i - 1]][c[i]] * k[i - 1] * (1.0 - k[i]) //前一次成功，后一次没成功 
				  + dis[c[i - 1]][d[i]] * (1.0 - k[i - 1]) * k[i] //前一次没成功，后一次成功 
				  + dis[c[i - 1]][c[i]] * (1.0 - k[i - 1]) * (1.0 - k[i]); //都没成功 
		for (int j = 0; j <= min(i, m); j++){
			f[i][j][0] = min(f[i - 1][j][0] + w1, f[i - 1][j][1] + w2);
			if (j > 0) f[i][j][1] = min(f[i - 1][j - 1][0] + w3, f[i - 1][j - 1][1] + w4);
		}
	}
	
	double ans = INF;
	for (int i = 0; i <= m; i++)
		ans = min(ans, (double)min(f[n][i][0], f[n][i][1]));
	printf("%0.2lf\n", ans);
	
	return 0;
}